Show that if $H$ is a subgroup of $S_n$ the symmetric group of order $n$, then either every member of $H$
is an even permutation or exactly half of the members are even.
I can see that if $a,b$ are arbitrary even elements in $H$, then their product and any combination of their products will be even, which implies $H$ is full of even permutations, but I'm not sure how to show the half-even half-odd part.
Best Answer
Let $H$ be a subgroup of $S_n$. If $H$ contains no odd permutations, then $H$ contains only even permutations, and we're done.
Otherwise, let $o\in H$ be an odd permutation and consider the function $f:H\rightarrow H$ that multiplies each element by $o$. Note that this function is bijective: it's injective, with an inverse function that multiplies each element by $o^{-1}$, and it's surjective, because for every element $h\in H$, we can find an element $h\cdot o^{-1}$ that $f$ maps into it.
Note that multiplying by an odd permutation changes odd permutations into even permutations and vice versa. It follows that $f$ is a bijection that perfectly pairs up the odd permutations in $H$ with the even permutations. Hence there are exactly as many odd permutations as even permutations in $H$.