Let $p$ be a prime and let $f(p)$ be the number of groups of order $31p^2$ up to isomorphism.

I have a homework problem which involves finding which choices for $p$ make $f(p)$ as large as possible. In order to solve the problem, I suspect that I will essentially need to classify groups of order $31p^2$. There will, of course be two cases to consider: either $p=31$ or $p \neq 31$.

If $p=31$, then we are looking for the number of groups of order $p^3$, hence $f(p)=5$ by this.

Now assume $p\neq31$. Then for a group $|G|=31p^2$ we have $n_p \equiv 1$ (mod $p$) and $n_p|31$, implying $n_p \in \{1,31\}$. So it seems there are two cases to consider.

If $n_p=31$, we have $31 \equiv 1$ (mod $p$) and hence $p|30$, so $p \in \{2,3,5\}$. Using OEIS, we see that (up to isomorphism) there are $4$ groups of order $31*2^2$, $4$ groups of order $31*3^2$, and $4$ groups of order $31*5^2$. Hence, since we are trying to maximize $f(p)$, we can assume WLOG $p \notin \{2,3,5\}$, which forces $n_p=1$.

If $n_p=1$, $G$ has a normal Sylow-p subgroup. But does this help at all?

I know the Fundamental Theorem of (finitely generated) Abelian Groups. So if it turns out that all the groups of order $31p^2$ are abelian (for a given $p$), then finding $f(p)$ will be stupidly easy.

This homework, by the way, coincides with our study for semidirect products, hence I wouldn't be surprised if semidirect products play a vital role in the solution.

  • $\begingroup$ Groups of order $p^2q$ with certain conditions on $p,q$ are abelian, see here. $\endgroup$ Nov 14, 2019 at 19:56
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    $\begingroup$ You seem to be doing well, so it is not clear why you have given up. The next thing to do is to consider what $n_{31}$ can be. $\endgroup$
    – Derek Holt
    Nov 14, 2019 at 19:57
  • $\begingroup$ When $n_p=1$ a group of order $31p^2$ is automatically a semidirect product. I tried to sketch a census of those. I am not experienced enough with this to feel overly confident, but will welcome a learning opportunity. $\endgroup$ Nov 14, 2019 at 20:30
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    $\begingroup$ Try calculating $f(31 \times 311^2)$ for example. $\endgroup$
    – Derek Holt
    Nov 14, 2019 at 20:36

1 Answer 1


Assume that $p\equiv1\pmod{31}$. I am fairly sure that these primes give you the largest number of non-isomorphic groups.

Assume first that the Sylow $p$-subgroup is isomorphic to $P=\Bbb{Z}_p^2$, that is, a $2$-dimensional vector space over the prime field of $p$ elements.

In this case $Aut(P)\cong GL_2(\Bbb{Z}_p)$. Because $31\mid p-1$, there is a multiplicative group $\mu_{31}$ of $31$st roots of unity in $\Bbb{Z}_p^*$. Let us fix a generator $g$ of $\mu_{31}$.

Consider the homomorphisms $\phi_j:C_{31}\to Aut(P)$ gotten by mapping a generator $c$ of $C_{31}$ to the diagonal matrix $\mathrm{diag}(g,g^j)$. Here $j$ takes values in the range $0\le j<31$. We can then form the semidirect product $$ G_j=P\rtimes_{\phi_j}C_{31}. $$ Observe that if $j>0$ and $j'$ is the multiplicative inverse of $j$ modulo $31$, i.e. $jj'\equiv1\pmod{31}$, then $\phi_j(c^{j'})=\mathrm{diag}(g^{j'},g)$ - a matrix conjugate to $\phi_{j'}(c)$. This implies that $G_j\cong G_{j'}$. On the other hand, if $j''\notin\{0,j,j'\}$ then it seems to me that $G_{j''}$ is not isomorphic to $G_j$ (see the next paragraph).

For if $j\neq0$ then $c$ does not commute with any non-identity element of $P$. This is because $\phi_j(c)$ does not have $1$ as an eigenvalue. The same applies to all non-trivial powers of $c$. It follows that there are no element of order $31p$ in $G_j$, so all the elements of $G_j\setminus P$ have order $31$. Thus $G_j$ has $p^2$ Sylow $31$-subgroups. All of those are conjugate to each other, and each contains two elements with an eigenvalue $g$ on $P$, namely the conjugates of $c$ and $c^{j'}$. The other eigenvalues of those elements are thus $g^j$ and $g^{j'}$ respectively. Any isomorphism $f:G_j\to G_{j''}$ would have to preserve this pair of eigenvalues, implying the claim $j''\in\{j,j'\}$.

Similarly, we see that $G_0$ is not isomorphic to any other $G_j$. This is because in $G_0$ we have elements of order $31p$ as any eigenvector of $\phi_0(c)$ belonging to eigenvalue $1$ commutes with $c$.

Let's take stock. $j=j'$ if and only if $j\equiv\pm1\pmod{31}$. The remaining $28$ choices of $j$ split into $14$ pairs $(j,j')$. Altogether we get $17$ non-abelian pairwise non-isomorphic semidirect product $(C_p\times C_p)\rtimes C_{31}$. In addition to two non-isomorphic abelian groups of order $31p^2$ we also have a semidirect product $C_{p^2}\rtimes C_{31}$ coming from embedding $C_{31}$ into $Aut(C_{p^2})\cong C_{p(p-1)}$.

Barring mistakes and/or oversights I arrived at twenty non-isomorphic groups of order $31p^2$ for any prime $p\equiv1\pmod{31}$.

The order of $GL_2(\Bbb{Z}_p)$ is $p(p-1)^2(p+1)$. The subgroups of order $p+1$ are cyclic, so I doubt we will get as many non-isomorphic semidirect products when $31\mid p+1$.

See this thread and others linked to it for more information about when semidirect products of two given groups are isomorphic.

  • $\begingroup$ Too busy to double check everything. Caveat reader. $\endgroup$ Nov 14, 2019 at 20:27
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    $\begingroup$ Yes I agree with your calculations for the case $p \equiv 1 \bmod 31$. I think you only get three groups when $p \equiv -1 \bmod 31$, only one of which is nonabelian. $\endgroup$
    – Derek Holt
    Nov 14, 2019 at 20:40
  • $\begingroup$ @JyrkiLahtonen +1 Wow, looks intimidating. I'm wondering, is there a way to show that $p \equiv 1$ (mod $31$) maximizes $f(p)$ without actually computing $f(p)=20$? I mean, I guess one could find that there are at least 6 isomorphic subgroups in this case, but is there an easy way to show that $f(p)$ will always be the same so long as we restrict $p \equiv 1$ (mod $31$)? $\endgroup$ Nov 16, 2019 at 18:36

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