UPSC 2024 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 finite group of order $mn$ , where $m$ and $n$ are prime numbers with $m>n$ . Show that $G$ has at most one subgroup of order $m$ .

Technique

Sylow III: $n_m\equiv 1\pmod m$ and $n_m\mid n$ ; the hypothesis $m>n$ forces $n_m=1$ .

Solution

Step 1 — Sylow context.

A subgroup of order $m$ in $G$ is a Sylow $m$ -subgroup (since $m\parallel |G|$ ). Let $n_m$ denote the number of Sylow $m$ -subgroups. By Sylow’s third theorem:

$n_m\equiv 1\pmod m$ , and
$n_m\mid n$ .

Step 2 — Combine the conditions.

Divisors of $n$ (prime) are $1$ and $n$ . If $n_m=n$ , then $n\equiv 1\pmod m$ , i.e., $m\mid(n-1)$ . Since $m>n>0$ , we have $n-1<m$ , so $m\nmid(n-1)$ unless $n-1=0$ , i.e., $n=1$ . But $n$ is prime, so $n\ge 2$ : contradiction.

Therefore $n_m=1$ .

Conclusion. $G$ has exactly one (hence at most one) subgroup of order $m$ .

Answer

\boxed{\;G\text{ has exactly one subgroup of order }m.\;}