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Thread: 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 $\displaystyle G$ be a group of order 42. The Sylow Theorem says that $\displaystyle G$ 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 $\displaystyle Z_3\trianglelefteq G$, so $\displaystyle Z_3Z_7\cong Z_3\times Z_7$ and is normal because its index is 2. We now look for homomorphisms $\displaystyle \phi_i:Z_2\to\mbox{Aut}(Z_3\times Z_7)$. We note that $\displaystyle \mbox{Aut}(Z_3\times Z_7)\cong Z_2\times Z_6$ and write $\displaystyle Z_2$ as $\displaystyle \langle~x~\rangle$, and $\displaystyle Z_2\times Z_6$ as $\displaystyle \langle~y~\rangle\times\langle~z~\rangle$. (Both have identity $\displaystyle 1$.)
    1. $\displaystyle \phi_0:x\mapsto(1,1)$ This is the trivial mapping and gives the only abelian group $\displaystyle G=Z_{42}$.
    2. $\displaystyle \phi_1:x\mapsto(1,z^3)$
    3. $\displaystyle \phi_2:x\mapsto(y,1)$
    4. $\displaystyle \phi_3:x\mapsto(y,z^3)$
    The last three mappings result in three non-abelian groups of order 42, namely $\displaystyle Z_{21}\rtimes_{\phi_i}Z_2$ for $\displaystyle i\in\{1,2,3\}$. My guess is that $\displaystyle \phi_1$ produces $\displaystyle Z_3\times D_{14}$, $\displaystyle \phi_2$ produces $\displaystyle Z_7\times D_6$, and $\displaystyle \phi_3$ produces $\displaystyle D_{42}$, 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 $\displaystyle \varphi:Z_3\to\mbox{Aut}(Z_7)$. We note that $\displaystyle \mbox{Aut}(Z_7)\cong Z_6$ and write $\displaystyle Z_3$ as $\displaystyle \langle~x~\rangle$ and $\displaystyle Z_6$ as $\displaystyle \langle~y~\rangle$. With this presentation, $\displaystyle \varphi:x\mapsto y^2$ is a valid homomorphism, so $\displaystyle Z_7\rtimes_{\varphi}Z_3\cong Z_7Z_3$. I will call this group $\displaystyle F_{21}$ because I think groups like this are called Frobenius groups. So now I want to look for homomorphisms $\displaystyle \psi_i:Z_2\to\mbox{Aut}(F_{21})$. Again writing $\displaystyle Z_2$ as $\displaystyle \langle~x~\rangle$, we have:
    1. $\displaystyle \psi_0:x\mapsto1$ This is the trivial mapping and gives $\displaystyle G=F_{21}\times Z_2$.
    2. $\displaystyle \psi_1:x\mapsto\,?$
    I don't know whether $\displaystyle ?$ can be anything valid because I don't understand the structure of $\displaystyle F_{21}$. How can I determine whether its automorphism group has any elements of order 2?

    Thanks in advance!
    Last edited by redsoxfan325; Aug 24th 2011 at 08:24 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 $\displaystyle G$ be a non-abelian group of order 42. The Sylow Theorem says that $\displaystyle G$ has a normal subgroup of order 7, and the number of Sylow-3 subgroups is either 1 or 7.

    If $\displaystyle n_3=1$, then $\displaystyle Z_3\trianglelefteq G$, so $\displaystyle Z_{21}=Z_3\times Z_7\cong Z_3Z_7$ and is normal because its index is 2. If we look for homomorphisms $\displaystyle \phi:Z_2\to\mbox{Aut}(Z_{21})$, we find the trivial one (which gives $\displaystyle Z_{42}$), or if $\displaystyle Z_2=\langle~x~\rangle$ and $\displaystyle \mbox{Aut}(Z_{21})\cong Z_2\times Z_6=\langle~y~\rangle\times\langle~z~\rangle$, then $\displaystyle \phi_1(x)=(1,z^3)$, $\displaystyle \phi_2(x)=(y,1)$, and $\displaystyle \phi_3(y,z^3)$. This gives us three non-abelian groups of order 42, namely $\displaystyle Z_{21}\rtimes_{\phi_i}Z_2$ for $\displaystyle i\in\{1,2,3\}$. 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 $\displaystyle n_3=7$, then we look for homomorphisms $\displaystyle \varphi:Z_3\to\mbox{Aut}(Z_7)$. If $\displaystyle Z_3=\langle~x~\rangle$ and $\displaystyle \mbox{Aut}(Z_7)=Z_6=\langle~y~\rangle$, then $\displaystyle \varphi(x)=y^2$ is a valid homomorphism, so $\displaystyle Z_7\rtimes_{\varphi}Z_3\cong Z_7Z_3$, which I will call $\displaystyle F_{21}$ because I think these are called Frobenius groups. So now I want to look for homomorphisms $\displaystyle \psi:Z_2\to\mbox{Aut}(F_{21})$. Aside from the trivial one, which gives $\displaystyle G=F_{21}\times Z_2$, I don't know whether there are any other ones because I don't understand the structure of $\displaystyle F_{21}$, 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 $\displaystyle Z_2\to\text{Aut}(Z_3Z_7)$. If $\displaystyle Z_3Z_7$ is abelian then we know that $\displaystyle \text{Aut}\left(Z_3Z_7)\cong\text{Aut}(\mathbb{Z}_ {21})\cong\left(\mathbb{Z}_{21}\right)^\times\cong \mathbb{Z}_{3}^\times\times\mathbb{Z}_7\times\cong \mathbb{Z}_2\times\mathbb{Z}_6$ which you can check has three subgroups of order two, thus ostensibly three possibilities. If $\displaystyle Z_3Z_7$ 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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  3. #3
    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 $\displaystyle Z_2\to\text{Aut}(Z_3Z_7)$. If $\displaystyle Z_3Z_7$ is abelian then we know that $\displaystyle \text{Aut}\left(Z_3Z_7)\cong\text{Aut}(\mathbb{Z}_ {21})\cong\left(\mathbb{Z}_{21}\right)^\times\cong \mathbb{Z}_{3}^\times\times\mathbb{Z}_7\times\cong \mathbb{Z}_2\times\mathbb{Z}_6$ which you can check has three subgroups of order two, thus ostensibly three possibilities. If $\displaystyle Z_3Z_7$ 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 $\displaystyle Z_3Z_7$ is abelian, then those three possibilities are the three groups I found in the first paragraph ($\displaystyle Z_{21}\rtimes_{\phi_i}Z_2$ for $\displaystyle i=1,2,3$). If $\displaystyle Z_3Z_7$ 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 $\displaystyle Z_3Z_7$ 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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