Modern Algebra — Proofs & PYQ

UPSC Maths Optional · Paper II · Full Proofs + PYQs 2000–2026

★ Most Repeated Themes Group Theory — Core Theorems Sylow Theory Ring Theory Field Theory & Irreducibility PYQ: Groups & Subgroups PYQ: Isomorphism & Quotient Groups PYQ: Sylow Theorems PYQ: Rings & Ideals PYQ: Polynomials & Irreducibility

★ Most Repeated Themes (2000–2026)

Sylow Theorems — appears every year

Prove a group of given order is cyclic / abelian / has specific structure using Sylow PYQ 22+ times
Find number of Sylow p-subgroups; conclude uniqueness (= normality) PYQ 18+ times
Group of order $pq$ ($p \lt q$ primes): classify via Sylow — abelian / cyclic (e.g. when $p \nmid (q-1)$) PYQ 12+ times

Isomorphism Theorems — high frequency

First isomorphism theorem application: identify quotient group PYQ 15+ times
Homomorphism proofs; kernel and image analysis PYQ 12+ times

Normal Subgroups & Quotient Groups

Prove $A_n$ is normal in $S_n$; describe $S_n/A_n\cong\mathbb{Z}_2$ PYQ 10+ times
Coset decompositions; Lagrange's theorem applications PYQ 20+ times

Ring Theory — Ideals & Fields

Prove ideals, maximal ideals, prime ideals; quotient ring structure PYQ 15+ times
Finite integral domain is a field PYQ 8+ times

Polynomial Irreducibility

Eisenstein criterion; irreducibility over $\mathbb{Q}$ PYQ 12+ times
Factorization in $\mathbb{Z}[x]$, mod p reduction PYQ 8+ times

§ Proofs — Group Theory

Lagrange's Theorem — Proof

If $H$ is a subgroup of a finite group $G$, then $|H|$ divides $|G|$.

Proof (Coset partition)

Step 1 Define left cosets $aH=\{ah:h\in H\}$ for $a\in G$.

Step 2 Cosets partition $G$: Define $a\sim b\iff a^{-1}b\in H$. This is an equivalence relation (reflexive: $a^{-1}a=e\in H$; symmetric: $a^{-1}b\in H\Rightarrow b^{-1}a=(a^{-1}b)^{-1}\in H$; transitive: $a^{-1}b,b^{-1}c\in H\Rightarrow a^{-1}c=(a^{-1}b)(b^{-1}c)\in H$). The equivalence class of $a$ is $aH$.

Step 3 All cosets have same size: Map $\phi:H\to aH$ by $\phi(h)=ah$ is a bijection (injective: $ah=ah'\Rightarrow h=h'$; surjective: obvious). So $|aH|=|H|$.

Step 4 $G$ is partitioned into cosets, each of size $|H|$. If there are $k$ cosets: $|G|=k|H|$. So $|H|\mid|G|$. $\blacksquare$

Corollaries: Order of any element divides $|G|$. Every group of prime order is cyclic.

First Isomorphism Theorem — Proof

If $\phi:G\to H$ is a group homomorphism, then $G/\ker\phi\cong\text{Im}(\phi)$.

Proof

Let $K=\ker\phi$. Define $\tilde\phi:G/K\to\text{Im}(\phi)$ by $\tilde\phi(aK)=\phi(a)$.

Well-defined: If $aK=bK\Rightarrow a^{-1}b\in K\Rightarrow\phi(a^{-1}b)=e\Rightarrow\phi(a)^{-1}\phi(b)=e\Rightarrow\phi(a)=\phi(b)$.

Homomorphism: $\tilde\phi(aK\cdot bK)=\tilde\phi(abK)=\phi(ab)=\phi(a)\phi(b)=\tilde\phi(aK)\tilde\phi(bK)$.

Injective: $\tilde\phi(aK)=e\Rightarrow\phi(a)=e\Rightarrow a\in K\Rightarrow aK=K=e_{G/K}$.

Surjective: For any $\phi(a)\in\text{Im}(\phi)$: $\tilde\phi(aK)=\phi(a)$. $\blacksquare$

Normal Subgroup Characterizations — Proof of Equivalences

$N\trianglelefteq G\iff gNg^{-1}=N\ \forall g\iff$ left cosets = right cosets $\iff G/N$ is a group.

Proof of main implication: gNg⁻¹ = N ⟹ G/N is a group

Define multiplication on $G/N$: $(aN)(bN)=abN$. Well-defined: If $aN=a'N$ and $bN=b'N$: $a'=an_1,b'=bn_2$. Then $a'b'=an_1bn_2=ab(b^{-1}n_1b)n_2=ab\cdot m\in abN$ (using $b^{-1}n_1b\in N$ by normality). So $a'b'N=abN$. Other group axioms follow from $G$. $\blacksquare$

Cayley's Theorem — Every Group is Isomorphic to a Permutation Group

Proof

For each $g\in G$, define $\lambda_g:G\to G$ by $\lambda_g(x)=gx$ (left multiplication). Each $\lambda_g$ is a bijection (inverse: $\lambda_{g^{-1}}$). The map $\phi:G\to S_G$ by $\phi(g)=\lambda_g$ is a group homomorphism: $\phi(gh)=\lambda_{gh}=\lambda_g\circ\lambda_h=\phi(g)\phi(h)$. Injective: $\ker\phi=\{g:\lambda_g=\text{id}\}=\{g:gx=x\ \forall x\}=\{e\}$. By first isomorphism theorem: $G\cong\text{Im}(\phi)\le S_G$. $\blacksquare$

Cyclic Groups — Classification

Every subgroup of a cyclic group is cyclic. If $G=\langle a\rangle$ has order $n$, then for each $d\mid n$: there is exactly one subgroup of order $d$, namely $\langle a^{n/d}\rangle$.

Proof

Let $H\le\langle a\rangle$. If $H=\{e\}$: cyclic. Otherwise: let $m$ be the smallest positive integer with $a^m\in H$. Claim $H=\langle a^m\rangle$. If $a^k\in H$: write $k=qm+r$, \(0\le r

§ Proofs — Sylow Theory

Sylow's First Theorem — Proof sketch

If $p^k\mid|G|$, then $G$ has a subgroup of order $p^k$. In particular, if $p^m\mid|G|$ but $p^{m+1}\nmid|G|$, there exists a subgroup of order $p^m$ (a Sylow $p$-subgroup).

Proof (Action on k-element subsets)

Let $|G|=p^m\cdot s$ with $\gcd(p,s)=1$. Let $G$ act on $\mathcal{X}=\{S\subset G:|S|=p^m\}$ by $g\cdot S=gS$. $|\mathcal{X}|=\binom{p^ms}{p^m}$. One can show $\gcd(|\mathcal{X}|,p)=1$ (key combinatorial fact). So not all orbits have size divisible by $p$. Some orbit $\mathcal{O}$ has $|\mathcal{O}|$ not divisible by $p$. For $S\in\mathcal{O}$: stabilizer $H=G_S=\{g\in G:gS=S\}$ has order $|G|/|\mathcal{O}|=p^m s/|\mathcal{O}|$. Since $p\nmid|\mathcal{O}|$: $p^m\mid|H|$. Also $|H|\le p^m$ (by embedding $H\hookrightarrow\text{bij}(S)$). So $|H|=p^m$. $\blacksquare$

Sylow's Second Theorem: All Sylow p-subgroups are conjugate

Sylow's Third Theorem: $n_p\equiv1\pmod p$ and $n_p\mid m$ — Proof

Proof of Third Theorem

Let $P$ be a fixed Sylow $p$-subgroup. Let $G$ act on $\mathcal{S}=\{\text{all Sylow }p\text{-subgroups}\}$ by conjugation. By 2nd theorem, this action is transitive: all are conjugate. So $n_p=|\mathcal{S}|=[G:N_G(P)]$. Since $P\le N_G(P)\le G$: $n_p=[G:N_G(P)]\mid[G:P]=m$. Now let $P$ act on $\mathcal{S}$ by conjugation. A fixed point $Q\in\mathcal{S}$: $PQP^{-1}=Q\Rightarrow P\le N_G(Q)\Rightarrow PQ\le G$. $|PQ|=|P||Q|/|P\cap Q|$. If $P\ne Q$: $PQ$ is not a Sylow $p$-subgroup unless $P=Q$. The unique fixed point of $P$ acting on $\mathcal{S}$ is $P$ itself. So $n_p=|\mathcal{S}|\equiv1\pmod p$. $\blacksquare$

§ Proofs — Ring Theory

Maximal Ideal ⟺ Quotient Ring is a Field — Proof

Proof

$(\Rightarrow)$ Let $M$ be maximal in commutative ring $R$. Let $a+M\ne0$ in $R/M$. Consider the ideal $M+Ra=\{m+ra:m\in M,r\in R\}\supseteq M$. By maximality of $M$: $M+Ra=R$. So $\exists m\in M,s\in R$ with $m+sa=1$. In $R/M$: $(s+M)(a+M)=sa+M=1-m+M=1+M$. So $a+M$ has inverse $s+M$. $R/M$ is a field.

$(\Leftarrow)$ If $R/M$ is a field: let $I\supsetneq M$ be an ideal. Then $\exists a\in I\setminus M$. In $R/M$, $a+M\ne0$, so invertible: $\exists b$ with $ab+M=M+1\Rightarrow ab-1\in M\subset I\Rightarrow 1\in I\Rightarrow I=R$. So $M$ is maximal. $\blacksquare$

Prime Ideal ⟺ Quotient Ring is an Integral Domain — Proof

Proof

$P$ is prime iff $ab\in P\Rightarrow a\in P$ or $b\in P$. In $R/P$: $(a+P)(b+P)=0+P\iff ab\in P\iff a\in P$ or $b\in P\iff a+P=0$ or $b+P=0$. So $R/P$ has no zero divisors iff $P$ is prime. $\blacksquare$

Finite Integral Domain is a Field — Proof

Proof

Let $D$ be a finite integral domain and $a\ne0$. Consider the map $\phi:D\to D$ by $\phi(x)=ax$. Injective: $ax=ay\Rightarrow a(x-y)=0\Rightarrow x-y=0$ (no zero divisors). Surjective: injective map on finite set is surjective. So $\exists b\in D$ with $ab=1$. So $a$ is a unit. Every nonzero element is a unit: $D$ is a field. $\blacksquare$

§ Proofs — Polynomial Irreducibility

Eisenstein's Criterion — Proof

Let $f(x)=a_nx^n+\cdots+a_0\in\mathbb{Z}[x]$. If a prime $p$ satisfies: $p\nmid a_n$, $p\mid a_i$ for \(i

Proof

By Gauss's lemma: irreducible over $\mathbb{Q}\iff$ primitive and irreducible over $\mathbb{Z}$. Suppose $f=gh$ in $\mathbb{Z}[x]$, $g=b_sx^s+\cdots+b_0$, $h=c_tx^t+\cdots+c_0$, $s+t=n$. Since $p\mid a_0=b_0c_0$ but $p^2\nmid a_0$: exactly one of $b_0,c_0$ is divisible by $p$. WLOG $p\mid b_0$ and $p\nmid c_0$. Since $p\nmid a_n=b_sc_t$: $p\nmid b_s$. Let $r$ be the smallest index with $p\nmid b_r$. Then $a_r=b_0c_r+b_1c_{r-1}+\cdots+b_rc_0$. Since $p\mid b_0,\ldots,b_{r-1}$ and $p\mid a_r$ (for \(r

§ PYQ — Groups & Subgroups (2000–2026)

PYQ 2024 — Order of Elements in S₄

List all elements of $S_4$, their cycle types, and orders. How many elements have order 4?

Solution

Cycle types and orders in $S_4$: (1): $e$ — order 1 (1 element). (2): transpositions — order 2 ($\binom42=6$ elements). (2,2): two disjoint transpositions — order 2 ($3$ elements: $(12)(34),(13)(24),(14)(23)$). (3): 3-cycles — order 3 ($2\binom43=8$ elements). (4): 4-cycles — order 4 ($3!=6$ elements). (3,1) same as (3). Total: $1+6+3+8+6=24=4!$. Elements of order 4: the six 4-cycles.

PYQ 2023 — Center of a Group

Find the center of $S_3$.

Solution

Center $Z(G)=\{g\in G:gx=xg\ \forall x\}$. In $S_3$: check if any non-identity element commutes with all. Take $(12)$: $(12)(123)=(132)(12)\ne(12)(123)=(23)$. Actually: $(12)(123)=(23)$ and $(123)(12)=(13)$. Not equal. So no non-identity element commutes with everything. $Z(S_3)=\{e\}$.

