Let $ G $ be a group of $ n \times n $ matrices. Suppose that some subset $ \{ g_j: 1 \leq j \leq n^2 \} $ of $ G $ is a basis for the space of all $ n \times n $ matrices. Furthermore suppose that the set $$ \{ \overline{g_i}: 1 \leq j \leq n^2 \} $$ is a group in $ \operatorname{PGL}_n $. What can we say about the size of $ G $? Must it be the case that $$ \lvert G\rvert\geq n^2 \prod p_i $$ where $ \prod p_i $ is the product of all distinct primes dividing $ n $?

When we take our field to be algebraically closed and characteristic 0 I think I have groups that saturate this bound for all $ n $. Is this bound even true? Can we do better? Mostly interested in doing everything over $ \mathbb{C} $ but could be cool to hear about other fields too.

  • $\begingroup$ The interesting question is over a year old, but back on the front page for some reason (and not because it's unanswered, since @FriederLadisch gave a very illuminating answer below), so I'll ask while it's here: where did this condition come from? $\endgroup$
    – LSpice
    Feb 13, 2023 at 15:52
  • 1
    $\begingroup$ @LSpice Back in November 2021 I was think a lot about what makes the Pauli matrices special, including the generalized ones used in quantum computing for prime power qudits. I decided that one property which especially stuck out to me was that they form a basis and simultaneously they (projectively) form a finite group. I decided to ask MO if they were the minimal such object/saturated a bound for such behavior. To my amazement by the next day I had a wonderful answer from Frieder Ladisch! I was looking over it this morning bc I'm about to post to MSE something related and I made a small edit. $\endgroup$ Feb 13, 2023 at 15:59

1 Answer 1


I consider the situation over $\mathbb{C}$. We can replace $G$ by the subgroup generated by the $g_j$'s, so assume that $G$ is generated by the $g_j$'s. That $G$ contains a spanning set of the space of all matrices yields that the center of $G$ only contains scalar matrices, and also that the inclusion $G \hookrightarrow \operatorname{GL}(n,\mathbb{C}) $ is an absolutely irreducible representation.
That the $g_j$'s modulo the center form a subgroup means that $|G: Z(G)| = n^2$. So $G$ has an irreducible representation of degree $n$ with $n^2 = |G:Z(G)|$. Such groups are called groups of central type. DeMeyer and Janusz (DeMeyer, F. R.; Janusz, G. J., Finite groups with an irreducible representation of large degree, Math. Z. 108, 145-153 (1969). ZBL0169.03502, Theorem 2) have shown that a finite group is of central type if and only if each Sylow $p$-subgroup $S$ of $G$ is of central type and $S\cap Z(G) =Z(S)$. Now a $p$-group always has a nontrivial center of order at least $p$. Thus the center of $G$ has an order divisible by all the primes involved in $n$, and your bound follows.
As you say, there are groups attaining the bound: An extraspecial $p$-group of order $p^{2k+1}$ has an irreducible, faithful representation of dimension $p^k$, and the image is a matrix group as in the question with $n=p^k$. For composite $n=p_1^{k_1}\dotsm p_m^{k_m}$, we take a direct product $P_1 \times \dotsb \times P_m$ , where each $P_i$ is extraspecial of order $p_i^{2k_i+1}$.

The situation should be the same over any field of characteristic zero, because the assumptions in the question yield that the given natural representation is absolutely irreducible, and also over fields such that the characteristic does not divide $n$. When the prime $p$ divides $n$, then my guess is that such a matrix group over a field of characteristic $p$ is not possible.

  • $\begingroup$ Everything in here looks great thanks so much! The one question I still have is could you give an example of the general result over (non algebraically closed) fields. For example could you give a 3 dimensional real representation of the extraspecial 3-group with $ 3^3=27 $ elements. I am a bit surprised that this exitsts. $\endgroup$ Nov 14, 2021 at 19:28
  • $\begingroup$ I'm not sure what you mean, but maybe there is a misunderstanding here: I do not claim that the extraspecial group with $3^3$ elements has such a representation over the reals. I claim that when $G$ is a group of $3\times 3$ matrices over some field as in your question, then that field must contain primitive 3rd roots of unity. $\endgroup$ Nov 14, 2021 at 19:47
  • $\begingroup$ You're right just a misunderstanding! Everything sounds good then, thanks again for the great answer! $\endgroup$ Nov 14, 2021 at 19:52
  • $\begingroup$ are all groups of central type 2-step nilpotent? $\endgroup$ Mar 14 at 21:20

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