Semi-direct product help....

Let me clarify. You are given the two groups $C_4$ and $C_{10}$.

To define a semi-direct product $C_{10} \rtimes C_4$, you need an action of $C_4$ on $C_{10}$ by automorphisms. Such an action is equivalent to a homomorphism $C_4 \rightarrow \mathrm{Aut}(C_{10})$. Make sure you understand that before proceeding. So, until an action is specified, you haven't specified a group - different actions give rise to potentially nonisomorphic semidirect products.

Then, given such a homomorphism $\theta$, the semi direct product $C_{10} \rtimes_\theta C_4$ (notice I have said 'the' - fixing an action determines the semidirect product) is given by elements $(g,h)$ (with $g \in C_{10}$ and $h \in C_4$) with multiplication $(g,h)(g',h') = (g\theta(h)(g'),hh')$. Notice that $\theta(h)$ is an automorphism of $C_{10}$ and we are applying it to $g'$ when it is permuted past $h$.

Therefore, your problem is to find all automorphisms $\theta\colon C_4 \rightarrow \mathrm{Aut}(C_{10})$ or equivalently, all automorphisms $\phi$ of $C_{10}$ such that $\phi^4 = 1$ (again, if you don't understand this stop and prove it or ask).

Once you have done that, (so that you have a list of homomorphisms $\theta_1, \ldots, \theta_n\colon C_4 \rightarrow \mathrm{Aut}(C_{10})$ you need to work out which of the semidirect products $C_{10} \rtimes_{\theta_i} C_4$ are isomorphic to each other.

Thanks but I'm still confused by the last part which I just cannot get my head around. I've tried working it out but I just can't for some reason

Be more specific.

You mean, you have found all homomorphisms $\theta\colon C_4 \rightarrow \mathrm{Aut}(C_{10})$?

Have you written presentations for the semidirect products? Is the problem that you don't know any ways of showing that groups are isomorphic or not?

I mean, you have 4 homomorphisms, 4 semi direct products and to answer the last question you can really just play around by hand to see which are isomorphic or not. This is much more basic group theory/modular arithmetic.

Yes the first part is the problem, finding the homomorphisms. Finding the distinct groups shouldn't be a problem, but I'm struggling on how to find the homomorphisms. I know that Aut(C10) is isomorphic to C4, but I'm not sure on how to proceed from there.

Ok, I can answer when you mention specifics.

This is the expanded version of the comment from my orginal post: I said that finding all homomorphisms $\theta\colon C_4 \rightarrow \mathrm{Aut}(C_{10})$ was equivalent to finding automorphisms $\phi$ of $C_{10}$ such that $\phi^4=1$.

The point is that $C_4$ is cyclic, i.e. generated by one element, say $x$. Therefore, any homomorphism $f\colon C_4 \rightarrow G$ out of $C_4$ is entirely determined by the image $f(g)$. So to specify a homomorphism $f\colon C_4 \rightarrow G$, you simply need to state an element $g \in G$ where you map $x$ to. However, this is only well defined as a homomorphism if $g^4=1$ since we have $x^4=1$ and therefore we need $f(x^4)=f(x)^4=g^4$ to be equal to $1$.

More generally, if a group is given by a presentation, then a homomorphism out of it is specified by where you send the generators to. If you have another group and choose some elements to send the generators - this is well defined as a homomorphism precisely if the relations are mapped to zero. To be fair, the universal property of free groups and this simple corollary for group presentations is seldom highlighted (or often even mentioned) in beginners group theory texts. I suppose you can think of it a bit like linear algebra - you define a linear map by specifying it on a basis. The danger with groups is that all vector spaces are free, if you specify images for basis vectors then you have a well defined linear map but with groups - you have to check that the relations are satisfied to get a well defined map.

Now, in this case, the group $G$ is $\mathrm{Aut}(C_{10})$ which is isomorphic to $C_4$. In particular, each automorphism $\phi \in \mathrm{Aut}(C_{10})$ satisfies $\phi^4=1$. So if you pick a generator $x$ for $C_4$, any choice of element $\phi \in \mathrm{Aut}(C_{10})$ defines a homomorphism by $x \mapsto \phi$. In other words, there are four such homomorphisms, each one sending your chosen element $x$ of $C_4$ to a different automorphism of $C_{10}$.

If you have more questions, please try to ask more precisely since I will explain in detail to help but you saying I don't understand and me typing everything will a) take me ages and b) probably won't help you.

OK thanks I get it now.

So basically you have find elements of $Aut(C_{10})$ such that $\phi^4 =1$ for all $\phi$ in $Aut(C_{10}$. But since $Aut(C_{10})$ is isomorphic to $C_4$ hence all elements of $C_4$ are homomorphisms $h: C_4 \mapsto Aut(C_{10})$, right?

OK I got this now. Do you have any resources for more questions on semi direct product construction?