PYQ 2022 — Group of Order 15 is Cyclic

Show that every group of order 15 is cyclic.

Solution

$|G|=15=3\cdot5$. Sylow: $n_5\equiv1\pmod5$ and $n_5\mid3\Rightarrow n_5=1$. So unique Sylow 5-subgroup $P_5\trianglelefteq G$. $n_3\equiv1\pmod3$ and $n_3\mid5\Rightarrow n_3=1$. So unique Sylow 3-subgroup $P_3\trianglelefteq G$. $|P_3|=3,|P_5|=5\Rightarrow P_3\cap P_5=\{e\}$. $P_3P_5\le G$, $|P_3P_5|=15\Rightarrow G=P_3P_5\cong P_3\times P_5\cong\mathbb{Z}_3\times\mathbb{Z}_5\cong\mathbb{Z}_{15}$ (since $\gcd(3,5)=1$). Cyclic. $\blacksquare$

PYQ 2021 — Coset Decomposition

Find all left cosets of $H=\{e,(12)\}$ in $S_3$.

Solution

$S_3=\{e,(12),(13),(23),(123),(132)\}$. $eH=H=\{e,(12)\}$. $(13)H=\{(13),(13)(12)\}=\{(13),(132)\}$. $(23)H=\{(23),(23)(12)\}=\{(23),(123)\}$. Three cosets, each of size 2. $[S_3:H]=3$. (Note: $H$ is NOT normal since left coset $(13)H=\{(13),(132)\}\ne H(13)=\{(13),(123)\}$.)

PYQ 2020 — Subgroup Generated by Two Elements

Show that $\mathbb{Z}$ is generated by any two coprime integers $m,n$.

Solution

$\langle m,n\rangle$ is a subgroup of $\mathbb{Z}$, hence of the form $d\mathbb{Z}$ for some $d\ge0$. Since $m,n\in\langle m,n\rangle=d\mathbb{Z}$: $d\mid m$ and $d\mid n\Rightarrow d\mid\gcd(m,n)=1\Rightarrow d=1$. So $\langle m,n\rangle=\mathbb{Z}$. $\blacksquare$

PYQ 2018 — Group of Even Integers

Show that the set of even integers forms a group under addition. Is it isomorphic to $(\mathbb{Z},+)$?

Solution

Even integers $2\mathbb{Z}$: closed under $+$, associative (inherited), identity $0$, inverses $-2n$. Group. Define $\phi:\mathbb{Z}\to2\mathbb{Z}$ by $\phi(n)=2n$. Bijective (injective: $2m=2n\Rightarrow m=n$; surjective: $2n\in2\mathbb{Z}$). Homomorphism: $\phi(m+n)=2(m+n)=2m+2n=\phi(m)+\phi(n)$. Isomorphism: $2\mathbb{Z}\cong\mathbb{Z}$. $\blacksquare$

PYQ 2015 — Abelian Groups of Order p²

Classify all abelian groups of order $p^2$ (p prime).

Solution

By fundamental theorem of finitely generated abelian groups: abelian groups of order $p^2$ are either $\mathbb{Z}_{p^2}$ (cyclic) or $\mathbb{Z}_p\times\mathbb{Z}_p$ (elementary abelian). These are the only two, up to isomorphism. They are non-isomorphic since $\mathbb{Z}_{p^2}$ has an element of order $p^2$ while all non-identity elements of $\mathbb{Z}_p\times\mathbb{Z}_p$ have order $p$.

§ PYQ — Isomorphism & Quotient Groups (2000–2026)

PYQ 2024 — First Isomorphism Theorem Application

If $\phi:\mathbb{Z}\to\mathbb{Z}_n$ is defined by $\phi(k)=k\pmod n$, find $\ker\phi$ and apply First Isomorphism Theorem.

Solution

$\phi$ is a ring (and group) homomorphism. $\ker\phi=\{k\in\mathbb{Z}:n\mid k\}=n\mathbb{Z}$. $\text{Im}(\phi)=\mathbb{Z}_n$ (surjective). By First Isomorphism Theorem: $\mathbb{Z}/n\mathbb{Z}\cong\mathbb{Z}_n$. $\blacksquare$

PYQ 2023 — Aₙ is Normal in Sₙ

Prove that $A_n$ (alternating group) is normal in $S_n$ and find $S_n/A_n$.

Solution

The sign homomorphism $\text{sgn}:S_n\to\{+1,-1\}$ (multiplicative) is surjective with $\ker(\text{sgn})=A_n$. So $A_n\trianglelefteq S_n$ (kernel of homomorphism). By First Isomorphism Theorem: $S_n/A_n\cong\{+1,-1\}\cong\mathbb{Z}_2$. Index $[S_n:A_n]=2$. $\blacksquare$

PYQ 2021 — Quotient Group ℤ/6ℤ

Describe the quotient group $\mathbb{Z}/6\mathbb{Z}$. List its elements, their orders, and identify the group.

Solution

Elements: $\{0+6\mathbb{Z},1+6\mathbb{Z},2+6\mathbb{Z},3+6\mathbb{Z},4+6\mathbb{Z},5+6\mathbb{Z}\}\cong\{0,1,2,3,4,5\}$. Under $+\pmod6$: cyclic of order 6, generated by $1+6\mathbb{Z}$. Orders: $o(0)=1,o(1)=6,o(2)=3,o(3)=2,o(4)=3,o(5)=6$. Isomorphic to $\mathbb{Z}_6$. $\blacksquare$

PYQ 2019 — Homomorphism Image

Find all group homomorphisms from $\mathbb{Z}_4$ to $\mathbb{Z}_6$.

Solution

A homomorphism $\phi:\mathbb{Z}_4\to\mathbb{Z}_6$ is determined by $\phi(1)$. We need $4\phi(1)\equiv0\pmod6\Rightarrow\phi(1)\in\{0,3\}$ (since $4k\equiv0\pmod6\iff 6\mid4k\iff3\mid2k\iff3\mid k$... actually: $\text{ord}(\phi(1))\mid\gcd(4,6)=2$. Elements of $\mathbb{Z}_6$ with order dividing 2: $\{0,3\}$). So two homomorphisms: $\phi(1)=0$ (trivial) and $\phi(1)=3$ (image = $\{0,3\}\cong\mathbb{Z}_2$).

§ PYQ — Sylow Theorems (2000–2026)

PYQ 2024 — Group of Order 35

Show every group of order 35 is cyclic.

Solution

$35=5\cdot7$. $n_7\equiv1\pmod7,n_7\mid5\Rightarrow n_7=1$. $n_5\equiv1\pmod5,n_5\mid7\Rightarrow n_5=1$. Both Sylow subgroups are normal and unique. $P_5P_7=G$, $P_5\cap P_7=\{e\}$. $G\cong\mathbb{Z}_5\times\mathbb{Z}_7\cong\mathbb{Z}_{35}$. Cyclic. $\blacksquare$

PYQ 2022 — Group of Order 45

Prove that every group of order 45 is abelian.

Solution

$45=3^2\cdot5$. $n_5\equiv1\pmod5,n_5\mid9\Rightarrow n_5\in\{1\}$ (since 6 not $\equiv1\pmod5$, and $1,9$: $9\equiv4\ne1$). So $n_5=1\Rightarrow P_5\trianglelefteq G$. $n_3\equiv1\pmod3,n_3\mid5\Rightarrow n_3=1\Rightarrow P_3\trianglelefteq G$. $|P_3|=9=3^2$: groups of order $p^2$ are abelian. $|P_5|=5$: cyclic, abelian. $G\cong P_3\times P_5$: product of abelian groups is abelian. $\blacksquare$

PYQ 2020 — Sylow Subgroups of S₄

Find the Sylow 2-subgroups of $S_4$ and their number.

Solution

$|S_4|=24=2^3\cdot3$. Sylow 2-subgroups have order 8. $n_2\equiv1\pmod2,n_2\mid3\Rightarrow n_2\in\{1,3\}$. The three Sylow 2-subgroups are the three copies of $D_4$ (dihedral group) in $S_4$: $\langle(1234),(13)\rangle$, $\langle(1324),(12)\rangle$, $\langle(1243),(14)\rangle$... more precisely, there are $n_2=3$ Sylow 2-subgroups, each isomorphic to $D_4$.

PYQ 2017 — Group of Order pq

Let $|G|=pq$ with $p \lt q$ primes and $p\nmid(q-1)$. Show $G\cong\mathbb{Z}_{pq}$.

Solution

$n_q\equiv1\pmod q,n_q\mid p\Rightarrow n_q=1$ (since $n_q=p$ would require $p\equiv1\pmod q\Rightarrow p\ge q$, contradiction). So $P_q\trianglelefteq G$. $n_p\equiv1\pmod p,n_p\mid q\Rightarrow n_p\in\{1,q\}$. If $n_p=q$: $q\equiv1\pmod p\Rightarrow p\mid(q-1)$. Contradicts hypothesis. So $n_p=1\Rightarrow P_p\trianglelefteq G$. Then $G\cong P_p\times P_q\cong\mathbb{Z}_p\times\mathbb{Z}_q\cong\mathbb{Z}_{pq}$. $\blacksquare$

§ PYQ — Rings & Ideals (2000–2026)

PYQ 2023 — Ideal Verification

Show that $I=n\mathbb{Z}$ is an ideal in $\mathbb{Z}$. When is it maximal?

Solution

Ideal: (i) $0\in I$. (ii) $na,nb\in I\Rightarrow n(a-b)\in I$. (iii) For any $r\in\mathbb{Z}$: $r\cdot na=n(ra)\in I$. So $I$ is an ideal. Maximal iff $n$ is prime: $\mathbb{Z}/n\mathbb{Z}\cong\mathbb{Z}_n$ is a field iff $n$ is prime. By the theorem: maximal ideal iff quotient is a field. $\blacksquare$

PYQ 2021 — $\mathbb{Z}[i]$ is a Principal Ideal Domain

Show that every ideal in the Gaussian integers $\mathbb{Z}[i]$ is principal.

Solution

$\mathbb{Z}[i]$ is a Euclidean domain with norm $N(a+bi)=a^2+b^2$. Every Euclidean domain is a PID (standard result). Proof sketch: given ideal $I\ne0$, take $d\in I$ with $N(d)$ minimal. For any $a\in I$: by division algorithm in Euclidean domain, $a=qd+r$ with \(N(r)

PYQ 2019 — Ring Homomorphism Image

If $\phi:\mathbb{Z}\to\mathbb{Z}_6$ and $\phi:\mathbb{Z}\to\mathbb{Z}_4$ are natural projections, find $\ker(\phi\times\phi')$.

Solution

Define $\psi:\mathbb{Z}\to\mathbb{Z}_6\times\mathbb{Z}_4$ by $\psi(n)=(n\bmod6,n\bmod4)$. $\ker\psi=\{n:\text{lcm}(6,4)\mid n\}=12\mathbb{Z}$. By CRT ($\gcd(6,4)=2\ne1$): image is not all of $\mathbb{Z}_6\times\mathbb{Z}_4$. Image = $\{(a,b):a\equiv b\pmod2\}$, of size 12.

PYQ 2016 — Units in Quotient Ring

Find all units in $\mathbb{Z}/12\mathbb{Z}$.

Solution

Units are elements coprime to 12: $\{1,5,7,11\}$ (i.e., $\gcd(k,12)=1$). $|\text{units}|=\phi(12)=4$. They form the group $(\mathbb{Z}/12\mathbb{Z})^\times\cong\mathbb{Z}_2\times\mathbb{Z}_2$ (Klein 4-group).

§ PYQ — Polynomials & Irreducibility (2000–2026)

PYQ 2024 — Eisenstein Application

Show $f(x)=x^4+x^3+x^2+x+1$ is irreducible over $\mathbb{Q}$.

Solution

Substitute $x\to x+1$: $f(x+1)=\dfrac{(x+1)^5-1}{x}=x^4+5x^3+10x^2+10x+5$. Apply Eisenstein with $p=5$: $5\nmid1$ (leading), $5\mid5,10,10,5$, $25\nmid5$. So $f(x+1)$ is irreducible, hence $f(x)$ is irreducible over $\mathbb{Q}$. $\blacksquare$

PYQ 2023 — Rational Root Test + Irreducibility

Show $f(x)=x^3-3x+1$ is irreducible over $\mathbb{Q}$.

Solution

By rational root theorem: possible rational roots are $\pm1$. $f(1)=1-3+1=-1\ne0$, $f(-1)=-1+3+1=3\ne0$. So $f$ has no rational roots. Since $\deg f=3$: no rational roots iff irreducible over $\mathbb{Q}$. $\blacksquare$

PYQ 2022 — Eisenstein for Cyclotomic-type

Show $p(x)=x^p-p$ is irreducible over $\mathbb{Q}$ for prime $p$.

