# Number of groups of order $31p^2$ (up to isomorphism)

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.

• Groups of order $p^2q$ with certain conditions on $p,q$ are abelian, see here. Nov 14, 2019 at 19:56
• 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. Nov 14, 2019 at 19:57
• 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. Nov 14, 2019 at 20:30
• Try calculating $f(31 \times 311^2)$ for example. Nov 14, 2019 at 20:36

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 $$31\mid p+1$$.

• 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. Nov 14, 2019 at 20:40
• @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 $31$)? Nov 16, 2019 at 18:36