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I am struggling with semidirect products and how they can be used to classify groups of a certain order. In particular, I need help with the nonabelian case. This is the problem I am working with..

Classify all groups of order $pq^2$ with $p$,$q$ primes, $p<q$, $p\nmid(q-1)$, and $p^2\nmid(q+1)$. Use can use the fact that $GL_2(\mathbb{Z}_q)$ has $(q^2-1)(q^2-q)$ elements.

Ok. So here's my thought process. I first considered when $G$ was abelian and applied the Fundamental Theorem of Finitely Generated Abelian Groups (FTFGAG) to obtain all abelian groups of this order. My results were: $\mathbb{Z}_{pq^2}$ and $\mathbb{Z}_{pq}\times\mathbb{Z}_q$.

Next considered when $G$ was nonabelian and applied Sylow's Theorem to determine how many Sylow $p$ and $q$ subgroups there were in $G$. I found the Sylow $q$-subgroups to be unique, and hence normal. I let the Sylow $q$-subgroup be called $H$. I let $K$ be any Sylow $p$-subgroup in $G$. Then by Lagrange, $H\cap K=1$. Next I showed that $G=HK$ and let $\varphi: K\rightarrow \text{Aut}(H)$ be a homomorphism. Then by applying theorem 12 from Dummit and Foote (sorry it didn't have a name :/), I got $G\cong H\rtimes_\varphi K$.

Now I just need to consider all isomorphisms of $H$. They are $\mathbb{Z_{q^2}}$ and $\mathbb{Z}_q\times\mathbb{Z}_q$.

Suppose $H=\mathbb{Z_{q^2}}$. Then $|\text{Aut}(H)|=q(q-1)$. Since $p\nmid q$ and $p\nmid q-1$, there does not exist an element of order $p$ in $\text{Aut}(H)$ by Lagrange. This means the only homomorphism is trivial. Therefore $H\rtimes_\varphi K\cong \mathbb{Z}_{q^2}\times \mathbb{Z}_p$. But this is abelian and contradicts my assumption that $G$ is nonabelian. Plus FTFGAG, already classified all abelian groups. Therefore $H=\mathbb{Z}_{q^2}$ does not result in a new group.


And here is where I start to get lost...

Suppose $H=\mathbb{Z_q}\times\mathbb{Z_q}$. Then $|\text{Aut}(H)|=(q^2-1)(q^2-q)$. I do not see how $p$ divides this...

Help with this last part would be much appreciated. Thanks.

Also here are some resources I have looked at: https://crazyproject.wordpress.com/2010/06/25/classify-the-groups-of-order-75/

http://www.math.purdue.edu/~lipman/5532011/order-p%5E2q.pdf

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  • $\begingroup$ Hint: $(q^2-1)(q^2-q)=q(q^2-1)(q-1)=q(q-1)^2(q+1)$. $\endgroup$ – Justin Benfield Oct 29 '16 at 7:19
  • $\begingroup$ So use the argument that $\exists$ Sylow $q$ subgroup in Aut($H$)? This would imply there to be an automorphism of order $q$. Right? Or does $p|q+1$? If so, I don't see that... $\endgroup$ – jm.byrnes Oct 29 '16 at 7:21
  • $\begingroup$ You have that $p^2$ does not divide $(q+1)$ by assumption, so either $p$ divides $(q+1)$ or $q(q-1)^2(q+1)$ is coprime to $p$. In particular, since $p$ is prime, the image of the homomorphism is either trivial or $\mathbb{Z}_p$. $\endgroup$ – Justin Benfield Oct 29 '16 at 7:24
  • $\begingroup$ Ok. I believe I see what you are saying. So based on my assumption $p|(q+1)$ or $p|(q(1+q)(q-1)^2)$. In particular, this would mean there would exist Sylow $p$-subgroups in Aut($H$). So all subgroups order $p$ in Aut($H$) are conjugate in Aut($H$) by the Sylow Theorem. $\endgroup$ – jm.byrnes Oct 29 '16 at 7:37
  • $\begingroup$ It's more powerful than that: remember that we are trying to find homomorphisms from $K$ (which is of order $p$), to $Aut(H)$ which is of order $q(q-1)^2(q+1)$. Now, my previous comment establishes that either $|Aut(H)|$ is coprime to $|K|$ (implying homomorphism must be trivial!) or $Aut(H)$ has a Sylow $p$-subgroup of order exactly $p$ (and none larger, since p^2 does not divide $|Aut(H)|$), therefore the possible images of the homomorphism are either the trivial group, or the cyclic subgroup $\mathbb{Z}_p$ (which incidentally is isomorphic to $K$). $\endgroup$ – Justin Benfield Oct 29 '16 at 7:41
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To give a proper answer, let us start with what you have already established (and a few observations):

