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
Let us use the mapping $$\sgn : G \to \mathbb{Z}_2$$ which, from Parity of a Permutation, is a homomorphism.

Then $$G / \mathrm{ker} \left({\sgn}\right) \cong \mathrm{Im} \left({\sgn}\right)$$ from the First Isomorphism Theorem.

The only possibilities for $$\mathrm{Im} \left({\sgn}\right)$$ are $$\left\{{0}\right\}$$ or $$\mathbb{Z}_2$$.

So either:
 * $$\left|{G / \mathrm{ker} \left({\sgn}\right)}\right| = 1$$, in which case all the permutations of $$G$$ are even, or
 * $$\left|{G / \mathrm{ker} \left({\sgn}\right)}\right| = 2$$, in which case exactly half of them are.