Solution

Apply Eisenstein with the prime $p$: $p\nmid1$ (leading coeff), $p\mid p$ (constant term), $p^2\nmid p$. All middle coefficients are 0, hence divisible by $p$. Eisenstein's criterion: $x^p-p$ is irreducible over $\mathbb{Q}$. $\blacksquare$

PYQ 2020 — Mod p Reduction Test

Show $f(x)=x^3+x+1$ is irreducible over $\mathbb{Q}$.

Solution

Rational root check: $\pm1$ are only candidates. $f(1)=3\ne0,f(-1)=-1\ne0$. No rational roots → irreducible (degree 3 with no rational root is irreducible over $\mathbb{Q}$). Alternatively: mod 2 gives $\bar f(x)=x^3+x+1\in\mathbb{F}_2[x]$. Check $\bar f(0)=1\ne0,\bar f(1)=1\ne0$ in $\mathbb{F}_2$. So $\bar f$ has no roots in $\mathbb{F}_2$, hence irreducible over $\mathbb{F}_2$ (degree 3). By mod-$p$ criterion: $f$ is irreducible over $\mathbb{Q}$. $\blacksquare$

PYQ 2016 — GCD in ℤ[i]

Find $\gcd(3+4i, 4+3i)$ in $\mathbb{Z}[i]$.

Solution

$N(3+4i)=25=5^2$, $N(4+3i)=25=5^2$. Use Euclidean algorithm: $(4+3i)/(3+4i)=(4+3i)(3-4i)/25=(12+16-9i+12i\cdot... $

$(4+3i)(3-4i)=12-16i+9i-12i^2=12+12+(-16+9)i=24-7i$. Divide by 25: quotient $\approx1$ (round: $q=1+0i$ or check $(24-7i)/25\approx(1,-0.28)\to q=1$). Remainder: $(4+3i)-1(3+4i)=1-i$. $N(1-i)=2$. Now $\gcd(3+4i,1-i)$: $(3+4i)/(1-i)=(3+4i)(1+i)/2=(3+4i+3i+4i^2)/2=(-1+7i)/2$. Not integer — take $q=-1+4i$/2... better: $(3+4i)=q(1-i)+r$. $(3+4i)/(1-i)=(3+4i)(1+i)/2=(-1+7i)/2$. Round to $q=(0+4i)$: $r=(3+4i)-(4i)(1-i)=(3+4i)-(4i+4)=(−1)$. So $r=−1$, $N(r)=1$. So $\gcd=1-i$ (up to units). $\gcd(3+4i,4+3i)=1+i$ (since $(1-i)$ and $(1+i)$ are associates: $1-i=-i(1+i)$).

PYQ 2013 — Unique Factorization

Factor $10=2\cdot5$ in $\mathbb{Z}[i]$.

Solution

$2=-i(1+i)^2$ (up to units: $N(1+i)=2$, $(1+i)^2=2i$, and $2=(-i)(2i)=-i\cdot2i=2$... so $2=-i(1+i)^2$, a prime associated to $(1+i)$ (ramified)). $5=(2+i)(2-i)$ (split: $N(2+i)=5$, $(2+i)$ and $(2-i)$ are primes). So $10=(-i)(1+i)^2(2+i)(2-i)$. The factorization in $\mathbb{Z}[i]$ is essentially $10=-i(1+i)^2(2+i)(2-i)$ (up to unit $-i$). $\blacksquare$

§ PYQ — Groups: Subgroups, Cosets, Normal Subgroups

PYQ 2024 — Group Axioms Verification

Show that $G=\{1,-1,i,-i\}$ forms a group under complex multiplication.

Solution

Closure: Product of any two elements stays in $G$: $i\cdot i=-1\in G$, $i\cdot(-1)=-i\in G$, etc. (all 16 products verified by Cayley table). Associativity: Inherited from $\mathbb{C}$. Identity: $1\in G$, and $1\cdot g=g\cdot1=g$ for all $g$. Inverses: $1^{-1}=1$, $(-1)^{-1}=-1$, $i^{-1}=-i\in G$, $(-i)^{-1}=i\in G$. All axioms satisfied, so $G$ is a group. Note $G\cong\mathbb{Z}_4$ (cyclic of order 4, generated by $i$). $\blacksquare$

PYQ 2023 — Subgroups of ℤ₁₂

Find all subgroups of $\mathbb{Z}_{12}$ and draw the subgroup lattice.

Solution

By Lagrange, subgroup orders must divide $12$. Divisors: 1, 2, 3, 4, 6, 12. Since $\mathbb{Z}_{12}$ is cyclic, it has exactly one subgroup of each order dividing 12: $\langle0\rangle=\{0\}$ (order 1), $\langle6\rangle=\{0,6\}$ (order 2), $\langle4\rangle=\{0,4,8\}$ (order 3), $\langle3\rangle=\{0,3,6,9\}$ (order 4), $\langle2\rangle=\{0,2,4,6,8,10\}$ (order 6), $\mathbb{Z}_{12}$ (order 12). Lattice: $\mathbb{Z}_{12}$ at top, $\langle2\rangle$ and $\langle3\rangle$ below it, $\langle4\rangle$ and $\langle6\rangle$ below those, $\{0\}$ at bottom. $\blacksquare$

PYQ 2022 — Lagrange's Theorem

State and prove Lagrange's theorem: the order of any subgroup $H$ of a finite group $G$ divides $|G|$.

Solution

For any $a\in G$, the left coset $aH=\{ah:h\in H\}$. Key facts: (1) $|aH|=|H|$ (map $h\mapsto ah$ is a bijection). (2) Two cosets are either equal or disjoint: if $aH\cap bH\ne\emptyset$, then $ah_1=bh_2$, so $b^{-1}a=h_2h_1^{-1}\in H$, giving $aH=bH$. (3) $G$ is the disjoint union of all distinct left cosets. If there are $k$ distinct cosets, then $|G|=k|H|$, so $|H|\mid|G|$. The index $[G:H]=k=|G|/|H|$. $\blacksquare$

PYQ 2021 — Prime Order Groups Are Cyclic

Show that every group of prime order $p$ is cyclic.

Solution

Let $|G|=p$ (prime) and pick any $g\ne e$. The subgroup $\langle g\rangle$ has order dividing $p$ (Lagrange). Since $g\ne e$, $|\langle g\rangle|\ne1$. As $p$ is prime, $|\langle g\rangle|=p$. Thus $\langle g\rangle=G$, so $G$ is cyclic. Consequence: groups of prime order are simple and abelian. $\blacksquare$

PYQ 2020 — Index-2 Subgroups Are Normal

Prove that any subgroup $H$ of index 2 in a group $G$ is normal.

Solution

$[G:H]=2$ means there are exactly two left cosets: $H$ and $G\setminus H$, and exactly two right cosets: $H$ and $G\setminus H$. For any $g\in H$: $gH=H=Hg$. For any $g\notin H$: the unique left coset other than $H$ is $gH=G\setminus H$, and similarly $Hg=G\setminus H$. Thus $gH=Hg$ for all $g\in G$, so $H\trianglelefteq G$. $\blacksquare$

PYQ 2019 — Center Z(G) Is Normal and Abelian

Show that the center $Z(G)=\{z\in G: zg=gz\ \forall g\in G\}$ is a normal abelian subgroup of $G$.

Solution

Subgroup: $e\in Z(G)$. If $z_1,z_2\in Z(G)$, then $(z_1z_2)g=z_1(z_2g)=z_1(gz_2)=(z_1g)z_2=(gz_1)z_2=g(z_1z_2)$, so $z_1z_2\in Z(G)$. If $z\in Z(G)$, then $z^{-1}g=(g^{-1}z)^{-1}=(zg^{-1})^{-1}=gz^{-1}$, so $z^{-1}\in Z(G)$. Abelian: Any two elements of $Z(G)$ commute with everything, hence with each other. Normal: For any $g\in G,z\in Z(G)$: $gzg^{-1}=z\in Z(G)$, so $gZ(G)g^{-1}=Z(G)$. $\blacksquare$

PYQ 2018 — A_n Is Normal in S_n of Index 2

Show that the alternating group $A_n$ is a normal subgroup of $S_n$ with $[S_n:A_n]=2$.

Solution

Define the sign homomorphism $\text{sgn}:S_n\to\{+1,-1\}$ by $\text{sgn}(\sigma)=(-1)^k$ where $k$ is the number of transpositions in any decomposition of $\sigma$ (well-defined). Then $A_n=\ker(\text{sgn})$, which is normal in $S_n$ as kernels of homomorphisms are always normal. Since $\text{sgn}$ is surjective (any transposition maps to $-1$), by the first isomorphism theorem $S_n/A_n\cong\{+1,-1\}\cong\mathbb{Z}_2$, so $[S_n:A_n]=2$. $\blacksquare$

PYQ 2017 — Conjugate Subgroups Are Isomorphic

Show that if $H\le G$, then $gHg^{-1}\le G$ and $H\cong gHg^{-1}$ for all $g\in G$.

Solution

Define $\phi:H\to gHg^{-1}$ by $\phi(h)=ghg^{-1}$. Homomorphism: $\phi(h_1h_2)=g(h_1h_2)g^{-1}=(gh_1g^{-1})(gh_2g^{-1})=\phi(h_1)\phi(h_2)$. Injective: $\phi(h_1)=\phi(h_2)\Rightarrow h_1=h_2$. Surjective: every element of $gHg^{-1}$ is $\phi(h)$ for some $h$. Hence $\phi$ is an isomorphism $H\cong gHg^{-1}$. That $gHg^{-1}\le G$ follows from it being the image of a subgroup under an inner automorphism. $\blacksquare$

PYQ 2016 — ℚ/ℤ Has All Elements of Finite Order; ℚ Has None

Show every nonzero element of $\mathbb{Q}/\mathbb{Z}$ has finite order, but no nonzero element of $\mathbb{Q}$ has finite order under addition.

Solution

In $\mathbb{Q}/\mathbb{Z}$: A coset $a/b+\mathbb{Z}$ (with $b>0$) satisfies $b\cdot(a/b+\mathbb{Z})=a+\mathbb{Z}=\mathbb{Z}$ (the identity), so the order divides $b$. Hence every element has finite order. In $\mathbb{Q}$: Suppose $q\ne0$ and $nq=0$ for some $n\in\mathbb{Z}^+$. Then $nq=0$ implies $q=0$ (since $\mathbb{Q}$ is a field, hence an integral domain). Contradiction. So no nonzero rational has finite order in $(\mathbb{Q},+)$. $\blacksquare$

PYQ 2015 — G/Z(G) Cyclic Implies G Abelian

Prove: if $G/Z(G)$ is cyclic, then $G$ is abelian.

Solution

Let $G/Z(G)=\langle xZ(G)\rangle$. Every element of $G$ has the form $x^m z$ for some $m\in\mathbb{Z},z\in Z(G)$. Take $a=x^m z_1,b=x^n z_2$. Then $ab=(x^m z_1)(x^n z_2)=x^m x^n z_1 z_2$ (since $z_1$ commutes with everything) $=x^{m+n}z_1 z_2=x^n x^m z_2 z_1=ba$. So $G$ is abelian. Corollary: $G/Z(G)$ is never a nontrivial cyclic group — it is either trivial (G abelian) or non-cyclic. $\blacksquare$

PYQ 2014 — Cauchy's Theorem

State and prove Cauchy's theorem: if prime $p$ divides $|G|$, then $G$ has an element of order $p$.

Solution

Consider $S=\{(a_1,\ldots,a_p)\in G^p: a_1a_2\cdots a_p=e\}$. Then $|S|=|G|^{p-1}$ (choose first $p-1$ freely, last determined). The cyclic group $\mathbb{Z}_p$ acts on $S$ by cyclic shift: $k\cdot(a_1,\ldots,a_p)=(a_{k+1},\ldots,a_p,a_1,\ldots,a_k)$ (this stays in $S$ since the product is unchanged cyclically). Fixed points are tuples $(a,a,\ldots,a)$ with $a^p=e$. The tuple $(e,e,\ldots,e)$ is always fixed. Since $p\mid|G|^{p-1}$ and orbits have size 1 or $p$, the number of fixed points $\equiv0\pmod{p}$. At least one fixed point is $(e,\ldots,e)$; since total is $\equiv0\pmod p$, there exists another fixed point $(a,\ldots,a)$ with $a\ne e$ and $a^p=e$. This $a$ has order $p$. $\blacksquare$

PYQ 2013 — p-Groups Have Non-Trivial Center

Prove that if $|G|=p^n$ ($p$ prime, $n\ge1$), then $Z(G)\ne\{e\}$.

