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

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.

• 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. – A Stasinski Oct 14 '18 at 10:13
• 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. – Derek Holt Oct 14 '18 at 14:24
• 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 ${\rm SL}(4,2) \cong A_8$.) – Derek Holt Oct 14 '18 at 14:47
• @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 $SL_n(Z/9)$. – YCor Oct 14 '18 at 17:02
• 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? – YCor Oct 15 '18 at 0:46

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.