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Thread: Semi Direct Product

  1. #1
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    Semi Direct Product

    the help on the last post does seem to have helped me a lot atleast upto the point where i can understand the basics.

    heres the next one, as usual any hints on how to approach the question would be greatly appreciated. i will attempt the questions with the hints

    (1)
    describe explicitly all homomorphisms

    $\displaystyle \varphi : C_4 \rightarrow Aut(C_5)$

    (2)
    For each such homomorphism $\displaystyle \varphi$ describe the semidirect product $\displaystyle C_5 \rtimes_{\varphi} C_4$ in terms of generators and relations.

    (3) How many distinct isomorphism types of groups of the form $\displaystyle C_5 \rtimes_{\varphi} C_4$ are there?
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  2. #2
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    Quote Originally Posted by joanne_q View Post

    (1)
    describe explicitly all homomorphisms

    $\displaystyle \varphi : C_4 \rightarrow Aut(C_5)$
    Suppose that $\displaystyle \left< a \right> = \mbox{C}_5$, this means we can think of $\displaystyle \mbox{C}_5$ as $\displaystyle \{ e,a,a^2,a^3,a^4\}$. Suppose that $\displaystyle \phi$ is an automorphism of $\displaystyle \mbox{C}_5$. The mapping $\displaystyle \phi$ is completely determined by $\displaystyle \phi(a)$ (be sure you understand why this is). Now $\displaystyle \phi (a)$ can take $\displaystyle 5$ possible values: $\displaystyle e,a,a^2,a^3,a^4$. It cannot be $\displaystyle e$ because since $\displaystyle \phi$ is an automophism it maps $\displaystyle e\mapsto e$. This means there are $\displaystyle 4$ possible values for $\displaystyle \phi(a)$: $\displaystyle a,a^2,a^3,a^4$. Let $\displaystyle \phi_1$ be the mapping determined by $\displaystyle \phi (a) = a$, then it means $\displaystyle \phi_1(e) = e \ \phi_1(a) = a \ \phi_1 (a^2) = a^2 \ \phi_1 (a^3) = a^3 \ \phi_1 (a^4) = a^4$, i.e. $\displaystyle \phi_1$ is the identity mapping. Let $\displaystyle \phi_2$ be the mapping determined by $\displaystyle \phi (a) = a^2$, then it means $\displaystyle \phi_2 (e) = e \ \phi_2 (a) = a^2 \ \phi_2 (a^2) = a^4 \ \phi_2 (a^3) = a \ \phi_2 (a^4) = a^3$. Let $\displaystyle \phi_3$ be the mapping determined by $\displaystyle \phi(a) = a^3$, then it means $\displaystyle \phi_3 (e) = e \ \phi_3 (a) = a^3 \ \phi_3 (a^2) = a\ \phi_3 (a^3) = a^4 \ \phi_4 (a^4) = a^2$. Finally let $\displaystyle \phi_4$ be $\displaystyle \phi_4 (e) = e \ \phi_4 (a) = a^4 \ \phi_4 (a^2) = a^3 \ \phi_4 (a^3) = a^2 \ \phi_4 (a^4) = a $. These $\displaystyle \phi_1 , \phi_2 , \phi_3 , \phi_4$ are all automorphism of $\displaystyle \left< a \right> $ and they form a group. Note that this group has an element of order $\displaystyle 4$, also note that $\displaystyle \mbox{ord} (\phi_1) = 1$ while $\displaystyle \mbox{ord} (\phi_2) = \mbox{ord} (\phi_4) = 4$ and $\displaystyle \mbox{ord}(\phi_3) = 2$. This means that $\displaystyle \mbox{Aut}(\left< a \right>)$ is a cyclic group generated by $\displaystyle \phi_2$ (or $\displaystyle \phi_4$). Thus, we can think of this automorphism group as $\displaystyle \{ \bold{1}, \phi_2, \phi_2^2, \phi_2^3 \}$. Now to complete this problem we can think of $\displaystyle \mbox{C}_4$ as $\displaystyle \{ e,b,b^2.b^3\}$ (where $\displaystyle b$) is its generator. If $\displaystyle \theta: \left< b \right> \mapsto \mbox{Aut}(\left< a\right>)$ is a homomorphism then it is completely determined by $\displaystyle \theta (b)$ thus $\displaystyle \theta(b) = \bold{1}, \phi_2,\phi_2^2 , \phi_2^3$, it cannot be $\displaystyle \bold{1}$ (the identity) thus there are just three possibilities $\displaystyle \theta(b) = \phi_2,\phi_2^2,\phi_2^3$. Do the similar computation done above to get your answer.
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