Suppose N is a normal subgroup of G and H is a subgroup of G, and G is the semi-direct product of N and H.

Then is there a nice way to prove that N(and)H = {1}

Thanks.

Printable View

- Aug 28th 2013, 09:43 PMakarimiSemi-Direct product proof
Suppose N is a normal subgroup of G and H is a subgroup of G, and G is the semi-direct product of N and H.

Then is there a nice way to prove that N(and)H = {1}

Thanks. - Aug 28th 2013, 09:48 PMDrexel28Re: Semi-Direct product proof
I'm confused by your question. I'm going to assume by your wording that you're considering groups of the form $\displaystyle G=N\rtimes_\varphi H$? Then, we identify $\displaystyle N$ with $\displaystyle N\times\{1\}$ inside $\displaystyle N\times H$ and similarly for $\displaystyle H$. In this context, your question should be easily solved.

If this is not what you mean, could you please elaborate? - Aug 28th 2013, 10:07 PMakarimiRe: Semi-Direct product proof
Hmm I'm not sure if it's the same as what you're saying but it could just be my lack of knowledge, this is my first time reading about semi-direct products but I have yet to see their connection to cross products.

The question I asked was part of a bigger Lemma: Let N <| G and H < G. Then G = N ><| H if and only if G = NH and N(and)H = {1}.

The part I'm having trouble with is using the fact that G = N ><| H to implicate that N(and)H = {1}.

I Hope I've made it a bit clearer. - Sep 4th 2013, 05:46 PMDevenoRe: Semi-Direct product proof
Hint: Show the mapping G → Nx|H given by nh → (n,h) is an isomorphism iff G = NH and N∩H = {e}.