Solution

Use the class equation: $|G|=|Z(G)|+\sum_{[a]\text{ non-central}}[G:C_G(a)]$. Each non-central term $[G:C_G(a)]$ divides $|G|=p^n$ and is $>1$, so each is divisible by $p$. Hence $p\mid|G|$ and $p\mid$ each non-central term, so $p\mid|Z(G)|$. In particular $|Z(G)|\ge p>1$, so $Z(G)\ne\{e\}$. $\blacksquare$

PYQ 2012 — GL(n,ℝ) Is a Group

Show that $GL(n,\mathbb{R})$, the set of invertible $n\times n$ real matrices, is a group under matrix multiplication.

Solution

Closure: If $A,B$ are invertible, then $(AB)^{-1}=B^{-1}A^{-1}$ exists, so $AB\in GL(n,\mathbb{R})$. Associativity: Matrix multiplication is associative. Identity: $I_n\in GL(n,\mathbb{R})$ since $\det(I_n)=1\ne0$. Inverses: If $A\in GL(n,\mathbb{R})$, then $A^{-1}$ exists and $\det(A^{-1})=1/\det(A)\ne0$, so $A^{-1}\in GL(n,\mathbb{R})$. All axioms satisfied. $\blacksquare$

PYQ 2011 — Kernel of Homomorphism Is Normal

Prove that if $\phi:G\to H$ is a group homomorphism, then $\ker(\phi)$ is a normal subgroup of $G$.

Solution

Let $K=\ker(\phi)=\{g\in G:\phi(g)=e_H\}$. Subgroup: $\phi(e_G)=e_H$ so $e_G\in K$. If $a,b\in K$: $\phi(ab)=\phi(a)\phi(b)=e_H$, so $ab\in K$; $\phi(a^{-1})=\phi(a)^{-1}=e_H^{-1}=e_H$, so $a^{-1}\in K$. Normal: For any $g\in G,k\in K$: $\phi(gkg^{-1})=\phi(g)\phi(k)\phi(g^{-1})=\phi(g)e_H\phi(g)^{-1}=e_H$. So $gkg^{-1}\in K$, giving $gKg^{-1}\subseteq K$. Thus $K\trianglelefteq G$. $\blacksquare$

PYQ 2010 — First Isomorphism Theorem

State and prove the First Isomorphism Theorem: $G/\ker(\phi)\cong\text{Im}(\phi)$.

Solution

Let $\phi:G\to H$ be a homomorphism, $K=\ker(\phi)$. Define $\bar\phi:G/K\to\text{Im}(\phi)$ by $\bar\phi(gK)=\phi(g)$. Well-defined: If $gK=g'K$, then $g'^{-1}g\in K$, so $\phi(g'^{-1}g)=e_H$, so $\phi(g)=\phi(g')$. Homomorphism: $\bar\phi((g_1K)(g_2K))=\bar\phi(g_1g_2K)=\phi(g_1g_2)=\phi(g_1)\phi(g_2)=\bar\phi(g_1K)\bar\phi(g_2K)$. Injective: $\bar\phi(gK)=e_H\Rightarrow\phi(g)=e_H\Rightarrow g\in K\Rightarrow gK=K$ (identity). Surjective: Every $\phi(g)\in\text{Im}(\phi)$ is $\bar\phi(gK)$. Hence $\bar\phi$ is an isomorphism. $\blacksquare$

PYQ 2009 — Quotient Group ℤ/nℤ ≅ ℤ_n

Show that $\mathbb{Z}/n\mathbb{Z}\cong\mathbb{Z}_n$.

Solution

Define $\phi:\mathbb{Z}\to\mathbb{Z}_n$ by $\phi(k)=k\bmod n$. This is a surjective ring (and group) homomorphism. Its kernel is $\{k\in\mathbb{Z}:k\equiv0\pmod n\}=n\mathbb{Z}$. By the First Isomorphism Theorem, $\mathbb{Z}/n\mathbb{Z}\cong\text{Im}(\phi)=\mathbb{Z}_n$. Explicitly the cosets are $0+n\mathbb{Z},1+n\mathbb{Z},\ldots,(n-1)+n\mathbb{Z}$, and addition of cosets mirrors addition mod $n$. $\blacksquare$

PYQ 2008 — Dihedral Group D_n

Show that the dihedral group $D_n$ has exactly $2n$ elements and describe its structure.

Solution

$D_n$ is the symmetry group of a regular $n$-gon. It consists of $n$ rotations $\{r^0,r^1,\ldots,r^{n-1}\}$ where $r$ is rotation by $2\pi/n$, and $n$ reflections $\{s,sr,sr^2,\ldots,sr^{n-1}\}$ where $s$ is a fixed reflection. Total: $2n$ elements. Relations: $r^n=e$, $s^2=e$, $srs^{-1}=r^{-1}$ (i.e., $sr=r^{-1}s$). So $D_n=\langle r,s\mid r^n=s^2=e,\, srs^{-1}=r^{-1}\rangle$. The rotations form a cyclic normal subgroup of index 2. For $n\ge3$, $D_n$ is non-abelian. $\blacksquare$

PYQ 2007 — S_3 Is Non-Abelian of Order 6

Show that $S_3$ is a non-abelian group of order 6 and list its subgroups.

Solution

$S_3$ consists of all permutations of $\{1,2,3\}$: $|S_3|=3!=6$. Elements: $e,(12),(13),(23),(123),(132)$. Non-abelian: $(12)(13)=(132)$ but $(13)(12)=(123)\ne(132)$. Subgroups: $\{e\}$ (order 1), $\{e,(12)\},\{e,(13)\},\{e,(23)\}$ (order 2, three of them), $\{e,(123),(132)\}=A_3\cong\mathbb{Z}_3$ (order 3), $S_3$ itself (order 6). Only $A_3$ and $S_3$ are normal (and $\{e\}$). $\blacksquare$

PYQ 2006 — Subgroups of Cyclic Groups Are Cyclic

Prove that every subgroup of a cyclic group is cyclic.

Solution

Let $G=\langle a\rangle$ and $H\le G$. If $H=\{e\}$, it's cyclic. Otherwise, let $m$ be the smallest positive integer with $a^m\in H$. Claim $H=\langle a^m\rangle$. For any $a^k\in H$, write $k=qm+r$ with \(0\le r

PYQ 2005 — HK Is a Subgroup iff HK = KH

Prove that if $H,K\le G$, then $HK\le G$ if and only if $HK=KH$.

Solution

($\Rightarrow$) Suppose $HK\le G$. For any $kh\in KH$, we have $(kh)^{-1}=h^{-1}k^{-1}\in HK$, and since $HK$ is a subgroup, $(h^{-1}k^{-1})^{-1}=kh\in HK$. Thus $KH\subseteq HK$. Similarly $HK\subseteq KH$, so $HK=KH$. ($\Leftarrow$) Suppose $HK=KH$. Then $e=ee\in HK$. For $h_1k_1,h_2k_2\in HK$: product $h_1(k_1h_2)k_2$; since $k_1h_2\in KH=HK$, write it as $h_3k_3$, giving $h_1h_3k_3k_2\in HK$. Inverses: $(hk)^{-1}=k^{-1}h^{-1}\in KH=HK$. So $HK\le G$. $\blacksquare$

§ PYQ — Permutation Groups, Rings & Fields

PYQ 2024 — Composition of Permutations and Sign

Write the permutation $(1\,2\,3\,4)(1\,3\,2)$ as a product of disjoint cycles and find its sign.

Solution

Compose right-to-left: apply $(1\,3\,2)$ first, then $(1\,2\,3\,4)$. Track each element: $1\xrightarrow{(132)}3\xrightarrow{(1234)}4$; $2\xrightarrow{(132)}1\xrightarrow{(1234)}2$; $3\xrightarrow{(132)}2\xrightarrow{(1234)}3$... wait, recompute: $3\xrightarrow{(132)}2\xrightarrow{(1234)}3$. Hmm, let me be careful. $(1\,3\,2)$: $1\to3,3\to2,2\to1$, fixes 4. $(1\,2\,3\,4)$: $1\to2,2\to3,3\to4,4\to1$. Composition (left to right = apply $(1\,3\,2)$ then $(1\,2\,3\,4)$): $1\to3\to4$; $2\to1\to2$; $3\to2\to3$; $4\to4\to1$. Result: $(1\,4)$, a single transposition. Sign $= (-1)^1 = -1$ (odd permutation). $\blacksquare$

PYQ 2023 — Every Permutation Is a Product of Disjoint Cycles

Prove that every permutation in $S_n$ can be written as a product of disjoint cycles, and this decomposition is unique up to order.

Solution

Existence: For $\sigma\in S_n$, take any $a_1\in\{1,\ldots,n\}$ and form the orbit $\{a_1,\sigma(a_1),\sigma^2(a_1),\ldots\}$. Since $\{1,\ldots,n\}$ is finite, this orbit is finite; say it returns to $a_1$ after $k$ steps, giving cycle $(a_1\,\sigma(a_1)\,\cdots\,\sigma^{k-1}(a_1))$. Repeat with elements not yet covered. The resulting cycles are disjoint and their product equals $\sigma$. Uniqueness: Disjoint cycles commute, and each cycle is determined by the orbit of any of its elements under $\sigma$, so the partition into orbits is unique. $\blacksquare$

PYQ 2022 — A_4 Has No Subgroup of Order 6

Prove that $A_4$ has no subgroup of order 6.

Solution

Suppose $H\le A_4$ with $|H|=6$. Then $[A_4:H]=12/6=2$, so $H\trianglelefteq A_4$. The elements of $A_4$ of order 2 are the three products of two disjoint transpositions: $(12)(34),(13)(24),(14)(23)$; these form the Klein four-group $V=\{e,(12)(34),(13)(24),(14)(23)\}$. Now $H$ must contain elements whose orders divide 6, but $A_4$ has 8 elements of order 3 (the 3-cycles). Since $|H|=6$, $H$ can contain at most 5 non-identity elements. If $H$ omits a 3-cycle $\sigma$, then $A_4=H\cup\sigma H$ (two cosets). Check: $A_4$ has exactly 3 elements of order 2; all must be in $H$ (as $H\trianglelefteq A_4$ and the conjugates of order-2 elements are order-2). So $V\subseteq H$. Then $H$ has the 4 elements of $V$ plus 2 more (of order 3). But 3-cycles come in inverse pairs: if $\sigma\in H$, then $\sigma^2=\sigma^{-1}\in H$. So $H$ must contain 0 or 2 (paired) 3-cycles. With $|H|=6$: 4 from $V$ + 2 from one 3-cycle pair. But the 3-cycles of $A_4$ split into conjugacy classes of size 4 each; $H$ cannot contain exactly 2 of them and remain normal (conjugation shuffles all four). Contradiction. Hence no such $H$ exists. $\blacksquare$

PYQ 2021 — Order of Element in S_n Is LCM of Cycle Lengths

Prove that the order of a permutation equals the LCM of the lengths of its disjoint cycles.

Solution

Let $\sigma=c_1c_2\cdots c_r$ (disjoint cycles of lengths $l_1,\ldots,l_r$). Disjoint cycles commute, so $\sigma^k=c_1^k c_2^k\cdots c_r^k$. Now $\sigma^k=e\iff c_i^k=e$ for all $i\iff l_i\mid k$ for all $i\iff\text{lcm}(l_1,\ldots,l_r)\mid k$. The smallest positive such $k$ is $\text{lcm}(l_1,\ldots,l_r)$. Hence $\text{ord}(\sigma)=\text{lcm}(l_1,\ldots,l_r)$. Example: $(1\,2\,3)(4\,5)\in S_5$ has order $\text{lcm}(3,2)=6$. $\blacksquare$

PYQ 2020 — ℤ[i] Is an Integral Domain

Show that the Gaussian integers $\mathbb{Z}[i]=\{a+bi:a,b\in\mathbb{Z}\}$ form an integral domain.

Solution

$\mathbb{Z}[i]\subset\mathbb{C}$, which is a field, hence an integral domain. More explicitly: $\mathbb{Z}[i]$ is a commutative ring with unity $1$. It has no zero divisors: if $(a+bi)(c+di)=0$ in $\mathbb{Z}[i]$, taking norms $N(a+bi)N(c+di)=N(0)=0$ where $N(x+yi)=x^2+y^2\ge0$. Since $N$ is multiplicative and takes non-negative integer values, $N(a+bi)=0$ or $N(c+di)=0$, which forces $a+bi=0$ or $c+di=0$. Hence $\mathbb{Z}[i]$ is an integral domain. $\blacksquare$

PYQ 2019 — Units in ℤ[i]

Find all units in $\mathbb{Z}[i]$.

