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Let G be a group. Let $D$ be the subgroup of $G$ generated by the elements of the form $ghg^{-1}h^{-1}$, where $g,h\in G$. Show that $D$ is a normal subgroup.

I am having trouble showing that $D$ is normal. Take $x\in G$. We want to show that $xghg^{-1}h^{-1}$ can be written as something like $aba^{-1}b^{-1}x$. Or equivalently, we can show that $xghg^{-1}h^{-1}x^{-1} \in D$. But I don't see how to proceed? I feel like we should rewrite the identity element but that didn't really work out.

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  • $\begingroup$ @anon I made a mistake. Let me edit the question... $\endgroup$ – 3x89g2 Jun 1 '15 at 7:41
  • $\begingroup$ By [result you've presumably proven], the intersection of subgroups is a subgroup. $\hspace{1.12 in}$ Therefore $D$ is a subgroup. $\;$ $\endgroup$ – user57159 Jun 1 '15 at 8:03
  • $\begingroup$ Conjugation is a group homomorphism. Try the 2nd approach. $\endgroup$ – j.p. Jun 1 '15 at 9:29
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    $\begingroup$ This is an excellent example of a situation where it is easier to prove a stronger result: prove instead that the subgroup in question is a characteristic subgroup of $G$. $\endgroup$ – Derek Holt Jun 1 '15 at 10:20
  • $\begingroup$ Indeed, it is not hard to show that for any $\phi \in \text{Aut}(G)$, that; $\phi([x,y]) = [\phi(x),\phi(y)]$. $\endgroup$ – David Wheeler Jun 2 '15 at 23:07
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Mathematicians have a cool notation to make this easier to parse:

They define:

$a^g := g^{-1}ag$ (this is just a short way to write conjugation).

Note that:

$(ab)^g = a^gb^g$ (this is key in what we are about to prove, so work out the details yourself).

Finally, they also define:

$[x,y] := xyx^{-1}y^{-1}$ (this saves a lot of time writing the funny inverse superscripts). From the product rule of conjugation above, we have:

$[x,y]^g = [x^g,y^g]$, in other words "the conjugate of a commutator is the commutator of the conjugates".

Now, an element of $D$ is a product:

$u = [x_1,y_1][x_2,y_2]\cdots[x_k,y_k]$

And we want to show $u^g \in D$ whenever $u$ is. So:

$u^g = ([x_1,y_1][x_2,y_2]\cdots[x_k,y_k])^g$

$= [x_1,y_1]^g[x_2,y_2]^g\cdots[x_k,y_k]^g$

$= [x_1^g,y_1^g][x_2^g,y_2^g]\cdots[x_k^g,y_k^g] \in D$, so $D$ is normal.

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Basically you have to show that the commutator subgroup, $D$ of $G$ is normal. So let $x\in G$, then you have to show for $u=ghg^{-1}h^{-1} \in D$, $xux^{-1}\in D$.

So $x^{-1}ux = u(u^{-1}x^{-1}ux) \in D$

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$$xghg^{-1}h^{-1}x^{-1}= xg(x^{-1}x)h(x^{-1}x)g^{-1}(x^{-1}x)h^{-1}x^{-1}=(xgx^{-1})(xhx^{-1})(xgx^{-1})^{-1}(xhx^{-1})^{-1}$$

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  • $\begingroup$ I think I got that right, its 3AM and my eyes are feeling fuzzy looking at the inverses. That is the idea anyway to show $xDx^{-1} \subseteq D$ so $D$ is normal. I think you can show it is a subgroup. I have faith in you. PS if someone with fresher eyes/more coffee sees something off, feel free to correct it. $\endgroup$ – CPM Jun 1 '15 at 8:07

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