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Math Help - 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

    \varphi : C_4 \rightarrow Aut(C_5)

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

    (3) How many distinct isomorphism types of groups of the form 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

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