How to deduce a formula (see below) for number of conjugacy classes in $\operatorname{GL}_n(F_2)$? (More generally F_q) ?
Is there some description of conjugacy classes or we just know how many of them but do not know how to describe them ?
Can someone send me a paper, please ?:
"Pairs of commuting matrices over a finite field"
Walter Feit and N. J. Fine
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.dmj/1077468920
FORMULA from OEIS
The number $a(n)$ of conjugacy classes in the group $\operatorname{GL}(n, q)$ is the coefficient of $t^n$ in the infinite product: product $k=1, 2, … (1-t^k)/(1-qt^k)$ – Noam Katz (noamkj(AT)hotmail.com), Mar 30 2001.
Simple observations
Clearly if characteristic polynom is different then matrices are not conjugated.
For each char. pol. it is easy to give a matrix with such char. pol. so we get (q-1)q^{n-1}
possibilities at least.
But if char. pol has a roots with multiplicities – then we may have several Jordan cells
and several conjugacy classes with same characteristic polynomial.
So it is not clear for me how to count them.
Diagonal elements of corresponding Jordan cells may not live in F_q but in some alg. extentsion
of it, these makes me completely puzzled – is it possible to control this ? Hardly…
Related and not so much but nice questions 🙂
Number of conjugacy classes in generic finite group?
Sizes of twisted conjugacy classes of $PSL(n,q)$
How many conjugacy classes of subgroups does GL(2,p) have?
Decomposition of GL(2,p) into irreducible representations
Products of Conjugacy Classes in S_n
The number of conjugacy classes and the order of the group
Arithmetic relations between degrees of irrchar and cardinals of conjugacy classes
Can we bound degrees of complex irreps in terms of the average conjugacy class size?
Symmetric group irreps in tensor products of exterior products of the standard representation
Best Answer
Edit: Unfortunately, the first version of this answer was wrong, so I had to rewrite it almost entirely.
So the formula you want to prove is: $$ \sum_{n=1}^\infty C_{n,q} t^n =\prod_{k=1}^\infty \frac{1-t^k}{1-qt^k} $$ where $C_{n,q}$ denotes the number of conjugacy classes in $\textrm{GL}_n(q)$.
Note that $$ \frac{1-t^k}{1-qt^k} =1+\sum_{i=1}^\infty (q^i-q^{i-1}) t^{ki} $$ As Paul Garrett notes in his answer, $q^i-q^{i-1}$ is the number of monic polynomials in $\mathbb F_q[x]$ of degree $i$ with non-vanishing constant term. One usual way of counting conjugagy classes of matrices in $\textrm{GL}_n(q)$ (or $\textrm{Mat}_n(\mathbb F_q)$, if one admits the eigenvalue $0$) is counting isomorphism classes of $\mathbb F_q[x]$-modules of dimension $n$, which by the structure theorem on f.g. modules over PID's are given by $$ \mathbb F_{q}[x]/(p_k(x)) \oplus \mathbb F_{q}[x]/(p_k(x)\cdot p_{k-1}(x)) \ldots \oplus F_{q}[x]/(p_k(x)\cdot p_{k-1}(x)\cdots p_{1}(x)) $$ where $p_1,\ldots,p_k$ are monic polynomials with constant term $\neq 0$ (uniquely determined by the isomorphism type of the module) such that $1\cdot \textrm{deg}(p_1)+2\cdot \textrm{deg}(p_2) \ \ldots + k \cdot \textrm{deg}(p_k) = n \textrm{ (the dimension of the module)}$. Note that some of the $p_i$ may very well be equal to $1$.
From the above considerations it is clear that we can parametrize isomorphism classes of $\mathbb F_q[x]$-modules by giving a sequence of monic polynomials which eventually becomes constant $=1$, by letting the module given above correspond to the sequence $$ (f_1,f_2,f_3,\ldots) =(p_k, p_{p-1},\ldots,p_1,1,1,\ldots) $$ Note that in this sequence, $f_m$ constributes $m\cdot \textrm{deg}(f_m)$ to the dimension of the module. It follows $$ \begin{array}{rcl} \sum_{n=1}^\infty C_{n,q} t^n &=& \prod_{m=1}^{\infty}\left(\sum_{i=0}^\infty \#\textrm{(choices for $f_m$ that contribute $mi$ to the dimension)}\cdot t^{mi}\right)\newline &=&\prod_{m=1}^{\infty} \left(1+\sum_{i=1}^\infty (q^i-q^{i-1}) t^{mi}\right) =\prod_{m=1}^\infty \frac{1-t^m}{1-qt^m} \end{array} $$