Why every commutator is an even permutation?

Question

I was reading the solution of proving that $S_4$ is a solvable group here Show that $S_{4}$ is a solvable group.

But I did not understand the statement " every commutator is an even permutation",could anyone clarify this for me please?

Answer 1

Hint Consider the homomorphism $\pi : S_n \to \{\pm 1\}$ that assigns a permutation to its sign, and apply it to a general commutator $g h g^{-1} h^{-1}$.

Answer 2

The sign homomorphism $\operatorname{sgn}:S_n\to C_2$ is onto a commutative group, so $\ker\operatorname{sgn}\supseteq[S_n,S_n]$.

Answer 3

In elementary terms, if $g$ can be written as a product of $a$ transpositions, and $h$ can be written as a product of $b$ transpositions, then $g^{-1} h^{-1} g h$ can be written as a product of $a + b + a + b$ = $2 (a + b)$ transpositions.