Cosets and Lagrange’s theorem

At a Glance

• Frequency: 5 sub-parts across 5 of 14 years (2019, 2023, 2024, 2025, 2026)
• Priority tier: T3
• Marks (count): 10 (4)
• Average solve time: ~6 min
• Difficulty mix: easy 2, medium 2
• Section: A | Dominant type: proof

Why This Chapter Matters

Coset and Lagrange questions appear in Section A compulsory (Q1) in four recent years, always for 10 marks. All questions use one of two weapons: (a) the First Isomorphism Theorem — the image order divides the domain order — or (b) the product-set formula $|HK|=|H||K|/|H\cap K|$ combined with $|HK|\le|G|$. Sylow III provides a third tool for questions about counting subgroups of prime order. Master these three tools and all four historical questions are solved in under 7 minutes each.

Minimum Theory

Lagrange’s theorem. If $H\le G$ (finite group), then $|H|$ divides $|G|$ and $|G|=[G:H]\cdot|H|$.

Index multiplication. For $K\le H\le G$: $[G:K]=[G:H]\cdot[H:K]$.

First Isomorphism Theorem. If $\phi:G\to G'$ is a homomorphism then $G/\ker\phi\cong\operatorname{Im}\phi$. So $|\operatorname{Im}\phi|$ divides $|G|$. For a surjection, $|G'|$ divides $|G|$.

Product-set formula. For finite subgroups $H,K\le G$: $|HK|=\frac{|H|\cdot|K|}{|H\cap K|}, \qquad |HK|\le|G|.$ These two combine to give $|H\cap K|\ge|H||K|/|G|$.

Sylow III. The number $n_p$ of Sylow $p$-subgroups satisfies $n_p\equiv1\pmod p$ and $n_p\mid[G:P]$.

The centre, and the lemma that makes Lagrange bite on it. The centre is $Z(G)=\{a\in G:\ ax=xa\ \text{for every }x\in G\}.$ It is a normal subgroup ($gzg^{-1}=z$ for $z\in Z(G)$), so $|Z(G)|$ divides $|G|$ and the quotient $G/Z(G)$ exists — both facts are needed together.

Lemma. If $G/Z(G)$ is cyclic, then $G$ is abelian (hence $Z(G)=G$).

Proof. Write $Z=Z(G)$ and suppose $G/Z=\langle gZ\rangle$. Any $a,b\in G$ have $a=g^iz_1$, $b=g^jz_2$ with $z_1,z_2\in Z$. Since central elements commute with everything and powers of $g$ commute with each other, $ab=g^{i+j}z_1z_2=g^{j}z_2g^{i}z_1=ba$. $\blacksquare$

The lemma is a contrapositive engine: it says $[G:Z(G)]$ can never be prime, because a prime-order quotient is cyclic and would force $G$ abelian, i.e. $[G:Z(G)]=1$. Translating a sentence about commuting elements into a statement about $|Z(G)|$ is the first move in any centre question.

Question Archetypes

sylow-counting (1 question(s); 2024)

Recognition Cues — $|G|=mn$ with $m,n$ prime and $m>n$; asked for “at most one subgroup of order $m$.”

Solution Template

1. Subgroups of order $m$ are precisely the Sylow $m$-subgroups.
2. Sylow III: $n_m\equiv1\pmod m$ and $n_m\mid n$ (since $[G:P]=n$).
3. Divisors of prime $n$ are $\{1,n\}$. Rule out $n_m=n$: that would need $n\equiv1\pmod m$, i.e. $m\mid(n-1)$; impossible since $m>n>n-1\ge0$.
4. Conclude $n_m=1$.

Worked Example

2024 Paper 2, 2024-P2-Q1a (10 marks)

$G$ has order $mn$, $m>n$ prime. Show $G$ has at most one subgroup of order $m$.

