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UPSC 2026 Maths Optional Paper 2 Q1a — Step-by-Step Solution

10 marks · Section A

Cosets and Lagrange's theorem · Algebra · asked 5× in 14 yrs · Read the full method →

Question

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

Technique

“No non-identity element commutes with every element of $G$” is, word for word, the statement $Z(G)=\{e\}$. So the question is not really about $G$ at all — it is about the possible orders of the centre. Lagrange restricts $|Z(G)|$ to the divisors $1,p,q,pq$ of $pq$; the two middle values are killed by the standard lemma “if $G/Z(G)$ is cyclic then $G$ is abelian”, because a group of prime order is cyclic. The trigger to recognise in the hall: order $=$ product of two primes $+$ a question about the centre $\Rightarrow$ Lagrange on $|Z(G)|$ plus the $G/Z(G)$ lemma.

Solution

Step 0 — Translate the statement into the language of the centre.

Recall the centre of $G$,

Then:

• ”$G$ is abelian” $\iff Z(G)=G$;
• “no non-identity element commutes with every element of $G$” $\iff Z(G)=\{e\}$.

So the assertion to be proved is exactly

Equivalently: the centre of a group of order $pq$ can never be a proper non-trivial subgroup. That is what we prove.

Step 1 — $Z(G)$ is a normal subgroup, so Lagrange applies and $G/Z(G)$ exists.

$e\in Z(G)$, so $Z(G)\neq\varnothing$. If $b\in Z(G)$ and $x\in G$, then from $bx=xb$, multiplying on the left and on the right by $b^{-1}$,

so $b^{-1}\in Z(G)$. And if also $a\in Z(G)$, then for every $x\in G$

so $ab^{-1}\in Z(G)$. Hence $Z(G)\le G$ by the subgroup criterion.

It is normal: for $g\in G$, $z\in Z(G)$, $gzg^{-1}=zgg^{-1}=z\in Z(G)$. Hence the quotient group $G/Z(G)$ is defined, and by Lagrange

By unique factorisation the divisors of $pq$ are exactly

(if $p=q$ this list reads $1,p,p^{2}$ — the middle two coincide, which changes nothing below). Therefore

The two extreme values are precisely the two alternatives in the statement. So the whole proof consists of eliminating the two middle values.

Step 2 — The key lemma: if $G/Z(G)$ is cyclic, then $G$ is abelian.

Proof of the lemma. Write $Z=Z(G)$ and suppose $G/Z=\langle gZ\rangle$ for some $g\in G$. Let $a,b\in G$ be arbitrary. Their cosets are powers of $gZ$, so there are integers $i,j$ with

that is,

Since $z_{1},z_{2}$ commute with everything, and powers of $g$ commute with each other,

As $a,b$ were arbitrary, $G$ is abelian. $\qquad\blacksquare$

(The lemma is genuinely needed in full: it is not enough to say ”$G/Z$ has prime order so $G$ has prime order” — that is false. The content is that cyclicity of the quotient lifts commutativity back to $G$.)

Step 3 — Eliminate $|Z(G)|=p$ and $|Z(G)|=q$.

Suppose, for contradiction, that $Z(G)$ is a proper non-trivial subgroup, i.e. $1<|Z(G)|<pq$. By Step 1 the index is

which is a prime number in either case.

A group of prime order is cyclic (any non-identity element generates a subgroup whose order divides the prime and exceeds $1$, hence equals the whole group). So $G/Z(G)$ is cyclic.

By the Lemma of Step 2, $G$ is abelian, and therefore $Z(G)=G$, i.e. $|Z(G)|=pq$. This contradicts $|Z(G)|<pq$.

Hence $|Z(G)|=p$ and $|Z(G)|=q$ are impossible.

Step 4 — Conclusion.

Only the two extreme values survive:

• $|Z(G)|=pq$: then $Z(G)=G$, i.e. $G$ is abelian;
• $|Z(G)|=1$: then $Z(G)=\{e\}$, i.e. no non-identity element of $G$ commutes with every element of $G$.

Exactly one of the two alternatives holds (they are mutually exclusive as soon as $pq>1$, which is automatic). $\qquad\blacksquare$

Step 5 — Two remarks the examiner will look for.

(i) The argument never assumes $p\neq q$. The only fact used about $pq$ is that its divisors are $1,p,q,pq$, so that every proper non-trivial divisor is prime. When $p=q$ the divisor list of $p^{2}$ is $1,p,p^{2}$ and the unique proper non-trivial index is $p$ — still prime, so Step 3 runs verbatim. Hence the statement and the proof are valid for $p=q$ as well, and no case-split on $p\overset{?}{=}q$ is needed.

(ii) Both alternatives genuinely occur, so neither can be dropped.

• $G=\mathbb{Z}_{pq}$ (cyclic) is abelian: the first alternative.
• $G=S_{3}$, of order $6=2\cdot 3$, has $Z(S_{3})=\{e\}$: the second alternative. Likewise the non-abelian group of order $21=3\cdot 7$.

Refining further (not required, but it is what the dichotomy is hiding): for $p=q$ a group of order $p^{2}$ is always abelian — the class equation forces $Z(G)\neq\{e\}$, and then Step 3 forces $Z(G)=G$. For $p<q$ a non-abelian group of order $pq$ exists iff $p\mid q-1$, and it is then unique up to isomorphism. So the second alternative is realised precisely by those groups.

Answer