# 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 TheDerivedSubgroupIsNormal 2013-03-22 16:04:39 2013-03-22 16:04:39 juanman (12619) juanman (12619) 8 juanman (12619) Proof msc 20A05 msc 20E15 msc 20F14