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According to A000001, the number of groups of order $375$ is $7$. How to figure them out?

$5$ groups can be found in this way: Because the number of $125$-order groups is $5$, we can make the direct product of those groups of order $125$ with $Z_3$ and then we get $5$ $375$-order groups.

But how to find the another two groups?

UPD: one of the rest of groups is $C_5^3\rtimes C_3$, and now the task is to find the last missing group.

UPD2: Now I think that a more profound comprehension (to Jyrki Lahtonen's answer) would be appreciated...

FFjet
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    one of them is $C_5^3\rtimes C_3$, where $C_3$ permutes the factors. – Rylee Lyman Sep 08 '19 at 04:37
  • The Sylow $5$-subgroup $P$ is necessarily normal, so the group is either a direct product $P\times C_3$ or a semi-direct product $P\rtimes C_3$. The latter case occurs when $P$ has (an) automorphism(s) of order three. For sure $GL_3(\Bbb{F}_5)$ has order divisible by three ($GL_2$ would be enough), implying that $C_5\times C_5\times C_5$ has such automorphisms. Not sure how many more groups that gives. – Jyrki Lahtonen Sep 08 '19 at 04:38
  • @RyleeLyman and what would be the last group? – FFjet Sep 08 '19 at 04:47
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    I think this result together with the observation that Sylow $3$-subgroups of $GL_3(\Bbb{F}_5)$ are cyclic of order three imply that there is up to isomorphism a single non-abelian semidirect product $(C_5^3)\rtimes C_3$ (=the one described in Rylee Lyman's comment). I suspect the missing group is a semidirect product of the group of strictly upper triangular matrices over $\Bbb{F}_5$ and $C_3$. – Jyrki Lahtonen Sep 08 '19 at 04:52
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    You now have a description of all $7$ isomorphism types of groups of order $375$, but you have not attempted to prove that there are no more than $7$. To do that you would have to prove that the automorphism groups of $C_5^3$ and the extraspecial groups of order $5^3$ have unique conjugacy classes of subgroups of order $3$ (see Jyrki Lahtonen's comment for${\rm Aut}(C_5^3)$), and that none of the other three types of groups of order $5^3$ have automorphisms of order $3$. – Derek Holt Sep 08 '19 at 09:54

1 Answers1

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It seems to me that the unitriangular group of order $5^3$, GAP ID $(5^3,3)$ has an automorphism of order three.

Let us denote $$ M(a,b,c)=\left(\begin{array}{ccc}1&a&c\\0&1&b\\0&0&1\end{array}\right). $$ Let us consider the mapping $$ \phi:M(a,b,c)\mapsto M(-b,a-b,c-ab+\frac{b(b-1)}2). $$ I brute force checked that $\phi$ is a homomorphism of the group $$U_3(K)=\{M(a,b,c)\mid a,b,c\in K\},$$ $K$ a field such that $1+1\neq0$. Furthermore, $\phi$ has order three.

This gives rise to a non-trivial semidirect product $U_3(K)\rtimes C_3$ with conjugation by a generator of the latter factor corresponding to $\phi$.

With $K=\Bbb{F}_5$ we get the missing seventh group $G$. Its Sylow $5$-subgroup is obviously isomorphic to $U_3(\Bbb{F}_5)$. The only other group of order $375$ with that property in the OP's list is the direct product $H=U_3(\Bbb{F}_5)\times C_3$. But that is not isomorphic to $G$. We easily see that matrices of the form $M(0,0,c)$ form the center of $U_3(K)$, and as $\phi(M(0,0,c))=M(0,0,c)$ this describes the center of $G$ as well. But $Z(H)$ has order $15$.

It is well known and easy to check that $(a,b)\mapsto (-b,a-b)$ is a linear transformation of $K^2$ of order three. The idea was to try and lift that to an automorphism of $U_3(K)$, and $\phi$ popped out. The reason why that might work is that $$U_3(K)/Z(U_3(K))\cong (K^2,+)$$ with the projection given by $M(a,b,c)\mapsto (a,b)$. Furthermore, $M(0,0,1)$ is the commutator of $M(1,0,0)$ and $M(0,1,0)$, so looking for suitable homomorphic images of $M(1,0,0)$ and $M(0,1,0)$ was a natural thing to try. This worked out, but is not really convincing.

I am ashamed to admit that I don't know what the obstructions to lifting an automorphism of $K^2$ to one of $U_3(K)$ are in general. A bit of searching lead me to this excellent post by Jack Schmidt describing the automorphisms of extraspecial $p$-groups (of which $U_3(\Bbb{F}_p)$ are the simplest examples). He also links to articles containing the details.

Jyrki Lahtonen
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