# Help with a proof: Normal subgroup

• Nov 27th 2013, 08:21 AM
Stormey
Help with a proof: Normal subgroup
Hi.
I need to prove:

$H\triangleleft G \Leftrightarrow \forall g\in G,\hspace{5} \forall h\in H:\hspace{5} g^{-1}hg\in H$

So, one direction is quite easy:

$\Rightarrow$ :
If $H$ is a subgroup of $G$, it follows that $\forall g\in G$: $gH=Hg$.
in particular, for every $h\in H$, $gh=hg$ , so $h=g^{-1}hg$, $\forall g\in G, \forall h\in H$.

but the second direction is not quite clear for me:
$\Leftarrow$ :
Since $\forall g\in G$, $\forall h\in H$: $g^{-1}hg\in H$, there exist some $h_0\in H$, such that $g^{-1}hg=h_0$
and then:
$gh_0=hg$

but how do I know that $h_0=h$?

• Nov 27th 2013, 10:12 AM
emakarov
Re: Help with a proof: Normal subgroup
Quote:

Originally Posted by Stormey
$H\triangleleft G \Leftrightarrow \forall g\in G,\hspace{5} \forall h\in H:\hspace{5} g^{-1}hg\in H$

So, one direction is quite easy:

$\Rightarrow$ :
If $H$ is a subgroup of $G$, it follows that $\forall g\in G$: $gH=Hg$.

You mean, if H is a normal subgroup.

Quote:

Originally Posted by Stormey
in particular, for every $h\in H$, $gh=hg$

$gH=Hg$ does not mean this. It means that sets gH and Hg are equal. Now, gH = {gh | h ∈ H} and Hg = {h'g | h' ∈ H}. For an arbitrary h ∈ H, the fact that gh ∈ {h'g | h' ∈ H} means that there exists some h' ∈ H such that gh = h'g. Similarly, for every h' ∈ H there exists an h such that gh = h'g.
• Nov 27th 2013, 10:52 AM
Stormey
Re: Help with a proof: Normal subgroup
Hi, emakarov.
thanks for the help.

Quote:

Originally Posted by emakarov
You mean, if H is a normal subgroup.

Yes, sorry. I meant normal.

Quote:

Originally Posted by emakarov
$gH=Hg$ does not mean this. It means that sets gH and Hg are equal. Now, gH = {gh | h ∈ H} and Hg = {h'g | h' ∈ H}. For an arbitrary h ∈ H, the fact that gh ∈ {h'g | h' ∈ H} means that there exists some h' ∈ H such that gh = h'g. Similarly, for every h' ∈ H there exists an h such that gh = h'g.

OK, I get now why it's wrong to say that $gH=Hg$ means gh=hg, so then how do I go from "there exists an h such that gh = h'g" to gh=hg for every g in G, and for every h in H?

**edit

scrap that last post.
I got it.
for some reason I thought that every normal subgroup is abelian...
• Nov 27th 2013, 11:04 AM
emakarov
Re: Help with a proof: Normal subgroup
Quote:

Originally Posted by Stormey
how do I go from "there exists an h such that gh = h'g" to gh=hg for every g in G, and for every h in H?

You don't because gH = Hg does not imply gh = hg for every h ∈ H and g ∈ G. But gh = hg is not required for the proof you need.
• Nov 27th 2013, 11:13 AM
Stormey
Re: Help with a proof: Normal subgroup
Thanks.
edited my last post.