Semi Direct Product

**joanne_q** · February 2nd 2008, 05:56 PM

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?

**ThePerfectHacker** · February 2nd 2008, 07:19 PM

Originally Posted by joanne_q

(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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