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Math Help - Classification of Non-Abelian Groups of Order 42

  1. #1
    Super Member redsoxfan325's Avatar
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    Classification of Non-Abelian Groups of Order 42

    I'm trying to do this methodically. Let G be a group of order 42. The Sylow Theorem says that G has a normal subgroup of order 7, and the number of Sylow-3 subgroups is either 1 or 7.

    If there is only one Sylow-3 subgroup, then Z_3\trianglelefteq G, so Z_3Z_7\cong Z_3\times Z_7 and is normal because its index is 2. We now look for homomorphisms \phi_i:Z_2\to\mbox{Aut}(Z_3\times Z_7). We note that \mbox{Aut}(Z_3\times Z_7)\cong Z_2\times Z_6 and write Z_2 as \langle~x~\rangle, and Z_2\times Z_6 as \langle~y~\rangle\times\langle~z~\rangle. (Both have identity 1.)
    1. \phi_0:x\mapsto(1,1) This is the trivial mapping and gives the only abelian group G=Z_{42}.
    2. \phi_1:x\mapsto(1,z^3)
    3. \phi_2:x\mapsto(y,1)
    4. \phi_3:x\mapsto(y,z^3)
    The last three mappings result in three non-abelian groups of order 42, namely Z_{21}\rtimes_{\phi_i}Z_2 for i\in\{1,2,3\}. My guess is that \phi_1 produces Z_3\times D_{14}, \phi_2 produces Z_7\times D_6, and \phi_3 produces D_{42}, but I don't know how to prove it. So my question is: What is the best way to determine whether these (#2, #3, and #4) are all unique up to isomorphism and, furthermore, whether they are isomorphic to more familiar groups?

    Next, if there are three Sylow-3 subgroups, then we look for homomorphisms \varphi:Z_3\to\mbox{Aut}(Z_7). We note that \mbox{Aut}(Z_7)\cong Z_6 and write Z_3 as \langle~x~\rangle and Z_6 as \langle~y~\rangle. With this presentation, \varphi:x\mapsto y^2 is a valid homomorphism, so Z_7\rtimes_{\varphi}Z_3\cong Z_7Z_3. I will call this group F_{21} because I think groups like this are called Frobenius groups. So now I want to look for homomorphisms \psi_i:Z_2\to\mbox{Aut}(F_{21}). Again writing Z_2 as \langle~x~\rangle, we have:
    1. \psi_0:x\mapsto1 This is the trivial mapping and gives G=F_{21}\times Z_2.
    2. \psi_1:x\mapsto\,?
    I don't know whether ? can be anything valid because I don't understand the structure of F_{21}. How can I determine whether its automorphism group has any elements of order 2?

    Thanks in advance!
    Last edited by redsoxfan325; August 24th 2011 at 09:24 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Classification of Non-Abelian Groups of Order 42

    Quote Originally Posted by redsoxfan325 View Post
    I'm trying to do this methodically. Let G be a non-abelian group of order 42. The Sylow Theorem says that G has a normal subgroup of order 7, and the number of Sylow-3 subgroups is either 1 or 7.

    If n_3=1, then Z_3\trianglelefteq G, so Z_{21}=Z_3\times Z_7\cong Z_3Z_7 and is normal because its index is 2. If we look for homomorphisms \phi:Z_2\to\mbox{Aut}(Z_{21}), we find the trivial one (which gives Z_{42}), or if Z_2=\langle~x~\rangle and \mbox{Aut}(Z_{21})\cong Z_2\times Z_6=\langle~y~\rangle\times\langle~z~\rangle, then \phi_1(x)=(1,z^3), \phi_2(x)=(y,1), and \phi_3(y,z^3). This gives us three non-abelian groups of order 42, namely Z_{21}\rtimes_{\phi_i}Z_2 for i\in\{1,2,3\}. What is the best way to determine whether these are all unique up to isomorphism and, furthermore, whether they are isomorphic to more familiar groups?

    Next, if n_3=7, then we look for homomorphisms \varphi:Z_3\to\mbox{Aut}(Z_7). If Z_3=\langle~x~\rangle and \mbox{Aut}(Z_7)=Z_6=\langle~y~\rangle, then \varphi(x)=y^2 is a valid homomorphism, so Z_7\rtimes_{\varphi}Z_3\cong Z_7Z_3, which I will call F_{21} because I think these are called Frobenius groups. So now I want to look for homomorphisms \psi:Z_2\to\mbox{Aut}(F_{21}). Aside from the trivial one, which gives G=F_{21}\times Z_2, I don't know whether there are any other ones because I don't understand the structure of F_{21}, so in particular I don't know whether its automorphism group has any elements of order 2.

    Thanks in advance!
    What a messy problem. Let's see if we can work on the second paragraph. So, we want to look for morphisms Z_2\to\text{Aut}(Z_3Z_7). If Z_3Z_7 is abelian then we know that \text{Aut}\left(Z_3Z_7)\cong\text{Aut}(\mathbb{Z}_  {21})\cong\left(\mathbb{Z}_{21}\right)^\times\cong \mathbb{Z}_{3}^\times\times\mathbb{Z}_7\times\cong  \mathbb{Z}_2\times\mathbb{Z}_6 which you can check has three subgroups of order two, thus ostensibly three possibilities. If Z_3Z_7 is nonabelian you can check that it has only one, up to conjugacy, subgroup of order 2[/tex] and thus there is one choice there.

    Can you work for there, and come back with what you find? All the four possibilities I mentioned are, in fact, nonisomorphic.
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    Super Member redsoxfan325's Avatar
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    Re: Classification of Non-Abelian Groups of Order 42

    Quote Originally Posted by Drexel28 View Post
    What a messy problem. Let's see if we can work on the second paragraph. So, we want to look for morphisms Z_2\to\text{Aut}(Z_3Z_7). If Z_3Z_7 is abelian then we know that \text{Aut}\left(Z_3Z_7)\cong\text{Aut}(\mathbb{Z}_  {21})\cong\left(\mathbb{Z}_{21}\right)^\times\cong \mathbb{Z}_{3}^\times\times\mathbb{Z}_7\times\cong  \mathbb{Z}_2\times\mathbb{Z}_6 which you can check has three subgroups of order two, thus ostensibly three possibilities. If Z_3Z_7 is nonabelian you can check that it has only one, up to conjugacy, subgroup of order 2[/tex] and thus there is one choice there.

    Can you work for there, and come back with what you find? All the four possibilities I mentioned are, in fact, nonisomorphic.
    Sorry for writing a somewhat messy question. I actually have done the work from there in my opening post. If Z_3Z_7 is abelian, then those three possibilities are the three groups I found in the first paragraph ( Z_{21}\rtimes_{\phi_i}Z_2 for i=1,2,3). If Z_3Z_7 is non-abelian, then this is where I get stuck because I'm not sure how to verify (short of a long and tedious calculation) that Z_3Z_7 has one subgroup of order 2 (up to conjugacy). Furthermore, what criteria do you use to determine that all four groups are non-isomorphic?
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  4. #4
    Super Member redsoxfan325's Avatar
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    Re: Classification of Non-Abelian Groups of Order 42

    Quote Originally Posted by Drexel28 View Post
    What a messy problem.
    I tried to clean up my original post a bit.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Re: Classification of Non-Abelian Groups of Order 42

    Quote Originally Posted by redsoxfan325 View Post
    I tried to clean up my original post a bit.
    I did this a while ago as part of a project. I'll try to find the paper for you and post it here. I believe what I did was you can discount all three of the assuming-abelian-case by element/order counting.
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