Help with a proof: Normal subgroup

Hi.

I need to prove:

So, one direction is quite easy:

:

If is a subgroup of , it follows that : .

in particular, for every , , so , .

but the second direction is not quite clear for me:

:

Since , : , there exist some , such that

and then:

but how do I know that ?

Thanks in advance.

Re: Help with a proof: Normal subgroup

Quote:

Originally Posted by

**Stormey**
So, one direction is quite easy:

:

If

is a subgroup of

, it follows that

:

.

You mean, if H is a *normal* subgroup.

Quote:

Originally Posted by

**Stormey** in particular, for every

,

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.

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** 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 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...

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.

Re: Help with a proof: Normal subgroup

Thanks.

edited my last post.