Show that every subgroup of $S_n$ has either every member as even or exactly half the members as even permutations.

Question

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.

Answer 1

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 $2$.

Answer 2

Here's another way, it also depends fundamentally on the fact that the sign of a permutation is well-defined and multiplicative, i.e. sign$(\sigma_1\sigma_2)=\text{sign}(\sigma_1)\text{sign}(\sigma_2)$:

Let $H$ be a subgroup of $S_n$.

If there are no odd permutations in $H$ then we're done.

If there is an odd permutation, say $\sigma\in H$, then consider the fact that $\sigma H=H$. Since "odd $\times$ odd $=$ even" and "odd $\times$ even $=$ odd" we have that there must be equal numbers of odd and even permutations in $H$.