You established that $G$ has normal Sylow $q$-subgroup $H$ of order $q^2$, and Sylow $p$-subgroup $K$ of order $p$ (note: This means $K\simeq\mathbb{Z}_p$).

Possibilities for $H$ are $H\simeq\mathbb{Z}_{q^2}$ and $H\simeq\mathbb{Z}_q\times\mathbb{Z}_q$.

For the former case, you have that $|Aut(\mathbb{Z}_{q^2})|=q(q-1)$ and we know that $p\nmid(q-1)$ by assumption, and also that $p\neq q$, therefore $|Aut(H)|$ is coprime to $|K|$ and thus $\phi(K)=\{e\}$ (homomorphism has trivial image).

You may not have known this, but if $\phi(K)$ is trivial, then the semidirect product, $H\rtimes_{\phi}K$ is none other than the direct product $H\times K$.

So the former case gives only $\mathbb{Z}_p\times\mathbb{Z}_{q^2}\simeq\mathbb{Z}_{pq^2}$. (The cyclic group)

For the 2nd case, we have that $|Aut(H)|=q(q-1)^2(q+1)$. Now we invoke the assumption $p^2\nmid (q+1)$ to conclude that either $p\mid (q+1)$ or $|Aut(H)|$ is coprime to $|K|$ (since again, $p\neq q$ and $p\nmid (q-1)$ by assumption). In the case that $p\nmid (q+1)$, then again the homomorphism has trivial image, and we get an abelian group $\mathbb{Z}_p\times(\mathbb{Z}_q\times\mathbb{Z}_q)\simeq\mathbb{Z}_{pq}\times\mathbb{Z}_q$. (The noncyclic abelian group)

Alternatively, if $p\mid (q+1)$ then not only do we get the groups above, but also a non-trivial semidirect product. Since $K=\mathbb{Z}_p$ and $p$ is prime, $K$ has no nontrivial proper subgroups, hence the only quotients are the trivial subgroup and all of $K$ and since quotients correspond to images of homomorphism because the kernel of said homomorphism is a normal subgroup, we know that only possible nontrivial image for $\phi$ is a subgroup isomorphic to $\mathbb{Z}_p$. Furthermore, since $p^2\nmid (q+1)$, we know that the sylow $p$-subgroup(s) of $Aut(H)$ are isomorphic to $\mathbb{Z}_p$ (and it turns out that it doesn't matter which one is the image of $\phi$ if there are more than one). The result is a nontrivial semidirect product $(\mathbb{Z}_q\times\mathbb{Z}_q)\rtimes_{\phi}\mathbb{Z}_p$.

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  • $\begingroup$ I understand this problem so much better now. Thank you for all of you clarifications. I already had the first half of this answer, but it was nice to verify. You explain things your work in a very clear manner. $\endgroup$ – jm.byrnes Oct 29 '16 at 8:46
  • $\begingroup$ Attempting to read quite a few advanced math books as an undergrad taught me just how frustrating and obfuscating unclear mathematical writing can be to deal with. So I've made a point of trying very hard to be thorough and clear as much as I can when I write mathematics that I intend for others to read. $\endgroup$ – Justin Benfield Oct 29 '16 at 9:00

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