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Suppose $R_1,R_2$ are finite unital commutative rings. Consider Heisenberg groups $H_3(R_1)$ and $H_3(R_2)$ (upper unitriangular marticies $3 \times 3$).

Proposition. If $R_1 \not\cong R_2$ (as rings) then $H_3(R_1) \not\cong H_3(R_2)$ (as multiplicative groups).

Is that true? It seem that if $R_1$ and $R_2$ have not isomorphic additive groups then $H_3(R_1) \not\cong H_3(R_2)$, since they have not isomorphic centers. But what about general case, or is that too broad?

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Theorem 1.13 of this paper (The model theory of unitriangular groups, Ann. Pure Appl. Logic 68(3), 1994, 225-261) by O. Belegradek says that the answer is positive even for infinite commutative rings.

However, Proposition 1.9 in the same paper asserts that this does not extend to the non-commutative (associative unital case). The counterexamples have the form $R_1=K\times K$, $R_2=K\times K^{\mathrm{op}}$, where $K$ is indecomposable and not isomorphic to $K^{\mathrm{op}}$.

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  • $\begingroup$ Actually before I fixed the tags, the only "ring" tag was "commutative rings", so the question was quite clearly for commutative rings (which should have been more explicit). I added the tag "ra-..." because Lie algebras are lurking behind... $\endgroup$ – YCor Dec 1 '16 at 23:00
  • $\begingroup$ @YCor, I never read tags. Anyway I give the answer in that case too. $\endgroup$ – Benjamin Steinberg Dec 1 '16 at 23:02
  • $\begingroup$ Yes I know: this was no criticism but rather an excuse, since editing the tags, I sort of hid the only hint that the question was only in the commutative case. But of course it's great your answer both: just I would rather formulate it now saying that the answer is positive but not its extension to the non-commutative case. $\endgroup$ – YCor Dec 1 '16 at 23:35
  • $\begingroup$ Probably not worth editing. Also finite is not important. $\endgroup$ – Benjamin Steinberg Dec 2 '16 at 2:07
  • $\begingroup$ I did it anyway :) btw you probably have in mind some finite ring not anti-isomorphic to itself? $\endgroup$ – YCor Dec 2 '16 at 2:12

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