Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I working on my thesis on semidirect products and splitting. I am trying to prove that if you assume that $G$ is a split extension of $N$ and $H$ then you can show that $G$ is a semidirect product of $N$ and $H$.

Let $0\xrightarrow{}N\xrightarrow{\alpha}G\xrightarrow{\beta}H\xrightarrow{}0$ be exact.

Define $\gamma :H\to G$ with $\beta\circ \gamma =id_{H}$

$N_{0}:=\alpha(N)$, $H_{0}:=\gamma(H)$

Show that $G=N_{0}H_{0}$ and that $N_{0}\cap H_{0}=1$

This is what I´ve done right now

I define $g\in N_{0}\cap H_{0}$ and with $H_{0}=\gamma(H)$ I have that $g=\gamma(h)$ since $g\in H_{0}$

At the same time $g\in N_{0}$ so that gives me $\beta(g)=id_{H}$.

So now I have that $\beta(\gamma(h))=id_{H}$ which is $\beta \circ \gamma$. That shows that only element in $N\cap H$ is the identity element.


Now to show that $g=nh$:

For $g\in G$ I can define $h:=\gamma(\beta(g))\in H_{0}$ and $n:=gh^{-1}$.

That gives me that $g=nh$. Now I have to show that $n\in N_{0}$.

With $n=g\gamma^{-1}(\beta(g))$ and $\beta(n)=id$ assuming $n\in N_{0}$

I can make the following equation:

$id=\beta(n)=\beta(g\gamma^{-1}(\beta(g)))$ and using that $\beta$ is homomorphic I get

$id=\beta(n)=\beta(g\gamma^{-1}(\beta(g)))$ = $\beta(g)\beta(\gamma^{-1}(\beta(g)))$

This is where I am now ... I dont know how to make the last step which gets me $\beta(g)\beta^{-1}(g)$ on the right side.

I appreciate any help and comments of my calculations. Thank you.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

I think I've some problems understanding your notation so allow me to prove the result in a different way.

As you prove you can write every $g \in G$ as a product $$g = n\gamma(\beta(g))$$ for some $h \in H$, now if you applies $\beta$ to $g$ what you get is $$\beta(g) = \beta(n \gamma(\beta(g))) = \beta(n)\beta(\gamma(\beta(g)))=\beta(n)\beta(g)$$ now a simple multiplication by $\beta(g)^{-1}$ on both sides proves that $$\text{id} = \beta(n)$$ which exactly states that $n \in N_0$.

Hope this helps.

share|improve this answer
    
well the problem is how do i get $β(γ^{-1}(β(g)))$ to become $\beta^{-1}(g)$. –  grendizer Dec 8 '12 at 11:44
    
Fine, give me one moment I change my answer, but allow me to use my notation. –  Giorgio Mossa Dec 8 '12 at 11:52
    
yes i fully follow what u are doing, but how does β(γ(β(g)))=β(g). –  grendizer Dec 8 '12 at 12:32
1  
oh, i think i know now. is it from what i stated earlier that $g=\gamma(h)$? –  grendizer Dec 8 '12 at 12:54
1  
@grendizer yes, that or more easily from the fact that $\beta \circ \gamma = \text{id}$, by splitting property. –  Giorgio Mossa Dec 8 '12 at 18:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.