Group of Permutations either All or Half Even

Theorem
Let $G$ be a group of permutations.

Then either exactly half of the permutations in $G$ are even, or they are all even.

Proof
From Parity Function is Homomorphism, the mapping:
 * $\operatorname{sgn} : G \to \Z_2$

is a homomorphism.

Then:
 * $G / \ker \left({\operatorname{sgn}}\right) \cong \operatorname{Im} \left({\operatorname{sgn}}\right)$

from the First Isomorphism Theorem.

The only possibilities for $\operatorname{Im} \left({\operatorname{sgn}}\right)$ are $\left\{{0}\right\}$ or $\Z_2$.

So either:
 * $\left|{G / \ker \left({\operatorname{sgn}}\right)}\right| = 1$

in which case all the permutations of $G$ are even, or:
 * $\left|{G / \ker \left({\operatorname{sgn}}\right)}\right| = 2$

in which case exactly half of them are.