Sylow III gives $n_m\in\{1,n\}$ and $n_m\equiv1\pmod{m}$. If $n_m=n$, then $m\mid(n-1)$; impossible since $m>n>n-1$. Hence $n_m=1$. $\boxed{\text{At most one subgroup of order }m.}\;\blacksquare$

order-counting-argument (1 question(s); 2025)

Recognition Cues — $o(H),o(K)>\sqrt{o(G)}$; prove nontrivial intersection.

Worked Example

2025 Paper 2, 2025-P2-Q1a (10 marks)

$H,K\le G$, $o(H),o(K)>\sqrt{o(G)}$. Show $H\cap K\ne\{e\}$.

Assume $|H\cap K|=1$. Then $|HK|=|H||K|>\sqrt{|G|}\cdot\sqrt{|G|}=|G|$. But $|HK|\le|G|$ — contradiction. Hence $|H\cap K|>1$.

$\boxed{H\cap K\ne\{e\}.}\;\blacksquare$

Proof of the product-set formula (needed to quote it): Map $H\times K\to HK$ by $(h,k)\mapsto hk$. Each element $g\in HK$ has exactly $|H\cap K|$ preimages $(ht, t^{-1}k)$ for $t\in H\cap K$. So $|H||K|=|HK|\cdot|H\cap K|$, giving $|HK|=|H||K|/|H\cap K|$.

subgroup-existence (1 question(s); 2016)

Worked Example

2016 Paper 2, 2016-P2-Q2b (10 marks, coset part)

Every nonzero element of $\mathbb{Z}_p$ ($p$ prime) generates $\mathbb{Z}_p$.

Let $a\ne0$ in $\mathbb{Z}_p$. By Lagrange, $|\langle a\rangle|$ divides $p$. Since $p$ is prime, $|\langle a\rangle|\in\{1,p\}$. Since $a\ne0$, $|\langle a\rangle|\ne1$, so $|\langle a\rangle|=p=|\mathbb{Z}_p|$, giving $\langle a\rangle=\mathbb{Z}_p$. $\blacksquare$

(Without Lagrange: $\gcd(a,p)=1$, so the $p$ multiples $0,a,2a,\ldots,(p-1)a$ are distinct mod $p$, exhausting $\mathbb{Z}_p$.)

onto-homomorphism (1 question(s); 2023)

Worked Example

2023 Paper 2, 2023-P2-Q1a (10 marks)

Does an onto homomorphism from $|G|=10$ to $|G'|=6$ exist?

If $\phi:G\twoheadrightarrow G'$ exists, then by the First Isomorphism Theorem $|G'|=[G:\ker\phi]$ divides $|G|=10$. But $6\nmid10$. No such homomorphism exists. $\blacksquare$

centre-order-dichotomy (1 question(s); 2026)

The tell is a question phrased entirely about elements, whose subject is really the order of the centre. Lagrange restricts $|Z(G)|$ to the divisors of $|G|$; the $G/Z(G)$-cyclic lemma kills the intermediate ones; only the two extremes survive, and those two extremes are the dichotomy the question states.

Recognition Cues

• $|G|=pq$ with $p,q$ prime (or $|G|=p^2$).
• Wording about commuting: “no non-identity element commutes with every element”, ”$G$ is abelian or …”, ”$Z(G)$ is trivial”.
• An “either … or …” conclusion with no third possibility offered.

Solution Template

1. Translate the sentence into the centre, at the top. ”$G$ is abelian” $\iff Z(G)=G$; “no non-identity element commutes with everything” $\iff Z(G)=\{e\}$. State what is now to be proved.
2. Show $Z(G)\trianglelefteq G$ so that Lagrange applies and $G/Z(G)$ exists.
3. List the divisors: $|Z(G)|\in\{1,p,q,pq\}$. The two extremes are the two alternatives; the whole proof is eliminating the middle.
4. Prove the lemma, do not quote it — this is where the marks are.
5. Eliminate: if $1<|Z(G)|<pq$ then $[G:Z(G)]$ is prime, hence $G/Z(G)$ is cyclic, hence $G$ is abelian, hence $Z(G)=G$ — contradiction.

