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 wish to ask following question. Kindly help me in following:

Let $p$ be an odd prime and $C_p$ denotes the cyclic group of order $p$. Let $G_1$ and $G_2$ be two groups both are isomorphic to $C_p \times C_p$. I wish to form a semi direct product (which is not a direct product) of $G_1$ and $G_2$ in which $G_1$ is normal so that I can produce a non abelian group of order $p^4$. To do this I need to define a non trivial homomorphism $\varphi:G_2 \rightarrow Aut(G_1)$. We know that $Aut(G_1) \cong GL(2,p)$ ($2 \times 2$ invertible matrices over the field $\mathbb{F}_p$). I know one non trivial homomorphism $\varphi: G_2 \rightarrow GL(2,p)$ which is defined as follows:

$(1,0) \mapsto \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right)$ and $(0,1) \mapsto \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$.

This will give me one non trivial semi direct product of $G_1$ and $G_2$.

I want to ask, Is there any other non trivial homomorphism $\varphi$ which gives me a non abelian group of order $p^4$ which is not isomorphic to what I have produced?

share|improve this question
    
The uniqueness of the nonabelian semidirect prodcut follows from the fact that a Sylow $p$-subgroup of ${\rm GL}(2,p)$ has order $p$. –  Derek Holt Sep 1 '13 at 12:37
add comment

1 Answer

This is not an answer, but too long for a comment : There are two facts that help to tell if two semi-direct products are isomorphic : Let $H$ and $K$ be groups.

  1. Let $\tau : K \to Aut(H)$ be a homomorphism, and $\sigma : K\to K$ an automorphism. Then $$ H\times_{\tau} K \cong H\times_{\tau\circ\sigma} K $$ [Just take $f(h,k) = (h,\sigma(k))$ to be the isomorphism.]

  2. Let $\tau_1$ and $\tau_2$ be monomorphisms from $K$ to $Aut(H)$ with $\tau_1(K) = \tau_2(K)$, then $$ H\times_{\tau_1} K \cong H\times_{\tau_2} K $$ [Take $\sigma = \tau_2\circ\tau_1^{-1}$ in 1.]

Now perhaps you can apply @DerekHolt's comment.

share|improve this answer
add comment

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.