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    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 C_{10} semi direct product with C_4, under the homomorphism h.

    So basically I will just explain how I've done everything. We know that C_{10}=(1, x, x^2,......,x^9|x^{10}=1) and that C_4=(1, y, y^2, y^3|y^4=1)

    Using the definition of the semi direct product you have the homomorphism h:C_4 \mapsto Aut(C_{10}).

    So you know that Aut (C_{10}) = (a_1, a_3, a_7, a_9) where a_i = x^i. Now using the generator a^3 you get that Aut(C_{10}) is isomorphic to C_4 = (1, b, b^2, b^3|b^4=1) where b= a^3 = x^3

    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 YX = (1,y)(x,1) = (h(y)x, y), using the definition of the semi-direct product. But the problem is what values of h(y) do I use? How do I work out what the possible values of h(y) for each YX ?
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    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.
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    (Original post by Mark85)
    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 :confused:
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    (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 :confused:
    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.
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    (Original post by Mark85)
    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.
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    (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.
    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.
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    (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 \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?
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    (Original post by Ultimate1)
    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?
    Well all elements of \mathrm{Aut}(C_{10}) (which is isomorphic to C_4) correspond bijectively with homomorphisms h\colon C_4 \mapsto Aut(C_{10}) and that correspondence is given by the homomorphism

    x \mapsto \phi

    for each \phi \in Aut(C_{10}) where x generates C_4.

    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 \phi \in \mathrm{Aut}(C_{10}).

    (Original post by Ultimate1)
    OK I got this now. Do you have any resources for more questions on semi direct product construction?
    What do you mean by questions on the semi-direct product construction?

    The construction is just a one liner:

    If \varphi\colon H \rightarrow \mathrm{Aut}(G) is an action of a group H on a group G by automorphisms, then the semi-direct product G \rtimes_\varphi H is the set of pairs \{(g,h)\mid g \in G, h \in H\} with multiplication given by (g,h)\cdot(g',h'):=(g\varphi(h)(  g), hh').

    Alternatively, if G and H are given by presentations G=\langle X\mid R \rangle and H=\langle Y \mid S \rangle then G \rtimes_\varphi H is given by G \rtimes_\varphi H = \langle X \cup Y \mid R \cup S, y_ix_jy_i^{-1}(\varphi(y_i)(x_j))^{-1} (y_i \in Y,x_j \in X)\rangle 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 \varphi.

    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.
 
 
 
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