Question

Prove that for every subgroup H of S_n, either all of the permutations in H are even, or exactly half are even.

Answer 1

Let $a$ be the number of odd permutations of H and $b$ denotes the number of even permutations. Then let $\{g_{1},g_{2},\cdots, g_{a}\}$ denotes the odd permutations. Then we consider $.S=\{g_{1}g_{1},g_{2}g_{1},g_{3}g_{1}\cdots g_{1}g_{a}\}$ . By closure property $S\subseteq H.$ Also $g_{1}g_{i}=g_{1}g_{j}\Rightarrow g_{i}=g_{j}.$ Hence, $|S|=a.$ $x,y$ odd permutations implies $xy$ even. Hence $S$ consists of even permutations. Hence $a\leq b [\because S\subseteq H]$ . Now even permutations of a subgroup always forms a group since, inverse of a even permutation is even and $x,y \ even \Rightarrow xy \ even.$ Also $identity$ is even permutation. Hence the even permutations are a subgroup of H. Hence by Lagrange, $b|a+b$ . This shows, $b|a.$ So either $a=0,$ or $a=b [\because b\geq a].$ Hence $b=|H| \ [case: a=0]\ or \ b=\frac{|H|}{2} \ [a=b].$