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Fix a positive integer $n$ and let $S$ be the set of $n$ by $n$ matrices with entries in $\mathbf{Z}_p$ (the $p$-adic integers) whose determinant is $p$. The group $G:=\mathrm{SL}_n(\mathbf{Z}_p)$ acts freely on $S$ via left multiplication.

Is it possible to write down an explicit list of representatives for the orbits of $G$ on $S$? If so, what is this list? If not, is there an algorithm for computing a complete list of orbit representatives?

For example, for $n=2$, I think that this is basically the "Hecke Operator at $p$" computation, and the $p+1$ orbits of $G$ on $S$ have representatives

$$ \left(\begin{matrix} p & b \\\ 0 & 1 \end{matrix}\right),\ b=0,1,\ldots, p-1\ \text{and}\ \left(\begin{matrix} 0 & -1 \\\ p & 0 \end{matrix}\right) $$

This is probably a question whose answer is well-known to many people, so I apologize in advance for my ignorance!

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Warning: I have not carefully checked the argument below.

First note that $\text{SL}_n(\mathbb{Z}_p)$ contains row operations of determinant $1$ (adding a multiple of a row to another row, transposing two rows and negating one of them, and multiplication by a diagonal matrix of determinant $1$). Given $M \in \mathcal{M}_n(\mathbb{Z}_p)$ of determinant $p$, we will row reduce it: if the entries in the first column are not all divisible by $p$, one of them is invertible, so by transposing and multiplying by a diagonal matrix we can place it in the first row and set it equal to $1$ to pivot on it. We can continue row reducing until arriving at a column where the only possible pivots are divisible by $p$ (this must eventually occur by the determinant condition), and then we just pivot on the one with minimal $p$-adic valuation (it must be precisely divisible by $p$ by the determinant condition). By multiplying by a diagonal matrix we can set this pivot to $p$, and proceed with row reduction.

The final result should be a matrix all of whose diagonal entries are $1$ except for a single diagonal entry of $p$ somewhere. In every column except the one containing $p$ all of the other entries are zero, and in the column containing $p$ the entries above it range from $0$ to $p-1$ and the entries below it are also zero. I think a tedious calculation will now confirm that if $A, B$ are distinct matrices in this form then $AB^{-1} \not \in \text{SL}_n(\mathbb{Z}_p)$. For $n = 2$ the representatives this algorithm gives are

$$\left[ \begin{array}{cc} p & 0 \\\ 0 & 1 \end{array} \right], \left[ \begin{array}{cc} 1 & b \\\ 0 & p \end{array} \right], b = 0, 1, ... p-1.$$

(If I'm not horribly mistaken, what you've written down are representatives for the orbits of the action by right multiplication.) This should generalize to a notion of Hermite normal form for matrices over a PID.

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To complement Qiaochu Yuan's answer, check out Lemma 9.3.2 in Goldfeld's book Automorphic Forms and L-functions for the group GL(n,R). On the two sides of (9.3.3) you can see the decompositions of the set of integral $n\times n$ matrices of determinant $N$ into double cosets and right cosets of $\mathrm{SL}_n(\mathbb{Z})$. The right hand side corresponds to your question, and yes, it gives rise to the $N$-th Hecke operator, see (9.3.5). Of course this is over $\mathbb{Z}$, but localizing at $p$ gives what you need: in that case $N$ is a $p$-power without loss of generality, since other prime powers are $p$-adic units.

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