UPSC 2023 Maths Optional Paper 2 Q2a — Step-by-Step Solution

15 marks · Section A

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

Question

Prove that a non-commutative group of order $2p$ , where $p$ is an odd prime, must have a subgroup of order $p$ .

Technique

Eliminate the other possible element orders allowed by Lagrange.

Solution

Strategy. Show $G$ has an element of order $p$ ; then the subgroup it generates has order $p$ .

Let $|G|=2p$ with $p$ an odd prime, and assume $G$ is non-abelian. By Lagrange’s theorem, the order of any element of $G$ divides $2p$ , so the possible orders are $1,\;2,\;p,\;2p$ .

Step 1 — Rule out “everyone has order 1 or 2”.

Suppose every non-identity element $g\in G$ satisfies $g^{2}=e$ , i.e. $g=g^{-1}$ . Then for any $x,y\in G$ ,

(xy)^{-1}=y^{-1}x^{-1}=yx,\qquad\text{but also}\qquad(xy)^{-1}=xy,

so $xy=yx$ and $G$ is abelian. This contradicts non-commutativity. Hence at least one element has order $p$ or $2p$ .

Step 2 — Rule out an element of order $2p$ .

If some $g\in G$ has order $2p$ , then $\langle g\rangle$ has $2p=|G|$ elements, so $G=\langle g\rangle$ is cyclic — in particular abelian. Again contradiction.

Step 3 — Conclude.

By Steps 1 and 2, some element $a\in G$ has order exactly $p$ . Then

H=\langle a\rangle=\{e,a,a^{2},\dots,a^{p-1}\}

is a subgroup of $G$ of order $p$ .

Answer

\boxed{\;G\text{ has a subgroup of order }p.\;}