# Help with a proof: Normal subgroup

• Nov 27th 2013, 07:21 AM
Stormey
Help with a proof: Normal subgroup
Hi.
I need to prove:

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

So, one direction is quite easy:

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

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

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

• Nov 27th 2013, 09:12 AM
emakarov
Re: Help with a proof: Normal subgroup
Quote:

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

So, one direction is quite easy:

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

You mean, if H is a normal subgroup.

Quote:

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

$\displaystyle gH=Hg$ 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, 09:52 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
$\displaystyle gH=Hg$ 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 $\displaystyle gH=Hg$ 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, 10:04 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, 10:13 AM
Stormey
Re: Help with a proof: Normal subgroup
Thanks.
edited my last post.