Solution

An element $u=a+bi\in\mathbb{Z}[i]$ is a unit iff $N(u)=a^2+b^2=1$ (since $N(uv)=N(u)N(v)=1$ requires both factors to be 1 for positive integers). The solutions to $a^2+b^2=1$ with $a,b\in\mathbb{Z}$ are $(a,b)\in\{(\pm1,0),(0,\pm1)\}$. So the units are $\{1,-1,i,-i\}$, forming a cyclic group of order 4 under multiplication. $\blacksquare$

PYQ 2018 — ℤ[√(−5)] Is Not a UFD

Show that $\mathbb{Z}[\sqrt{-5}]$ is not a unique factorization domain.

Solution

Consider $6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$. The norm is $N(a+b\sqrt{-5})=a^2+5b^2$. We have $N(2)=4,N(3)=9,N(1\pm\sqrt{-5})=6$. Each of $2,3,1+\sqrt{-5},1-\sqrt{-5}$ is irreducible (their norms have no factorizations into norms of non-units: units have norm 1, and there's no element of norm 2 or 3 since $a^2+5b^2=2$ or $3$ has no integer solutions). Yet $6$ has two distinct factorizations into irreducibles: $2\cdot3$ and $(1+\sqrt{-5})(1-\sqrt{-5})$. (One checks $2\nmid(1+\sqrt{-5})$ in this ring.) Hence $\mathbb{Z}[\sqrt{-5}]$ is not a UFD. $\blacksquare$

PYQ 2017 — Every Euclidean Domain Is a PID

Prove that every Euclidean domain is a principal ideal domain.

Solution

Let $R$ be a Euclidean domain with norm $\nu$, and let $I\trianglelefteq R$ be a nonzero ideal. Among all nonzero elements of $I$, pick $d$ with minimal $\nu(d)$. Claim $I=\langle d\rangle$. Clearly $\langle d\rangle\subseteq I$. For any $a\in I$, use the Euclidean algorithm: $a=qd+r$ with $r=0$ or $\nu(r)<\nu(d)$. Since $a,qd\in I$, we have $r=a-qd\in I$. By minimality of $\nu(d)$, $r=0$. So $a=qd\in\langle d\rangle$, giving $I=\langle d\rangle$. Hence every ideal is principal. $\blacksquare$

PYQ 2016 — Every PID Is a UFD

Prove that every principal ideal domain is a unique factorization domain.

Solution

Existence: Suppose $a\in R$ is not a unit and not zero. If $a$ is irreducible, done. Otherwise $a=a_1b_1$ (non-units). Continue factoring; this terminates because in a PID, ascending chains of ideals $\langle a\rangle\subseteq\langle a_1\rangle\subseteq\cdots$ must stabilize (PIDs are Noetherian). Uniqueness: In a PID, every irreducible is prime (since $\langle p\rangle$ is maximal when $p$ is irreducible — PID implies every nonzero prime ideal is maximal). If $p\mid ab$, then $p\mid a$ or $p\mid b$. Uniqueness of factorization then follows by standard induction on the number of prime factors, using the prime property. $\blacksquare$

PYQ 2015 — ℤ[x] Is UFD But Not PID

Show $\mathbb{Z}[x]$ is a UFD but not a PID.

Solution

UFD: $\mathbb{Z}$ is a UFD, and if $R$ is a UFD then $R[x]$ is a UFD (Gauss's theorem). Hence $\mathbb{Z}[x]$ is a UFD. Not PID: Consider the ideal $I=\langle 2,x\rangle=\{2f(x)+xg(x):f,g\in\mathbb{Z}[x]\}$. This is the set of polynomials with even constant term. Suppose $I=\langle h(x)\rangle$. Since $2\in I$, $h\mid2$, so $h\in\{1,-1,2,-2\}$. If $h=\pm1$, then $I=\mathbb{Z}[x]$, but $1\notin I$ (constant term of any element of $I$ is even). If $h=\pm2$, then $I=\langle2\rangle$, but $x\in I$ and $x$ is not divisible by 2 in $\mathbb{Z}[x]$. Contradiction. So $I$ is not principal, and $\mathbb{Z}[x]$ is not a PID. $\blacksquare$

PYQ 2014 — Eisenstein Criterion

Use Eisenstein's criterion to show $f(x)=x^4+4x+2$ is irreducible over $\mathbb{Q}$.

Solution

Apply Eisenstein with prime $p=2$: coefficients of $f(x)=x^4+0\cdot x^3+0\cdot x^2+4x+2$. Check: $p=2\nmid1$ (leading coefficient); $2\mid0,2\mid0,2\mid4,2\mid2$ (all non-leading coefficients divisible by 2); $p^2=4\nmid2$ (constant term not divisible by 4). All Eisenstein conditions satisfied. By Eisenstein's criterion, $f(x)$ is irreducible over $\mathbb{Q}$. $\blacksquare$

PYQ 2013 — Characteristic of a Field

Show that the characteristic of a field is either 0 or a prime $p$.

Solution

Let $F$ be a field. The characteristic is the smallest $n>0$ with $n\cdot1=0$ (or 0 if no such $n$ exists). Suppose $\text{char}(F)=n>0$ and $n=ab$ with \(1

PYQ 2012 — ℤ_p Is a Field iff p Is Prime

Prove that $\mathbb{Z}_p$ is a field if and only if $p$ is prime.

Solution

($\Rightarrow$) If $\mathbb{Z}_p$ is a field, it has no zero divisors. If $p=ab$ with \(1

PYQ 2011 — Ideals of ℤ Are nℤ

Prove that every ideal of $\mathbb{Z}$ is of the form $n\mathbb{Z}$ for some $n\ge0$.

Solution

$\mathbb{Z}$ is a Euclidean domain (with norm $\nu(n)=|n|$), hence a PID, so every ideal is principal: $I=\langle d\rangle=d\mathbb{Z}$ for some $d\in\mathbb{Z}$. We can take $d\ge0$ (since $\langle d\rangle=\langle -d\rangle$). Setting $n=|d|$ gives $I=n\mathbb{Z}$. The zero ideal is $0\mathbb{Z}=\{0\}$. $\blacksquare$

PYQ 2010 — Every Finite Integral Domain Is a Field

Prove that every finite integral domain is a field.

Solution

Let $R$ be a finite integral domain and $a\ne0$. Consider the map $\phi:R\to R$ defined by $\phi(r)=ar$. This is injective: if $ar=as$, then $a(r-s)=0$, and since $a\ne0$ and $R$ has no zero divisors, $r=s$. An injective map from a finite set to itself is surjective. So there exists $b\in R$ with $ab=1$. Hence $a$ is a unit, and since $a$ was arbitrary nonzero, $R$ is a field. $\blacksquare$

PYQ 2009 — Maximal Ideal M: R/M Is a Field

Prove that an ideal $M$ of a commutative ring $R$ is maximal if and only if $R/M$ is a field.

Solution

By the correspondence theorem, ideals of $R/M$ correspond to ideals of $R$ containing $M$. $M$ is maximal $\iff$ the only ideals of $R$ containing $M$ are $M$ and $R$ $\iff$ the only ideals of $R/M$ are $\{0\}$ and $R/M$ $\iff$ $R/M$ has no proper nonzero ideals $\iff$ $R/M$ is a field (for commutative rings with unity: a ring is a field iff it has no proper nonzero ideals). $\blacksquare$

PYQ 2008 — Prime Ideal P: R/P Is an Integral Domain

Prove that an ideal $P$ of a commutative ring $R$ is prime if and only if $R/P$ is an integral domain.

Solution

$P$ prime means: $ab\in P\Rightarrow a\in P$ or $b\in P$. In $R/P$: $(a+P)(b+P)=0+P\iff ab\in P\iff a\in P$ or $b\in P\iff a+P=0$ or $b+P=0$. So $R/P$ has no zero divisors. Also $R/P$ is commutative with unity $1+P\ne0+P$ (since $1\notin P$ for proper prime ideals). Hence $R/P$ is an integral domain. The converse is the same chain of iff's. $\blacksquare$

PYQ 2007 — R[x]/⟨p(x)⟩ Is a Field if p Is Irreducible

Show that if $F$ is a field and $p(x)\in F[x]$ is irreducible, then $F[x]/\langle p(x)\rangle$ is a field.

Solution

Since $F$ is a field, $F[x]$ is a PID. In a PID, an element is irreducible $\iff$ it generates a maximal ideal (because irreducible $\Rightarrow$ prime in PID $\Rightarrow$ prime ideal $\langle p\rangle$; and for PIDs, prime ideals are maximal among proper ideals). Since $\langle p(x)\rangle$ is maximal, $F[x]/\langle p(x)\rangle$ is a field. Concretely: for any nonzero $f+\langle p\rangle$, since $\gcd(f,p)=1$ (as $p$ is irreducible and $p\nmid f$), by Bezout in $F[x]$: $\exists g,h$ with $fg+ph=1$, so $(f+\langle p\rangle)(g+\langle p\rangle)=1+\langle p\rangle$. $\blacksquare$

PYQ 2006 — GCD in ℤ[i] by Euclidean Algorithm

Find $\gcd(11+3i,\,1+8i)$ in $\mathbb{Z}[i]$ using the Euclidean algorithm.

Solution

$\mathbb{Z}[i]$ is Euclidean with norm $N$. Divide: $(11+3i)/(1+8i)=(11+3i)(1-8i)/N(1+8i)=(11-88i+3i-24i^2)/(1+64)=(11+24+(3-88)i)/65=(35-85i)/65=(7/13)+(-17/13)i$. Round to nearest Gaussian integer: $q=1-i$ (since $7/13\approx0.54\to1$, $-17/13\approx-1.31\to-1$). Remainder: $r=(11+3i)-(1-i)(1+8i)=(11+3i)-(1+8i-i-8i^2)=(11+3i)-(1+8+7i)=(11+3i)-(9+7i)=2-4i$. Now $\gcd(1+8i,2-4i)$: $(1+8i)/(2-4i)=(1+8i)(2+4i)/(4+16)=(2+4i+16i+32i^2)/20=(2-32+(4+16)i)/20=(-30+20i)/20=-3/2+i$. Round: $q=-1+i$. Remainder: $r=(1+8i)-(-1+i)(2-4i)=(1+8i)-(-2+4i+2i-4i^2)=(1+8i)-(-2+4+6i)=(1+8i)+(2-4-6i)=(1+8i)+(-2-6i)=-1+2i$. Continue: $\gcd(2-4i,-1+2i)$: $(2-4i)/(-1+2i)=(2-4i)(-1-2i)/(1+4)=(-2-4i+4i+8i^2)/5=(-2-8)/5=-2$. Exact: $q=-2$, $r=(2-4i)-(-2)(-1+2i)=(2-4i)-(2-4i)=0$. So $\gcd=|-1+2i|=\sqrt5$, i.e., $\gcd=1-2i$ (or associate $-1+2i$). Verify: $N(-1+2i)=5$, and $N(11+3i)=130=5\cdot26$, $N(1+8i)=65=5\cdot13$. $\blacksquare$

PYQ 2005 — Factorization in ℤ[i]: 2 = (1+i)(1−i)

Show that 2 is not prime in $\mathbb{Z}[i]$ and find its factorization into primes.

Solution

In $\mathbb{Z}[i]$: $2=(1+i)(1-i)$. Check: $(1+i)(1-i)=1-i^2=1+1=2$. $N(1+i)=1^2+1^2=2$, which is prime in $\mathbb{Z}$. In a Euclidean domain, an element with prime norm is itself prime (irreducible). So $1+i$ is a Gaussian prime. Note $1-i=-i(1+i)$, so $1-i$ is an associate of $1+i$. Thus $2=-i(1+i)^2$ (up to units). The rational prime 2 ramifies in $\mathbb{Z}[i]$ (it becomes a square of a prime times a unit). Since 2 is not irreducible in $\mathbb{Z}[i]$, it is not prime there. $\blacksquare$

§ PYQ — Group Actions, Sylow Theory, Series

PYQ 2024 — Class Equation

State the class equation $|G|=|Z(G)|+\sum [G:C_G(a)]$ and use it for a group of order $p^2$.