Worked Example

2026 Paper 2, 2026-P2-Q1a (10 marks)

Let $G$ be a group of order $pq$, $p$ and $q$ primes. Show that either $G$ is abelian or no non-identity element commutes with every element of $G$.

Source: analysis/solutions/2026-P2-Q1a.md

Step 1 — Translate. The claim is exactly $Z(G)=G$ or $Z(G)=\{e\}$: the centre of a group of order $pq$ can never be a proper non-trivial subgroup.

Step 2 — Lagrange. $Z(G)\le G$ and is normal, so $|Z(G)|$ divides $pq$ and $G/Z(G)$ is defined. By unique factorisation the divisors of $pq$ are $1,p,q,pq$ (if $p=q$ this reads $1,p,p^2$ — the middle two coincide, which changes nothing).

Step 3 — Eliminate the middle. Suppose $1<|Z(G)|<pq$. Then $[G:Z(G)]=pq/|Z(G)|\in\{p,q\}$ is prime. A group of prime order is cyclic, so $G/Z(G)$ is cyclic, so by the Lemma $G$ is abelian, so $Z(G)=G$ and $|Z(G)|=pq$ — contradicting $|Z(G)|<pq$.

Step 4 — Conclude. Only $|Z(G)|=pq$ (so $G$ is abelian) and $|Z(G)|=1$ (so no non-identity element is central) survive. $\blacksquare$

Two remarks that finish the answer. (i) The proof never assumes $p\ne q$: it uses only that every proper non-trivial divisor of $pq$ is prime, which holds for $p=q$ too, so no case-split is needed. (ii) Both alternatives genuinely occur — $\mathbb{Z}_{pq}$ is abelian, while $S_3$ (order $6$) and the non-abelian group of order $21$ have trivial centre — so neither can be dropped.

Common Traps

• Quoting the $G/Z(G)$-cyclic lemma. It is the marked step in a 10-mark question. Prove it in three lines.
• ”$G/Z$ has prime order, so $G$ has prime order.” False, and it is not what the lemma says. The content is that cyclicity of the quotient lifts commutativity back to $G$.
• Assuming $p\ne q$. The question does not say it. One line (remark (i)) covers $p=q$ and costs nothing; assuming distinctness is assuming a hypothesis you were not given.
• Not translating the sentence at the start. “No non-identity element commutes with every element” must become $Z(G)=\{e\}$ on line one, or the rest of the proof is about a different object than the question.

Common Traps (all archetypes)

• Direction of divisibility. For a surjection, $|G'|$ divides $|G|$ (not the other way). For an injection, $|G|$ divides $|G'|$.
• Sylow: two conditions, not one. Using only $n_p\mid[G:P]$ without checking the $\equiv1\pmod p$ condition (or vice versa) leaves the argument incomplete.
• Product-set formula needs a proof or citation. In a 10-mark proof question, stating $|HK|=|H||K|/|H\cap K|$ without justification loses 3–4 marks. Give the one-line preimage-counting proof.
• Primality is crucial. The result “every nonzero element generates” requires $p$ prime; for composite $n$, elements with $\gcd(a,n)>1$ do not generate $\mathbb{Z}_n$.

Marks-Aware Writing

All four questions are 10 marks. Each proof needs: (a) the theorem invoked (named), (b) its application to the specific numerical data, (c) the case analysis ruling out the unwanted value, and (d) a boxed conclusion with $\blacksquare$. Stating the theorem without applying it to the numbers earns ≤3 marks.

Practice Set

• 2019-P2-Q1a (10 m) — — Hint: index multiplication; use a bijection between cosets of $K$ in $G$ and cosets of $K$ in $H$ paired with cosets of $H$ in $G$.