the derived subgroup is normal

We are going to prove:
”The derived subgroup (or commutator subgroup) $[G,G]$ is normal in $G$ ”

Proof:
We have to show that for each $x\\in[G,G]$ , $gxg^{-1}$ it is also in $[G,G]$ .

Since $[G,G]$ is the subgroup generated by the all commutators in $G$ , then for each $x\\in[G,G]$ we have $x=c_{1}c_{2}\\cdots c_{m}$ –a word of commutators– so $c_{i}=[a_{i},b_{i}]$ for all $i$ .

Now taking any element of $g\\in G$ we can see that

$\\displaystyle g[a_{i},b_{i}]g^{-1}$	$\\displaystyle=$	$\\displaystyle ga_{i}b_{i}a_{i}^{-1}b_{i}^{-1}g^{-1}$
	$\\displaystyle=$	$\\displaystyle ga_{i}g^{-1}gb_{i}g^{-1}ga_{i}^{-1}g^{-1}gb_{i}^{-1}g^{-1}$
	$\\displaystyle=$	$\\displaystyle(ga_{i}g^{-1})(gb_{i}g^{-1})(ga_{i}g^{-1})^{-1}(gb_{i}g^{-1})^{-1}$
	$\\displaystyle=$	$\\displaystyle[ga_{i}g^{-1},gb_{i}g^{-1}],$

that is

g[a_{i},b_{i}]g^{-1}=[ga_{i}g^{-1},gb_{i}g^{-1}]

so a conjugation of a commutator is another commutator, thenfor the conjugation

$\\displaystyle gxg^{-1}$	$\\displaystyle=$	$\\displaystyle gc_{1}c_{2}\\cdots c_{m}g^{-1}$
	$\\displaystyle=$	$\\displaystyle gc_{1}g^{-1}gc_{2}g^{-1}g\\cdots g^{-1}gc_{m}g^{-1}$
	$\\displaystyle=$	$\\displaystyle(gc_{1}g^{-1})(gc_{2}g^{-1})\\cdots(gc_{m}g^{-1})$

is another word of commutators, hence $gxg^{-1}$ is in $[G,G]$ which in turn implies that $[G,G]$ is normal in $G$ , QED.