Solution

G acts on itself by conjugation. The orbits are conjugacy classes. Elements of $Z(G)$ form singleton classes. For $a\notin Z(G)$, the class size is $[G:C_G(a)]>1$. So $|G|=|Z(G)|+\sum_{[a]\text{ non-central}}[G:C_G(a)]$. For $|G|=p^2$: Each non-central term divides $p^2$ and is $>1$, so divisible by $p$. Thus $p\mid|Z(G)|$, so $|Z(G)|\in\{p,p^2\}$. If $|Z(G)|=p$, then $|G/Z(G)|=p$, so $G/Z(G)$ is cyclic, hence $G$ is abelian (standard result), contradicting \(|Z(G)|=p

PYQ 2023 — Conjugacy Classes of S_4

Find all conjugacy classes of $S_4$ and their sizes.

Solution

Conjugacy classes in $S_n$ correspond to cycle types. For $S_4$ ($|S_4|=24$): (1) Identity $e$: cycle type $(1)(1)(1)(1)$, size 1. (2) Transpositions $(ab)$: cycle type $2+1+1$, count $\binom{4}{2}=6$. (3) 3-cycles $(abc)$: cycle type $3+1$, count $\frac{4\cdot3\cdot2}{3}=8$. (4) 4-cycles $(abcd)$: cycle type $4$, count $\frac{4!}{4}=6$. (5) Products of two disjoint transpositions $(ab)(cd)$: cycle type $2+2$, count $\frac{1}{2}\binom{4}{2}\binom{2}{2}=3$. Total: $1+6+8+6+3=24$. ✓ Five conjugacy classes. $\blacksquare$

PYQ 2022 — Sylow's First Theorem

State Sylow's First Theorem and outline its proof.

Solution

Statement: If $p^k\mid|G|$, then $G$ has a subgroup of order $p^k$. In particular, for $p^n\parallel|G|$ (highest power), there exists a Sylow $p$-subgroup of order $p^n$. Proof sketch: Let $|G|=p^n m$ with $p\nmid m$. Consider $\mathcal{S}=\{S\subseteq G:|S|=p^n\}$, acted on by $G$ by left multiplication. $|\mathcal{S}|=\binom{p^n m}{p^n}$. One shows $p\nmid|\mathcal{S}|$ (by Lucas' theorem or direct computation). So some orbit has size not divisible by $p$. If $S$ is in such an orbit, its stabilizer $G_S=\{g:gS=S\}$ has $|G_S|=|G|/|\text{orbit}|$. Analysis shows $|G_S|=p^n$, so $G_S$ is a Sylow $p$-subgroup. $\blacksquare$

PYQ 2021 — Sylow's Second Theorem

State and prove Sylow's Second Theorem: all Sylow $p$-subgroups are conjugate.

Solution

Statement: Any two Sylow $p$-subgroups $P,Q$ of $G$ are conjugate: $Q=gPg^{-1}$ for some $g\in G$. Proof: Let $P$ act on the set of left cosets $G/Q$ by left multiplication. The number of fixed points $\equiv|G/Q|\pmod p$. Since $|G/Q|=[G:Q]=m$ (the $p'$-part of $|G|$), $p\nmid m$, so there is at least one fixed point: some coset $gQ$ with $P\cdot gQ=gQ$, i.e., $p\cdot gQ=gQ$ for all $p\in P$, meaning $g^{-1}Pg\subseteq Q$. Since $|g^{-1}Pg|=|P|=|Q|=p^n$, we get $g^{-1}Pg=Q$, i.e., $Q=gPg^{-1}$. $\blacksquare$

PYQ 2020 — Sylow's Third Theorem

State Sylow's Third Theorem: $n_p\equiv 1\pmod p$ and $n_p\mid m$.

Solution

Let $|G|=p^n m$, $p\nmid m$. Let $n_p$ = number of Sylow $p$-subgroups. $n_p\mid m$: $G$ acts on Sylow $p$-subgroups by conjugation; the action is transitive (Second Theorem). So $n_p=[G:N_G(P)]$. Since $P\subseteq N_G(P)$, we have $|N_G(P)|\ge p^n$, so $n_p=[G:N_G(P)]\le m$. Also $n_p\mid|G|$; combined with the fact that $n_p\mid m$ follows from $n_p=[G:N_G(P)]$ and $p^n\mid|N_G(P)|$. $n_p\equiv1\pmod p$: Let $P$ act on the set of all Sylow $p$-subgroups by conjugation. The fixed points are Sylow subgroups $Q$ normalized by $P$. One shows the only such $Q$ is $P$ itself (if $P$ normalizes $Q$, then $PQ$ is a subgroup; $|PQ|=|P||Q|/|P\cap Q|$ is a power of $p$ contained in $G$, forcing $P=Q$). So there's exactly 1 fixed point, and $n_p=|\text{fixed points}|+\text{(sum of orbit sizes}\ge2)\equiv 1\pmod p$. $\blacksquare$

PYQ 2019 — Groups of Order 15 Are Cyclic

Classify all groups of order 15.

Solution

$15=3\cdot5$. Let $n_3$ = number of Sylow 3-subgroups: $n_3\equiv1\pmod3$ and $n_3\mid5$, so $n_3\in\{1,5\}$. Let $n_5$ = number of Sylow 5-subgroups: $n_5\equiv1\pmod5$ and $n_5\mid3$, so $n_5=1$. So there's a unique (hence normal) Sylow 5-subgroup $P_5\cong\mathbb{Z}_5$. If $n_3=5$, count elements: 5 subgroups of order 3 contribute $5\times2=10$ elements of order 3, plus 4 elements of order 5 from $P_5$, plus identity = 15. But then no room for the 5 Sylow 3-subgroups to not overlap, contradiction. So $n_3=1$, giving unique normal $P_3\cong\mathbb{Z}_3$. Then $G=P_3\times P_5\cong\mathbb{Z}_3\times\mathbb{Z}_5\cong\mathbb{Z}_{15}$. All groups of order 15 are cyclic. $\blacksquare$

PYQ 2018 — Classification of Groups of Order 8

List all groups of order 8 up to isomorphism.

Solution

There are exactly 5 groups of order $8=2^3$: (1) $\mathbb{Z}_8$ — cyclic, has element of order 8. (2) $\mathbb{Z}_4\times\mathbb{Z}_2$ — has element of order 4 but not 8; two elements of order 2. (3) $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$ — elementary abelian, all non-identity elements have order 2. (4) $D_4$ — dihedral group of square; non-abelian, has element of order 4. (5) $Q_8=\{\pm1,\pm i,\pm j,\pm k\}$ — quaternion group; non-abelian, unique element of order 2, all non-identity elements have order 4. Distinguished: $D_4$ has non-normal subgroup of order 2; $Q_8$ has all subgroups normal. $\blacksquare$

PYQ 2017 — Classification of Groups of Order 12

Classify all groups of order 12 up to isomorphism.

Solution

$12=4\cdot3=2^2\cdot3$. Sylow analysis: $n_3\in\{1,4\}$. If $n_3=4$, counting gives 8 elements of order 3 plus at least 4 from Sylow 2-subgroups — exactly 12, forcing unique Sylow 2-subgroup. Five groups of order 12: (1) $\mathbb{Z}_{12}\cong\mathbb{Z}_4\times\mathbb{Z}_3$ (cyclic). (2) $\mathbb{Z}_2\times\mathbb{Z}_6\cong\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3$ (abelian, non-cyclic). (3) $D_6\cong D_3\times\mathbb{Z}_2\cong S_3\times\mathbb{Z}_2$ (dihedral of hexagon). (4) $A_4$ (alternating group; $n_3=4$, no subgroup of order 6). (5) $\text{Dic}_3\cong\mathbb{Z}_3\rtimes\mathbb{Z}_4$ (dicyclic group). $\blacksquare$

PYQ 2016 — Group of Order p² Is Abelian

Prove that every group of order $p^2$ ($p$ prime) is abelian.

Solution

By the class equation argument (see PYQ 2024 above): $G$ is a $p$-group so $Z(G)\ne\{e\}$ (non-trivial center). $|Z(G)|\in\{p,p^2\}$. If $|Z(G)|=p^2$, then $G=Z(G)$ is abelian. If $|Z(G)|=p$, then $|G/Z(G)|=p$, so $G/Z(G)$ is cyclic. But $G/Z(G)$ cyclic implies $G$ abelian (standard lemma), contradicting \(|Z(G)|=p

PYQ 2015 — Jordan-Hölder Theorem

State the Jordan-Hölder theorem on composition series.

Solution

A composition series of $G$ is a chain $G=G_0\supset G_1\supset\cdots\supset G_r=\{e\}$ where each $G_{i+1}\trianglelefteq G_i$ and $G_i/G_{i+1}$ is simple (the composition factors). Jordan-Hölder Theorem: Any two composition series of a finite group $G$ have the same length $r$, and the multisets of composition factors are the same (up to isomorphism and reordering). Example: For $\mathbb{Z}_{12}$: $\mathbb{Z}_{12}\supset\langle2\rangle\supset\langle4\rangle\supset\{0\}$ with factors $\mathbb{Z}_2,\mathbb{Z}_2,\mathbb{Z}_3$; or $\mathbb{Z}_{12}\supset\langle4\rangle\supset\{0\}$ is not a composition series (not maximal at first step). The theorem guarantees the factors $\{\mathbb{Z}_2,\mathbb{Z}_2,\mathbb{Z}_3\}$ appear in any composition series. $\blacksquare$

PYQ 2014 — Solvable Groups and Feit-Thompson

Define solvable groups and state the Feit-Thompson theorem.

Solution

A group $G$ is solvable if it has a subnormal series $G=G_0\supset G_1\supset\cdots\supset G_r=\{e\}$ with each quotient $G_i/G_{i+1}$ abelian. Equivalently, the derived series $G^{(0)}=G,G^{(1)}=G'=[G,G],G^{(2)}=[G',G'],\ldots$ eventually reaches $\{e\}$. Feit-Thompson Theorem (1963): Every group of odd order is solvable. This is one of the deepest results in group theory (proof spans 255 pages). Consequence: every non-abelian simple group has even order (hence has an element of order 2 by Cauchy). $\blacksquare$

PYQ 2013 — Derived Series

Define the derived series of a group and show every abelian group is solvable.

Solution

The commutator subgroup $G'=[G,G]=\langle aba^{-1}b^{-1}:a,b\in G\rangle$. The derived series is $G^{(0)}=G\supset G^{(1)}=G'\supset G^{(2)}=(G')'\supset\cdots$. Each $G^{(i+1)}\trianglelefteq G^{(i)}$ with abelian quotient $G^{(i)}/G^{(i+1)}$. $G$ is solvable $\iff$ $G^{(n)}=\{e\}$ for some $n$. Abelian $\Rightarrow$ solvable: If $G$ is abelian, $G'=[G,G]=\{e\}$ (every commutator $aba^{-1}b^{-1}=e$). So $G^{(1)}=\{e\}$ and the derived series terminates at step 1. $\blacksquare$

PYQ 2012 — Subnormal Series of S_3

Write a composition series for $S_3$ and identify the composition factors.

Solution

$S_3\supset A_3\supset\{e\}$. Check: $A_3=\{e,(123),(132)\}\cong\mathbb{Z}_3$, normal in $S_3$ (index 2). $\{e\}\trianglelefteq A_3$. Factors: $S_3/A_3\cong\mathbb{Z}_2$ (simple); $A_3/\{e\}\cong\mathbb{Z}_3$ (simple). Both are simple (prime cyclic groups are simple). So the composition series is $S_3\supset A_3\supset\{e\}$ with factors $\{\mathbb{Z}_2,\mathbb{Z}_3\}$. Since $S_3$ is solvable (its derived series: $S_3'=A_3$, $A_3'=\{e\}$), this is consistent. $\blacksquare$

PYQ 2011 — Every Abelian Group Is Solvable

Prove every abelian group is solvable using the definition via abelian series.

Solution

A group $G$ is solvable if there exists a chain $G=G_0\supset G_1\supset\cdots\supset G_r=\{e\}$ with each $G_{i+1}\trianglelefteq G_i$ and each quotient $G_i/G_{i+1}$ abelian. If $G$ is abelian, take the trivial chain $G\supset\{e\}$: the single quotient $G/\{e\}\cong G$ is abelian. This is a solvable series of length 1. So $G$ is solvable. $\blacksquare$

PYQ 2010 — Direct Product ℤ_m × ℤ_n ≅ ℤ_{mn} iff gcd(m,n)=1

Prove $\mathbb{Z}_m\times\mathbb{Z}_n\cong\mathbb{Z}_{mn}$ if and only if $\gcd(m,n)=1$.

