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:
 * $\sgn : G \to \Z_2$

is a homomorphism.

Then:
 * $G / \map \ker \sgn \cong \Img \sgn$

from the First Isomorphism Theorem.

The only possibilities for $\Img \sgn$ are $\set 0$ or $\Z_2$.

So either:
 * $\order {G / \map \ker \sgn} = 1$

in which case all the permutations of $G$ are even, or:
 * $\order {G / \map \ker \sgn} = 2$

in which case exactly half of them are.