# Non-isomorphic Heisenberg groups over rings

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?

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 $K^{\mathrm{op}}$.

• 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...
– YCor
Dec 1 '16 at 23:00
• @YCor, I never read tags. Anyway I give the answer in that case too. Dec 1 '16 at 23:02
• 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.
– YCor
Dec 1 '16 at 23:35
• Probably not worth editing. Also finite is not important. Dec 2 '16 at 2:07
• I did it anyway :) btw you probably have in mind some finite ring not anti-isomorphic to itself?
– YCor
Dec 2 '16 at 2:12