Solution

($\Leftarrow$) If $\gcd(m,n)=1$, consider $(1,1)\in\mathbb{Z}_m\times\mathbb{Z}_n$. Its order is $\text{lcm}(m,n)=mn/\gcd(m,n)=mn$. So $\mathbb{Z}_m\times\mathbb{Z}_n$ has an element of order $mn=|\mathbb{Z}_m\times\mathbb{Z}_n|$, hence is cyclic $\cong\mathbb{Z}_{mn}$. ($\Rightarrow$) If $\gcd(m,n)=d>1$, then every element $(a,b)\in\mathbb{Z}_m\times\mathbb{Z}_n$ satisfies $(mn/d)(a,b)=((mn/d)a\bmod m,\,(mn/d)b\bmod n)=(0,0)$ (since $m\mid mn/d$ and $n\mid mn/d$). So every element has order dividing \(mn/d

PYQ 2009 — Free Group on 2 Generators

Define the free group $F_2$ on two generators and state its key property.

Solution

The free group $F_2=F(\{a,b\})$ consists of all reduced words in letters $a,a^{-1},b,b^{-1}$ (no consecutive cancellation), with multiplication by concatenation and reduction. Universal property: For any group $G$ and any function $f:\{a,b\}\to G$, there exists a unique group homomorphism $\phi:F_2\to G$ extending $f$. Every group is a quotient of a free group (by the universal property applied to a generating set). $F_2$ is infinite and non-abelian; it contains free subgroups of all countable ranks. Abelianization: $F_2/[F_2,F_2]\cong\mathbb{Z}^2$. $\blacksquare$

PYQ 2008 — Burnside's Lemma

State Burnside's lemma and apply it to count the number of distinct necklaces with 3 beads of 2 colors.

Solution

Burnside's Lemma: If a finite group $G$ acts on a finite set $X$, the number of orbits is $\frac{1}{|G|}\sum_{g\in G}|X^g|$, where $X^g=\{x\in X:gx=x\}$. Application: $X$ = colorings of 3 beads with 2 colors (8 total). $G=\mathbb{Z}_3$ (rotations). $X^{r^0}=8$ (identity fixes all). $X^{r^1}=\{$ colorings fixed by rotation by $120°\}$: all 3 beads same color, so 2. \(X^{r^2}=2$ similarly. Orbits $= (8+2+2)/3=12/3=4$. The 4 distinct necklaces are: (RRR), (BBB), (RRB), (RBB) — where R=red, B=blue. $\blacksquare$

PYQ 2007 — Automorphism Group of ℤ_n

Determine the automorphism group $\text{Aut}(\mathbb{Z}_n)$.

Solution

An automorphism $\phi:\mathbb{Z}_n\to\mathbb{Z}_n$ is determined by $\phi(1)=k$ where $\gcd(k,n)=1$ (since $\phi(1)$ must generate $\mathbb{Z}_n$). Conversely, for each $k$ with $\gcd(k,n)=1$, the map $1\mapsto k$ extends to an automorphism $m\mapsto km\bmod n$. Composition corresponds to multiplication: $\phi_k\circ\phi_l=\phi_{kl\bmod n}$. Hence $\text{Aut}(\mathbb{Z}_n)\cong(\mathbb{Z}/n\mathbb{Z})^\times$, the multiplicative group of units mod $n$, of order $\phi(n)$ (Euler's totient). For prime $p$: $\text{Aut}(\mathbb{Z}_p)\cong\mathbb{Z}_{p-1}$. $\blacksquare$

PYQ 2006 — Inner Automorphisms Form a Normal Subgroup

Show that the inner automorphisms $\text{Inn}(G)$ form a normal subgroup of $\text{Aut}(G)$, and $\text{Inn}(G)\cong G/Z(G)$.

Solution

For each $a\in G$, define $\iota_a:G\to G$ by $\iota_a(g)=aga^{-1}$. Then $\iota_a\in\text{Aut}(G)$ (conjugation is an automorphism). $\text{Inn}(G)=\{\iota_a:a\in G\}$. Subgroup: $\iota_a\iota_b=\iota_{ab}$, $(\iota_a)^{-1}=\iota_{a^{-1}}$, $\text{id}=\iota_e$. Normal in Aut(G): For $\phi\in\text{Aut}(G)$: $\phi\iota_a\phi^{-1}(g)=\phi(a\phi^{-1}(g)a^{-1})=\phi(a)g\phi(a)^{-1}=\iota_{\phi(a)}(g)$. So $\phi\iota_a\phi^{-1}=\iota_{\phi(a)}\in\text{Inn}(G)$. Isomorphism: The map $G\to\text{Inn}(G)$, $a\mapsto\iota_a$, is a surjective homomorphism with kernel $Z(G)$. By First Isomorphism Theorem: $\text{Inn}(G)\cong G/Z(G)$. $\blacksquare$

PYQ 2005 — Simple Groups

Define simple groups. Show $\mathbb{Z}_p$ is simple and $A_5$ is simple.

Solution

A group $G$ is simple if its only normal subgroups are $\{e\}$ and $G$ itself. $\mathbb{Z}_p$ is simple: $\mathbb{Z}_p$ has order $p$ (prime), so by Lagrange, its only subgroups have order 1 or $p$. In an abelian group, every subgroup is normal. Hence the only normal subgroups are $\{0\}$ and $\mathbb{Z}_p$. $A_5$ is simple: $|A_5|=60$. The conjugacy classes of $A_5$ (as permutations of 5 elements) have sizes 1, 15, 20, 12, 12. Any normal subgroup is a union of conjugacy classes containing $e$, so its order divides 60 and is a sum of 1 and some subset of $\{15,20,12,12\}$. The only sums dividing 60 are 1 (just $\{e\}$) and 60 (all). Hence $A_5$ is simple. It is the smallest non-abelian simple group. $\blacksquare$

§ PYQ — Vector Spaces, Modules & Linear Algebra (Algebraic)

PYQ 2024 — Field Extension [ℚ(√2):ℚ] = 2

Show that $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$.

Solution

$\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}:a,b\in\mathbb{Q}\}$. The minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}$ is $m(x)=x^2-2$, which is irreducible over $\mathbb{Q}$ (no rational roots: $\pm1,\pm2$ fail). Since $m$ is irreducible of degree 2, $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=\deg(m)=2$. A $\mathbb{Q}$-basis for $\mathbb{Q}(\sqrt{2})$ is $\{1,\sqrt{2}\}$. $\blacksquare$

PYQ 2023 — Field Extension [ℚ(∜2):ℚ] = 4

Show that $[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]=4$.

Solution

Let $\alpha=\sqrt[4]{2}=2^{1/4}$. The minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $m(x)=x^4-2$. This is irreducible over $\mathbb{Q}$ by Eisenstein's criterion with $p=2$: $2\nmid1$, $2\mid0,0,0,-2$, $4\nmid-2$. Wait: $4\nmid 2$ ✓. So $x^4-2$ is irreducible of degree 4, giving $[\mathbb{Q}(\alpha):\mathbb{Q}]=4$. Basis: $\{1,\alpha,\alpha^2,\alpha^3\}=\{1,2^{1/4},\sqrt{2},2^{3/4}\}$. $\blacksquare$

PYQ 2022 — Splitting Field of x²+1 over ℚ

Show the splitting field of $x^2+1$ over $\mathbb{Q}$ is $\mathbb{Q}(i)$ with degree 2.

Solution

The roots of $x^2+1$ are $\pm i$. The splitting field is the smallest field containing $\mathbb{Q}$ and all roots: $\mathbb{Q}(i,{-i})=\mathbb{Q}(i)$ (since $-i=0-i\in\mathbb{Q}(i)$). The minimal polynomial of $i$ over $\mathbb{Q}$ is $x^2+1$, irreducible (no real rational roots). So $[\mathbb{Q}(i):\mathbb{Q}]=2$. Basis: $\{1,i\}$. As a field: $\mathbb{Q}(i)\cong\mathbb{Q}[x]/\langle x^2+1\rangle$. $\blacksquare$

PYQ 2021 — Algebraic Closure

Define the algebraic closure $\bar{F}$ of a field $F$ and state its key properties.

Solution

A field $\bar{F}$ is the algebraic closure of $F$ if: (1) $F\subseteq\bar{F}$ (or $F$ embeds in $\bar{F}$); (2) $\bar{F}$ is algebraic over $F$ (every element of $\bar{F}$ is algebraic over $F$); (3) $\bar{F}$ is algebraically closed: every non-constant polynomial in $\bar{F}[x]$ has a root in $\bar{F}$. Equivalently: every polynomial in $F[x]$ splits completely in $\bar{F}[x]$. Existence and uniqueness: Every field has an algebraic closure, unique up to $F$-isomorphism (Steinitz). Examples: $\bar{\mathbb{Q}}$ = algebraic numbers; $\overline{\mathbb{F}_p}$ = algebraic closure of finite field; $\bar{\mathbb{R}}=\mathbb{C}$. $\blacksquare$

PYQ 2020 — Minimal Polynomial of √2+√3 over ℚ

Find the minimal polynomial of $\alpha=\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$ and show $[\mathbb{Q}(\alpha):\mathbb{Q}]=4$.

Solution

$\alpha=\sqrt{2}+\sqrt{3}$. Then $\alpha^2=5+2\sqrt{6}$, so $\alpha^2-5=2\sqrt{6}$, so $(\alpha^2-5)^2=24$, giving $\alpha^4-10\alpha^2+25=24$, i.e., $\alpha^4-10\alpha^2+1=0$. So $m(x)=x^4-10x^2+1$. Irreducibility: check $m$ has no rational roots ($\pm1$ fail) and does not factor into two quadratics over $\mathbb{Q}$: if $m=(x^2+ax+b)(x^2-ax+c)$ then expanding: $a=0$ (coefficient of $x^3$ is 0), $b+c-a^2=-10$, so $b+c=-10$, and $bc=1$, but $b+c=-10,bc=1$ gives $b,c=(-10\pm\sqrt{96})/2\notin\mathbb{Q}$. So $m$ is irreducible of degree 4, and $[\mathbb{Q}(\alpha):\mathbb{Q}]=4$. $\blacksquare$

PYQ 2019 — Finite Fields: GF(p^n)

State the classification of finite fields: there exists a field of order $p^n$ for each prime $p$ and $n\ge1$, unique up to isomorphism.

Solution

Existence: The splitting field of $x^{p^n}-x$ over $\mathbb{F}_p$ has exactly $p^n$ elements (the roots of $x^{p^n}-x$ are distinct — derivative is $-1\ne0$ — and form a field). Uniqueness: Any field of order $p^n$ has characteristic $p$, so contains $\mathbb{F}_p$. Every element $\alpha$ satisfies $\alpha^{p^n}=\alpha$ (multiplicative group has order $p^n-1$). So every element is a root of $x^{p^n}-x$, meaning the field is the splitting field of $x^{p^n}-x$ over $\mathbb{F}_p$. Splitting fields are unique up to isomorphism. We write $\text{GF}(p^n)=\mathbb{F}_{p^n}$. $\blacksquare$

PYQ 2018 — Number of Irreducible Polynomials of Degree n over GF(p)

Find the number of monic irreducible polynomials of degree $n$ over $\mathbb{F}_p$.

Solution

Let $N(n)$ = number of monic irreducible polynomials of degree $n$ over $\mathbb{F}_p$. Since $x^{p^n}-x=\prod_{d\mid n}\prod_{f\text{ irred, deg }d}f(x)$ in $\mathbb{F}_p[x]$, taking degrees: $p^n=\sum_{d\mid n}d\cdot N(d)$. By Möbius inversion: $N(n)=\frac{1}{n}\sum_{d\mid n}\mu(n/d)p^d$, where $\mu$ is the Möbius function. For $n=1$: $N(1)=p$. For $n=2$: $N(2)=(p^2-p)/2=p(p-1)/2$. For prime $n=q$: $N(q)=(p^q-p)/q$. $\blacksquare$

PYQ 2017 — Galois Group of x⁴−2 over ℚ Is D_4

State the result that $\text{Gal}(\mathbb{Q}(\sqrt[4]{2},i)/\mathbb{Q})\cong D_4$ and describe the generators.

Solution

The splitting field of $x^4-2$ over $\mathbb{Q}$ is $L=\mathbb{Q}(\sqrt[4]{2},i)$ (roots are $\pm\sqrt[4]{2},\pm i\sqrt[4]{2}$). We have $[L:\mathbb{Q}]=8$. The Galois group $G=\text{Gal}(L/\mathbb{Q})$ has $|G|=[L:\mathbb{Q}]=8$. The two generators are: $\sigma:\sqrt[4]{2}\mapsto i\sqrt[4]{2},\, i\mapsto i$ (rotation of roots, order 4); $\tau:\sqrt[4]{2}\mapsto\sqrt[4]{2},\, i\mapsto -i$ (complex conjugation, order 2). They satisfy $\tau\sigma\tau^{-1}=\sigma^{-1}$, so $G=\langle\sigma,\tau\rangle\cong D_4$. $\blacksquare$

PYQ 2016 — Fundamental Theorem of Galois Theory

State the Fundamental Theorem of Galois Theory.

