OK so I basically can't find any semidirect product calculations online and my notes are a bit iffy, so I used the definitions and an example from notes to try and understand it. I get the jist of it but I kind of confused by a few parts.
So basically I have to find all the isomorphivally distinc groups of semi direct product with , under the homomorphism h.
So basically I will just explain how I've done everything. We know that and that
Using the definition of the semi direct product you have the homomorphism .
So you know that where . Now using the generator you get that is isomorphic to where
Now here is where the confusion pops up for me. I know that to find all the groups you have to work out the product , using the definition of the semidirect product. But the problem is what values of do I use? How do I work out what the possible values of for each ?

Ultimate1
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 31032013 17:27

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 31032013 20:12
Let me clarify. You are given the two groups and .
To define a semidirect product , you need an action of on by automorphisms. Such an action is equivalent to a homomorphism . 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 , the semi direct product (notice I have said 'the'  fixing an action determines the semidirect product) is given by elements (with and ) with multiplication . Notice that is an automorphism of and we are applying it to when it is permuted past .
Therefore, your problem is to find all automorphisms or equivalently, all automorphisms of such that (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 you need to work out which of the semidirect products are isomorphic to each other.Last edited by Mark85; 31032013 at 20:27. 
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 01042013 21:59
(Original post by Mark85)
Let me clarify. You are given the two groups and .
To define a semidirect product , you need an action of on by automorphisms. Such an action is equivalent to a homomorphism . 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 , the semi direct product (notice I have said 'the'  fixing an action determines the semidirect product) is given by elements (with and ) with multiplication . Notice that is an automorphism of and we are applying it to when it is permuted past .
Therefore, your problem is to find all automorphisms or equivalently, all automorphisms of such that (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 you need to work out which of the semidirect products are isomorphic to each other. 
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 02042013 12:40
(Original post by Ultimate1)
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
You mean, you have found all homomorphisms ?
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.Last edited by Mark85; 02042013 at 18:20. 
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 02042013 20:30
(Original post by Mark85)
Be more specific.
You mean, you have found all homomorphisms ?
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. 
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 03042013 01:07
(Original post by Ultimate1)
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.
This is the expanded version of the comment from my orginal post: I said that finding all homomorphisms was equivalent to finding automorphisms of such that .
The point is that is cyclic, i.e. generated by one element, say . Therefore, any homomorphism out of is entirely determined by the image . So to specify a homomorphism , you simply need to state an element where you map to. However, this is only well defined as a homomorphism if since we have and therefore we need to be equal to .
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 is which is isomorphic to . In particular, each automorphism satisfies . So if you pick a generator for , any choice of element defines a homomorphism by . In other words, there are four such homomorphisms, each one sending your chosen element of to a different automorphism of .
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.Last edited by Mark85; 03042013 at 01:12. 
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 03042013 21:49
(Original post by Mark85)
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 was equivalent to finding automorphisms of such that .
The point is that is cyclic, i.e. generated by one element, say . Therefore, any homomorphism out of is entirely determined by the image . So to specify a homomorphism , you simply need to state an element where you map to. However, this is only well defined as a homomorphism if since we have and therefore we need to be equal to .
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 is which is isomorphic to . In particular, each automorphism satisfies . So if you pick a generator for , any choice of element defines a homomorphism by . In other words, there are four such homomorphisms, each one sending your chosen element of to a different automorphism of .
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.
So basically you have find elements of such that for all in . But since is isomorphic to hence all elements of are homomorphisms , right?
OK I got this now. Do you have any resources for more questions on semi direct product construction? 
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 04042013 00:58
(Original post by Ultimate1)
OK thanks I get it now.
So basically you have find elements of such that for all in . But since is isomorphic to hence all elements of are homomorphisms , right?
for each where generates .
The reason being that each such map is a well defined homomorphism because there is only one relation to check and it holds for all .
(Original post by Ultimate1)
OK I got this now. Do you have any resources for more questions on semi direct product construction?
The construction is just a one liner:
If is an action of a group H on a group G by automorphisms, then the semidirect product is the set of pairs with multiplication given by .
Alternatively, if G and H are given by presentations and then is given by where the unions are disjoint.
At least the second description here might look complicated but it is very simple to remember once you see that it is just basically a direct product but instead of the two factors commuting, when elements of H are permuted past elements of G  they act on them via the given action which is encoded by .
If you mean about the idea of semidirect products then it goes roughly and briefly as follows:
The idea is about breaking down groups into smaller pieces. If you find a normal subgroup N inside a group G, it is natural to ask whether G is a direct product of N. The first thing to check is whether N has a complement in G (i.e. a subgroup H such that HN = G and H and N trivially intersect). If such a subgroup H exists and is normal in G then G is indeed the direct product of N and H. However, if H exists but isn't normal, then G is still the semidirect product of H and N where H acts on N by conjugation (notice that if H acted on N trivially by conjugation then this would imply H normal)
More generally, one can talk about group extensions which are all the ways you can make a group G from two other groups N,Q such that N is (isomorphic to) a normal subgroup of G and Q is the quotient of the image of N in G. Group extensions are a fairly general thing and so one normally talks about different special cases. Semidirect products (often called split extensions in this context) are basically the most simple type of extension and are the easiest to classify. To begin classifying other basic classes of extensions is best done in the language of group cohomology. You can learn the basics in any book on homological algebra or any decent graduate group theory text. I say graduate text but as far as I can garner  the 'graduate' in 'graduate' maths texts is misleading; it generally refers to American graduate syllabus which overlaps with what would probably be covered at undergrad here. I may be wrong though  I never did a maths degree, that is just the impression I get from reading books and lecture notes online etc.
Also, you can just type all of these terms into google and find a wealth of information. For instance, just typing 'semidirect product' into google brings up wikipedia, some math sites and a whole host of lecture notes on the first page. You really don't have to look too far to find these things.
When you say in your OP that you can't find any examples of calculations online, you are sort of missing the point: For the purposes of what you are doing, you really just need the definition of a semidirect product  the rest of the details are just very basic standard elementary group theory (and in the case of cyclic groups  elementary number theory). You probably won't find exact examples of the things you want since in a book or in lecture notes  there would be no need to put so much detail since once they got to the stage of defining a semi direct product, the book would assume you were comfortable with the basic notion of a homomorphism etc.
Your basic problem, like most on here, is that you need to be able to formulate more precisely what exactly you don't know and then be able to do a simple web search or look through a book to find the things you don't remember and then insert them into the new definition. You aren't expected to remember everything but part of the point about being taight a new definition and being given exercises on it is that you go back a revise the concepts and definitions that you have already covered and solidify those. IMHO, that is the only way you really understand anything in maths  when you have revisted it for the umpteenth time in new contexts and examples.Last edited by Mark85; 04042013 at 01:14.
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