I have the following problem: I have to describe up to isomorphism the semidirect product $C_6 \rtimes C_2$, where $C_6$ denotes cyclic group of order six. I think I have to use external semidirect products.

So first I search for group $AutC_6$ and I find 2 automorphisms:

  1. $\sigma_0 = id_{C_6}$,
  2. $\sigma_1 = [x \rightarrow x^5]$.

To describe all semidirect products $C_6 \rtimes_{\phi} C_2$ I find all homomorphisms $\phi: C_2 \rightarrow AutC_6$. I look only at generator in $C_2 = <a>$.

First of all, I see that I can set $\phi(a) = \sigma_0$, so my $\phi$ sends all to identity. That $\phi$ is a proper homomorphism so by that I have my first external semidirect product.

Then I set $\phi(a) = \sigma_1$. I see that such $\phi$ is a proper homomorphism since

$id = \phi(e) = \phi(a a) = \phi(a) \circ \phi(a) = [x \rightarrow x^{25}] = [x\rightarrow x]$,

as we are in $C_6$.

Am I doing everything all right? Does that mean that there are two semidirect product? What should I do with the 'up to isomorphism part'?

Thanks in advance

  • 1
    $\begingroup$ You have found two semi-direct products that are distinct because in abelian groups a non-direct semi-direct product is not isomorphic to the direct product. The only thing left to do is check that there are no more order 2 automorphisms of $C_{6}$ and I think there aren't. So these should be the only two non-isomorphic semi-direct products of $C_{6}$ and $C_{2}$. $\endgroup$ – Siddharth Venkatesh May 3 '14 at 18:29
  • $\begingroup$ Thanks for your comment. Could you explain what does it mean that ' in abelian groups a non-direct semi-direct product is not isomorphic to the direct product'? By Euler's totient function there are only two automorphisms of $C_6$, so in this case I think I have faound all of them, but in some imagined situation what would happen, if I have found more order 2 automorphisms? Would that be enough to state that two semidirect products given by these two automorphisms of order two are isomorphic? $\endgroup$ – michael May 3 '14 at 19:27
  • 1
    $\begingroup$ Yeah, if you had found other automorphisms of order 2, you would have to figure out if the resulting semi-direct products are isomorphic or not. As for the other question, in the two semi-direct products that you found, the homomorphism $\phi: C_{2} \rightarrow \text{Aut}(C_{6})$ in the first is trivial and is non-trivial in the second. If the groups, were non-abelian, this would not necessarily imply that the resulting semi-direct products were non-isomorphic but when both the groups are abelian, it does. $\endgroup$ – Siddharth Venkatesh May 3 '14 at 19:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.