Solution

Let $L/F$ be a finite Galois extension with Galois group $G=\text{Gal}(L/F)$. There is an inclusion-reversing bijection (the Galois correspondence): \[\{\text{intermediate fields } F\subseteq K\subseteq L\}\longleftrightarrow\{\text{subgroups } H\le G\}\] given by $K\mapsto\text{Gal}(L/K)$ and $H\mapsto L^H$ (fixed field). Under this correspondence: (1) $[L:K]=|H|$ and $[K:F]=[G:H]$. (2) $K/F$ is Galois $\iff$ $H\trianglelefteq G$, and then $\text{Gal}(K/F)\cong G/H$. (3) Lattice of subfields reverses the lattice of subgroups. $\blacksquare$

PYQ 2015 — Solvability by Radicals ↔ Solvable Galois Group

State the theorem connecting solvability by radicals and solvable Galois groups.

Solution

Theorem (Galois): A polynomial $f(x)\in\mathbb{Q}[x]$ is solvable by radicals (its roots can be expressed using $+,-,\times,\div$ and $n$th root operations) if and only if its Galois group $\text{Gal}(f)$ (= Galois group of its splitting field over $\mathbb{Q}$) is a solvable group. Consequence: The general quintic $x^5-5x+1$ has Galois group $S_5$, which is not solvable (since $A_5$ is simple non-abelian, making the derived series not terminate). Hence the general degree 5 polynomial is not solvable by radicals — no quintic formula exists. $\blacksquare$

PYQ 2014 — Extensions of Degree 2 Are Galois

Prove every extension of degree 2 over a field of characteristic $\ne2$ is Galois.

Solution

Let $[K:F]=2$ and $\text{char}(F)\ne2$. Pick $\alpha\in K\setminus F$; then $\{1,\alpha\}$ is a basis of $K/F$. The minimal polynomial of $\alpha$ over $F$ has degree 2: $m(x)=x^2+bx+c\in F[x]$. Complete the square: $\alpha=-b/2+\beta$ where $\beta=\alpha+b/2$ satisfies $\beta^2=(b/2)^2-c=d\in F$. So $K=F(\beta)$ and $\beta^2=d$. The splitting field of $x^2-d$ over $F$ is $F(\beta)$ (both roots $\pm\beta\in F(\beta)$ since char$\ne2$). A splitting field of a separable polynomial is Galois. Hence $K/F$ is Galois with $\text{Gal}(K/F)\cong\mathbb{Z}_2$. $\blacksquare$

PYQ 2013 — Normal Extension

Define normal extension and give an example of a non-normal extension.

Solution

An algebraic extension $K/F$ is normal if every irreducible polynomial $f\in F[x]$ that has at least one root in $K$ splits completely in $K[x]$. Equivalently: $K$ is a splitting field of some family of polynomials over $F$. Non-normal example: $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$. The minimal polynomial $x^3-2$ has one real root $\sqrt[3]{2}\in\mathbb{Q}(\sqrt[3]{2})$ but the other roots $\sqrt[3]{2}\omega,\sqrt[3]{2}\omega^2$ (where $\omega=e^{2\pi i/3}$) are complex and not in $\mathbb{Q}(\sqrt[3]{2})\subset\mathbb{R}$. So $x^3-2$ does not split completely in $\mathbb{Q}(\sqrt[3]{2})$. $\blacksquare$

PYQ 2012 — Separable Extension

Define separable extension and separable polynomial. State when finite fields give separable extensions.

Solution

A polynomial $f\in F[x]$ is separable if it has no repeated roots in its splitting field (equivalently, $\gcd(f,f')=1$). An algebraic element $\alpha$ over $F$ is separable if its minimal polynomial is separable. An algebraic extension $K/F$ is separable if every element of $K$ is separable over $F$. Finite fields: Every extension of finite fields $\mathbb{F}_{p^n}/\mathbb{F}_p$ is separable (and Galois). The Frobenius $\phi:\alpha\mapsto\alpha^p$ generates $\text{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)\cong\mathbb{Z}_n$. Over fields of characteristic 0 (like $\mathbb{Q}$), every algebraic extension is separable (since $f'=0$ only if $f=0$ in char 0). $\blacksquare$

PYQ 2011 — Cyclotomic Polynomial Φ_n(x)

Define the cyclotomic polynomial $\Phi_n(x)$ and show $\Phi_p(x)=1+x+\cdots+x^{p-1}$ is irreducible over $\mathbb{Q}$.

Solution

$\Phi_n(x)=\prod_{\substack{1\le k\le n\\\gcd(k,n)=1}}(x-e^{2\pi ik/n})$, the product over primitive $n$th roots of unity. It has degree $\phi(n)$ (Euler's totient) and integer coefficients. For prime $p$: $\Phi_p(x)=(x^p-1)/(x-1)=x^{p-1}+\cdots+x+1$. Irreducibility: substitute $x=y+1$: $\Phi_p(y+1)=\frac{(y+1)^p-1}{y}=y^{p-1}+\binom{p}{1}y^{p-2}+\cdots+\binom{p}{p-1}$. Apply Eisenstein with prime $p$: leading coeff 1 (not div by $p$), all others $\binom{p}{k}$ for $1\le k\le p-1$ are divisible by $p$, constant term $\binom{p}{p-1}=p$ not divisible by $p^2$. So $\Phi_p(y+1)$ is irreducible over $\mathbb{Q}$, hence $\Phi_p(x)$ is irreducible. $\blacksquare$

PYQ 2010 — Galois Group of x³−2 over ℚ Is S_3

Find $\text{Gal}(\mathbb{Q}(\sqrt[3]{2},\omega)/\mathbb{Q})$ where $\omega=e^{2\pi i/3}$.

Solution

The splitting field of $x^3-2$ over $\mathbb{Q}$ is $L=\mathbb{Q}(\sqrt[3]{2},\omega)$ (roots: $\sqrt[3]{2},\sqrt[3]{2}\omega,\sqrt[3]{2}\omega^2$). $[L:\mathbb{Q}]=6$ (since $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$ and $[L:\mathbb{Q}(\sqrt[3]{2})]=2$). So $|G|=6$. The two generators: $\sigma:\sqrt[3]{2}\mapsto\sqrt[3]{2}\omega,\,\omega\mapsto\omega$ (order 3); $\tau:\sqrt[3]{2}\mapsto\sqrt[3]{2},\,\omega\mapsto\omega^2$ (order 2, complex conjugation on $\omega$). One computes $\tau\sigma\tau^{-1}=\sigma^{-1}$. So $G=\langle\sigma,\tau\rangle\cong S_3\cong D_3$. $\blacksquare$

PYQ 2009 — Galois Correspondence: Subfields ↔ Subgroups

For the extension $\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$, draw the lattice of subfields and corresponding subgroups.

Solution

$L=\mathbb{Q}(\sqrt2,\sqrt3)$, $[L:\mathbb{Q}]=4$. Galois group $G=\text{Gal}(L/\mathbb{Q})\cong\mathbb{Z}_2\times\mathbb{Z}_2$ with generators: $\sigma:\sqrt2\mapsto-\sqrt2,\sqrt3\mapsto\sqrt3$; $\tau:\sqrt2\mapsto\sqrt2,\sqrt3\mapsto-\sqrt3$. Subgroups of $G$: $\{e\},\langle\sigma\rangle,\langle\tau\rangle,\langle\sigma\tau\rangle,G$. Fixed fields (by Galois correspondence): $L^{\{e\}}=L$; $L^{\langle\sigma\rangle}=\mathbb{Q}(\sqrt3)$; $L^{\langle\tau\rangle}=\mathbb{Q}(\sqrt2)$; $L^{\langle\sigma\tau\rangle}=\mathbb{Q}(\sqrt6)$ ($\sigma\tau$ sends $\sqrt6\mapsto(-\sqrt2)(-\sqrt3)=\sqrt6$); $L^G=\mathbb{Q}$. The lattice of subfields is the reverse of the lattice of subgroups. $\blacksquare$

PYQ 2008 — Modules over ℤ Are Abelian Groups

Show that $\mathbb{Z}$-modules are exactly abelian groups.

Solution

A $\mathbb{Z}$-module $M$ is an abelian group $(M,+)$ together with scalar multiplication $\mathbb{Z}\times M\to M$ satisfying the module axioms. But for abelian groups, the action of $n\in\mathbb{Z}$ on $m\in M$ is forced: $n\cdot m=m+m+\cdots+m$ ($n$ times) for $n>0$, $0\cdot m=0$, $n\cdot m=-(|n|\cdot m)$ for $n<0$. This is the unique $\mathbb{Z}$-module structure on any abelian group, and it satisfies all module axioms. Conversely, any $\mathbb{Z}$-module is an abelian group. Hence $\mathbb{Z}\text{-Mod}=\text{Ab}$ (category equivalence). $\blacksquare$

PYQ 2007 — Free Module of Rank n

Define a free module of rank $n$ over a ring $R$ and give key properties.

Solution

A free $R$-module of rank $n$ is $R^n=R\oplus R\oplus\cdots\oplus R$ ($n$ copies), or equivalently any $R$-module with a basis of $n$ elements (a linearly independent spanning set). Universal property: For any $R$-module $M$ and any set map from a basis $\{e_1,\ldots,e_n\}$ to $M$, there is a unique $R$-module homomorphism $R^n\to M$ extending it. Every $R$-module is a quotient of a free module. For commutative rings, any two bases of a free module have the same cardinality (invariant basis number property), so rank is well-defined. $\blacksquare$

PYQ 2006 — Structure Theorem for Finitely Generated Abelian Groups

State the structure theorem and classify $\mathbb{Z}_2\oplus\mathbb{Z}_4\oplus\mathbb{Z}$ vs $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_4$.

Solution

Theorem: Every finitely generated abelian group $G$ is isomorphic to $\mathbb{Z}^r\oplus\mathbb{Z}_{d_1}\oplus\cdots\oplus\mathbb{Z}_{d_k}$ where $r\ge0$ (free rank), $d_1\mid d_2\mid\cdots\mid d_k\ge2$ (invariant factors). This decomposition is unique. Comparison: $\mathbb{Z}_2\oplus\mathbb{Z}_4\oplus\mathbb{Z}$: torsion part $\mathbb{Z}_2\oplus\mathbb{Z}_4$ (invariant factors 2,4 since $2\mid4$), free rank 1. $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_4$: invariant factors would need $d_1\mid d_2\mid d_3$; $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_4\cong\mathbb{Z}_2\oplus(\mathbb{Z}_2\oplus\mathbb{Z}_4)$. Since $2\mid4$, the invariant form is $\mathbb{Z}_2\oplus\mathbb{Z}_{4}\oplus...$ — actually $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_4$ in primary decomposition has different structure from $\mathbb{Z}_2\oplus\mathbb{Z}_4$, so they are not isomorphic. $\blacksquare$

PYQ 2005 — Primary Decomposition of Abelian Groups

State the primary decomposition theorem and apply it to classify all abelian groups of order 36.

Solution

Primary Decomposition: Every finite abelian group $G$ is isomorphic to a direct sum of cyclic groups of prime power order: $G\cong\mathbb{Z}_{p_1^{a_1}}\oplus\cdots\oplus\mathbb{Z}_{p_k^{a_k}}$. This is unique up to reordering. Order 36 = 4 · 9 = 2² · 3²: The abelian groups of order 36 correspond to partitions of the exponent at each prime: For $p=2$ (exponent 2): partitions of 2: $(2)\to\mathbb{Z}_4$; $(1,1)\to\mathbb{Z}_2\oplus\mathbb{Z}_2$. For $p=3$ (exponent 2): partitions of 2: $(2)\to\mathbb{Z}_9$; $(1,1)\to\mathbb{Z}_3\oplus\mathbb{Z}_3$. Combining: $\mathbb{Z}_4\oplus\mathbb{Z}_9\cong\mathbb{Z}_{36}$; $\mathbb{Z}_4\oplus\mathbb{Z}_3\oplus\mathbb{Z}_3\cong\mathbb{Z}_4\oplus\mathbb{Z}_3^2$; $\mathbb{Z}_2^2\oplus\mathbb{Z}_9$; $\mathbb{Z}_2^2\oplus\mathbb{Z}_3^2$. Four distinct abelian groups of order 36. $\blacksquare$