Commutator group normal in G

Question

$G$ is a group with $e$ as it's identity element. The commutator group $[G,G] = <[x,y]|x,y \in G>$ such that $[x,y]=x^{-1}y^{-1}xy \: \: \: \forall x,y \in G$

Show that $[G,G]\triangleleft G$

Solution: If $[G,G]\triangleleft G$, then $g^{-1}[x,y]g$ must belong to [G,G].

$$g^{-1}[x,y]g$$ $$g^{-1}x^{-1}y^{-1}xyg$$

I am stuck here. I am not sure how to proceed since we have three elements...

Thanks for the help!

score 1 · Answer 1 · answered Dec 08 '18 at 03:44

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HINT :

$$g^{-1}x^{-1}y^{-1}xyg=g^{-1}x^{-1}gg^{-1}y^{-1}gg^{-1}xgg^{-1}yg$$

Let $g^{-1}xg=u $ and $g^{-1}yg=v $. Then $$g^{-1}x^{-1}y^{-1}xyg=u^{-1}v^{-1}uv .$$

answered Dec 08 '18 at 03:44

cqfd

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Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$... – Arturo Magidin Dec 08 '18 at 03:55

Commutator group normal in G

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