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, then for 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.

Title	the derived subgroup is normal
Canonical name	TheDerivedSubgroupIsNormal
Date of creation	2013-03-22 16:04:39
Last modified on	2013-03-22 16:04:39
Owner	juanman (12619)
Last modified by	juanman (12619)
Numerical id	8
Author	juanman (12619)
Entry type	Proof
Classification	msc 20A05
Classification	msc 20E15
Classification	msc 20F14