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I need to decide how many semi-direct products $H\rtimes Q$ can be constructed for $H=C_{42}, Q=C_{3}$ where $C_{n}$ denotes the cyclic group of order n.

I know I need to be finding homomorphisms $\theta:C_{3}\rightarrow {\rm Aut}(C_{42})$, where ${\rm Aut}(G)$ denotes the automorphism group of G. Since ${\rm Aut}(C_{42})=C_{12}$ I've reduced this to finding homomorphisms $\theta:C_{3}\rightarrow C_{12})$.

Now, here is where I'm struggling to understand the material. From what I can gather, I need to map the generator of $C_3$ (which has order 3) to a generator of $C_{42}$ via an automorphism of order 3 also. So calling $C_3=\langle a|a^3=1\rangle, C_{42}=\langle b|b^{42}=0\rangle$ and ${\rm Aut}(C_{42})=\langle\mu |\mu^{12}=id\rangle$ (where $\mu$ is an automorphism which generates ${\rm Aut}(C_{42})$).

So if I'm getting this right, I need to find a $\mu:b\rightarrow b^k$ such that $\mu^3:b\rightarrow b^{k^3}=b$, so that the automorphism $\mu$ has order 3. So surely this means that $k^3 \equiv 1 \bmod 42$ then?

This is where I get stuck. Could somebody please tell me if I'm doing this right so far, and if not how I should be approaching this instead?

Many thanks.

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The semidirect product sign in LaTeX is \rtimes: $N \rtimes H$. Also, in LaTeX it's better to write a presentation of a group using $\langle$ (typed as \langle) instead of $<$ and $\rangle$ (typed as \rangle) instead of $>$. – KCd Nov 6 '12 at 15:16
Thanks or the tip. – Traxter Nov 6 '12 at 15:20
up vote 2 down vote accepted

You are definitely looking in the right direction, but some of the things you are saying are a bit confusing. Basically, you have outlined two ways to solve the problem simultaneously, and both of them are correct! I will try to distinguish between the two approaches, and for every approach we'll see how to go further.

First approach. The first approach is exactly what you said in the second paragraph. To find all the homomorphisms $\theta \colon C_3 \to {\rm Aut}(C_{42})$, you can use the fact that ${\rm Aut}(C_{42})$ is isomorphic to $C_{12}$. To solve the problem, you need to find all homomorphisms $\varphi \colon C_3 \to C_{12}$ and any single isomorphism $\psi \colon C_{12} \to {\rm Aut}(C_{42})$. Then you can get all the possible homomorphisms $\theta \colon C_3 \to {\rm Aut}(C_{42})$ as compositions $\theta=\psi \circ \varphi$.

Second approach. But that is not what you started to do in paragraph 3 and further. What you did there is point out a slightly different approach, which is shorter and easier than the long way from the first approach. The only confusing thing is this phrase that you "need to map the generator of $C_3$ (which has order 3) to a generator of $C_{42}$ via an automorphism of order $3$ also". What you probably mean is that you need to map the generator $a$ of $C_3$ to an automorphism $\mu \in {\rm Aut}(C_{42})$ such that $\mu^3 = {\rm Id}$.

Note that $\mu$ doesn't need to have order $3$. It can also have order $1$. There's no harm in that, because the homomorphic image of $C_3$ in ${\rm Aut}(C_{42})$ can be either isomorphic to $C_3$ or trivial.

So what you need to do is find all the possible $\mu \in {\rm Aut}(C_{42})$ such that $\mu^3 = 1$. Then for every such $\mu$ you will have a homomorphism $\theta \colon C_3 \to {\rm Aut}(C_{42})$ such that $\theta(a) = \mu$, and thus you will find all such homomorphisms.

Again, as you correctly pointed out, if $\mu \in {\rm Aut}(C_{42})$, then $\mu$ acts by taking some $k$-th power of each element: $\mu(x) = x^k$ for every $x \in C_{42}$. Saying that $\mu^3=1$ is the same as saying that $k^3 \equiv 1 \mod 42$. So yes, all that is left to do is find all such $k$ (up to equivalence modulo 42). For every such $k$ there is a $\mu \in {\rm Aut}(C_{42})$ such that $\mu(b)=b^k$ and a homomorphism $\theta \colon C_3 \to {\rm Aut}(C_{42})$ such that $\theta(a) = \mu$.

Wow, this is already quite a text. To sum up, you're doing fine. I would suggest finishing with the second approach first, and then if you're up for it, doing the first approach too and making sure that they ultimately give you the same result.

UPDATE: actually, this is still not the whole problem. After you've found all the homomorphisms $\theta \colon C_3 \to {\rm Aut}(C_{42})$, each such homomorphism will give you a semidirect product $G= C_{42} \rtimes_\theta C_3$. But be careful! If you have two different homomorphisms $\theta_1, \theta_2 \colon C_3 \to {\rm Aut}(C_{42})$, then the two semidirect products $G_1 = C_{42} \rtimes_{\theta_1} C_3$ and $G_2 = C_{42} \rtimes_{\theta_2} C_3$ can still be isomorphic! If such a thing happens, then the number of non-isomorphic semidirect products will be less then the number of homomorphisms $\theta$ that you found earlier. So, good luck figuring out how many different products there really are (hint: not so many).

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I really appreciate your response, it's starting to make sense in my head a bit now. I think it's just one of those things where I got confused with it and the wrong thing stuck. I'm going to read over what you wrote a few times and attempt the different approaches. Again, a huge thanks for putting so much effort into your reply. – Traxter Nov 6 '12 at 21:22
Dan, I've just been through doing the second approach (it's the one closest to the method my lecturer is using) and it's pretty much cleared up all the doubts I had surrounding this. – Traxter Nov 6 '12 at 21:37

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