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/


  • $\begingroup$ 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... $\endgroup$ – math123456789 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. $\endgroup$ – math123456789 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

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 $(\mathbb{Z}_q\times\mathbb{Z}_q)\rtimes_{\phi}\mathbb{Z}_p$.

  • $\begingroup$ 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. $\endgroup$ – math123456789 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.