Group of Permutations either All or Half Even
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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.
$\blacksquare$