Show that $D$ is a normal subgroup.

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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