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What is the conjugacy class of subgroups of order $p$ in $GL(n,p)$? (Are all subgroups of order $p$ conjugate in $GL(n,p)$?)

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You mean "What are the conjugacy classes of subgroups of order $p$ in $GL(n,p)$? –  joriki Apr 14 '11 at 6:55
Take a look at the possible subspaces of $\mathbb{F}_p^n$ fixed by an element of order $p$ in $\mathrm{GL}(n, p)$. Do they all have the same dimension? –  j.p. Apr 14 '11 at 7:12
@Alex: Should (s)he compute the order of $\mathrm{GL}(n, p)$ just to get to know the group, or does the order of the group help in any way for the question? –  j.p. Apr 14 '11 at 7:16
@Alex: I'm very curious about this. How does the order help considering that there is also a cyclic group of the same order? –  j.p. Apr 14 '11 at 8:18
@jug I'm sorry, I was oversimplifying things. Please ignore my comments. –  Alex B. Apr 14 '11 at 13:36
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2 Answers

up vote 5 down vote accepted

Every subgroup of order p in GL(n,p) is conjugate to a subgroup generated by a matrix in Frobenius normal form. Such a matrix is a direct sum of companion matrices of the polynomials (x−1)k where 1 ≤ k ≤ p, and is determined up to conjugacy by the partition of n defined by 1 ≤ k1 ≤ k2 ≤ … ≤ km ≤ p where n = k1 + k2 + … + km and all such partitions other than 1 ≤ 1 ≤ … ≤ 1 define subgroups of order p (k1 = k2 = … = kn = 1 defines the trivial subgroup).

These ki are calculated from the subgroup by considering the dimensions of subspaces fixed by the subgroup (eigenspaces of eigenvalue 1) and the associated generalized eigenspaces (nullspaces of those polynomials (x−1)k), so this is just a more detailed version of @jug's comment.

If we let I(n,p) be the number of conjugacy classes of subgroups of GL(n,p) of order p, then we have the bijective but slightly messy formula:

I(n,p) = number of partitions (disregarding order) of n into positive integers no greater than p. In particular, no, GL(n,p) contains non-conjugate subgroups of order p as soon as n ≥ 3 if p is odd, and as soon as n ≥ 4 if p = 2.

For p = 2, one gets the simple formula I(n,2) = floor(n/2).

I believe this is also equal to NrPartitions(n+p,p)−1, the number of partitions of n+p into at most p things (and subtracting one to remove the trivial subgroup). This should follow by dual Young tableaux or otherwise by general combinatorics.

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(Same as Jack's, different normal form!)

If $A\in\operatorname{GL}(n,p)$ has order $p$, then $A^p-I=0$. It follows that $A$ satisfies the polynomial $X^p-1$, so the minimal polynomial $m_A$ divides $X^p-1=(X-1)^p$. Since every eigenvalue of $A$ is a root of $m_A$, we conclude that the only eigenvalue of $A$ (in any algebraic closure of $\mathbb F_p$), is $1$.

It follows at once that both the Jordan form $J$ of $A$ and a matrix conjugating $A$ to $J$ are in fact both elements of $\operatorname{GL}(n,p)$. If we want to count the elements of $\{A\in\operatorname{GL}(n,p):A^p=I\}$ up to conjugation in $\operatorname{GL}(n,p)$, then, it is enough to count the conjugacy classes of matrices of $\operatorname{GL}(n,p)$ in Jordan canonical form with minimal polynomial dividing $(X-1)^p$.

This is a pretty simple excercise. The answer is: the number of partitions of $n$ with parts not larger than $p$.

Hmm, this is counting conjugacy classes of matrices and not of subgroups...*

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In general conjugacy classes of elements and conjugacy classes of cyclic subgroups are similar (the latter are called rational conjugacy classes). For GL (and p-elements) and "big" groups (like symmetric groups) they are the same. In other words, all the generators of a subgroup of order p have the same generalized eigenspaces, so are all conjugate, so every conjugacy class (of elements of order p) is rational. –  Jack Schmidt Apr 14 '11 at 21:09
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