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Let $p$ be a prime; $\mathbb{F}_{p}$ is the field with $p$ elements and $\mathbb{F}_{p}[t]$ the ring of polynomials in $t$ over $\mathbb{F}_{p}$.

Does $\mathrm{SL}_{n}(\mathbb{Z}/p^{2})$ have the same number of conjugacy classes as $\mathrm{SL}_{n}(\mathbb{F}_{p}[t]/t^{2})$?

When $p$ does not divide $n$ this follows from a theorem of P. Singla (see this paper). Note that the case when $p$ divides $n$ in this paper has a gap (see Section 5 here). In fact, when $p$ does not divide $n$, we have the stronger statement that the number of irreducible characters of degree $d$ is the same for both groups, for every $d$. However, we do not know the answer to the question in the title in general when $p$ divides $n$.

One can check that the answer is yes when $p=n=2$ (10 conjugacy classes) and for $p=n=3$ (127 conjugacy classes), using GAP (the $n=2$ case can also be done by hand), but for $n=4$, $p=2$ I don't know the answer, mainly because the only way I know to create the group over $\mathbb{F}_{p}[t]/t^{2}$ in GAP is via generators, and this seems to be very computationally inefficient.

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    $\begingroup$ No, this is not true for $\mathrm{SL}_2$ and $p=2$, modulo third powers. However, it may be true for all powers, as long as $p$ is sufficiently large. $\endgroup$
    – A Stasinski
    Commented Oct 14, 2018 at 10:13
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    $\begingroup$ I did a Magma calculation for the case $n=4$, $p=2$. Both groups have the structure $2^{1+14}.A_{8}$, and have $240$ conjugacy classes, but they are not isomorphic. $\endgroup$
    – Derek Holt
    Commented Oct 14, 2018 at 14:24
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    $\begingroup$ In the case $n=4,p=2$, the extraspecial normal subgroups are isomorphic. The groups are nonisomorphic because ${\rm SL}_4(F_2[t]/\langle t^2 \rangle)$ is a split extension $2^{1+14}:{\rm SL}(4,2)$ whereas ${\rm SL}_4(Z/4)$ is a nons[plit extension. (Note ${\rm SL}(4,2) \cong A_8$.) $\endgroup$
    – Derek Holt
    Commented Oct 14, 2018 at 14:47
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    $\begingroup$ @GeoffRobinson to do it by hand, one checks that in $SL_n(Z/p^2)$, the matrix $e_{12}(1)$ has no lift in $SL_n(Z/p^2)$ of order $p$. This argument works for all $p\neq 3$ (including $p=2$). For $p=3$ and $n\ge 3$, the matrix $(I+E_{12}+E_{23})$ has no lift of order 3 in $SL_n(Z/9)$. $\endgroup$
    – YCor
    Commented Oct 14, 2018 at 17:02
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    $\begingroup$ The answer to the question seems to be (part of) the main theorem in Singla arxiv.org/abs/1101.3696. I don't see any assumption that $p$ should not divide $n$ there, do I miss something? $\endgroup$
    – YCor
    Commented Oct 15, 2018 at 0:46

2 Answers 2

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Oh, I did not know about this ongoing discussion on math overflow. Amri pointed out to me about this discussion today morning only. As I discussed with you in a private communication, I don't know how to fix this at the moment. Moreover in this recent article of mine with M Hassain, we show that $SL_n$ for $p \mid n$ even for $n=2$ behaves pretty differently as compared to $GL_n$ (see Theorem 1.2). For example Corollary 1.3 of this article shows that the complex group algebras of $SL_2(Z/2^{2r} Z)$ are not isomorphic to $SL_2(F_2[t]/(t^{2r}) )$ for any $r > 1$. This is weaker than conjugacy class question for such groups, if one is interested in that, but still quite interesting given that corresponding group algebra of $GL_2$ are isomorphic.

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I would have preferred to not answer my own question, but here it goes. Yes, the two groups have the same number of conjugacy classes and in fact, the groups $\mathrm{SL}_{n}(W_{2}(\mathbb{F}_{q}))$ and $\mathrm{SL}_{n}(\mathbb{F}_{q}[t]/t^{2})$, for $q$ a power of a prime $p$ dividing $n$, have the same number of irreducible characters of dimension $d$ for every integer $d>1$. This is proved here (arXiv link here).

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