Commutator group in the center of a group

[AllRelative]

Homework Statement

[G,G] is the commutator group.
Let $H\triangleleft G$ such that $H\cap [G,G]$ = {e}. Show that $H \subseteq Z(G)$.

Homework Equations

The Attempt at a Solution

In the previous problem I showed that $G$ is abelian iif $[G,G] = {e}$. I also showed that $[G,G]\triangleleft G$.

***
I am unsure what $(H\cap [G,G])$ represents.
Is $(H\cap [G,G])$ = $[H,H]$,
or is it equal to {$[x,y] \in H | x,y \in G$}??
***
I began my solution like this:
$\forall [a,b] \in (H\cap [G,G])$, we have that $a^{-1}b^{-1}ab = e$
$\Rightarrow ab = ba$

something missing

$\Rightarrow H\subseteq Z(G)$.

 

[fresh_42]

I began my solution like this:
$\forall [a,b] \in (H\cap [G,G])$, we have that $a^{-1}b^{-1}ab = e \Rightarrow ab = ba$

That's the wrong direction. You started with an element which is the identity and want to show what? That $e\in Z(G)$?

Skip the direction. What can you say about $g^{-1}hgh\,$?

 

[AllRelative]

You're right, I started the wrong way haha.

So $g^{-1}hg \in G$ because of the normality of $H$ in $G$.

$g^{-1}h^{-1}gh$ = $g'h$ with $g'=g^{-1}h^{-1}g$.

 

[fresh_42]

You're right, I started the wrong way haha.

So $g^{-1}hg \in G$ because of the normality of $H$ in $G$.

$g^{-1}h^{-1}gh$ = $g'h$ with $g'=g^{-1}h^{-1}g$.

So $g^{-1}hg \in H$ because of the normality of $H$ in $G$.

$g^{-1}h^{-1}gh$ = $h'h$ with $h'=g^{-1}h^{-1}g$.

 

[AllRelative]

So $g^{-1}hg \in H$ because of the normality of $H$ in $G$.

$g^{-1}h^{-1}gh$ = $h'h$ with $h'=g^{-1}h^{-1}g$.

Sorry I still don't see it...

We have an $h' = g^{-1}h^{-1}g$.
We can write $(g^{-1}h^{-1}g)h = [g,h] = h'h$ which belongs to $H$.

I assume we have to show somehow that $h'h = hh'$ which would make $H$ abelian and therefore a subset of $Z(G)$.

But $h'h = hh'$
$(g^{-1}h^{-1}g)h = h(g^{-1}h^{-1}g)$

I don't think we can much with that last expression...

 

[fresh_42]

Sorry I still don't see it...

We have an $h' = g^{-1}h^{-1}g$.
We can write $(g^{-1}h^{-1}g)h = [g,h] = h'h$ which belongs to $H$.

I assume we have to show somehow that $h'h = hh'$ which would make $H$ abelian and therefore a subset of $Z(G)$.

But $h'h = hh'$
$(g^{-1}h^{-1}g)h = h(g^{-1}h^{-1}g)$

I don't think we can much with that last expression...

We already have $[g,h] \in [G,H] \subseteq [G,G]$ and $[g,h]=h'h \in H$. This is $[g,h]\in H \cap [G,G]=\{\,e\,\}\,.$
So $[g,h]=g^{-1}h^{-1}gh=e$.

Everything needed is already here.