Show that every subgroup of $S_n$ has either every member as even or exactly half the members as even permutations.
ATTEMPT:
Consider the homomorphism $\phi:S_n\to \Bbb Z_2$
as \begin{cases}
1 & \text{if h is even}\\-1&\text{if h is odd}\end{cases}.
Now if $H$ is a subgroup of $G$ so is $\phi(H)$
The only subgroups of $\Bbb Z_2$ are $\{1\},\{1,-1\}$.
If $\phi(H)=1$ then $H$ has all permutations to be even.
If $\phi(H)=\{1,-1\}$ then Number of even permutations $=$Number of elements of $S_n$ which are mapped to $1=\phi(H)=$Half of order of $\Bbb Z_2$ .
Though there exist ways to prove it,I want to solve the problem using this procedure.
Please tell if this way is correct or not.
Best Answer
Instead of noting that $\phi(H)$ is a subgroup (of $\Bbb Z_2$), rather let $\psi\colon H\to \Bbb Z_2$ be the restriction of $\phi$ to $H$ and use that $\psi^{-1}(0)$ is a subgroup of $H$ of index either $1$ or $2$.