According to Wikipedia, the Heisenberg group modulo $p$, where $p$ is an odd prime, has the presentation $$H(\mathbb{F}_p)=\langle x,y,z\mid x^p=y^p=z^p=1, \ xz=zx, \ yz=zy, \ z=xyx^{-1}y^{-1}\rangle.$$ I could even derive it, but the proof seems to work modulo any integer, not just an odd prime. Why should $p$ be an odd prime? (If it works modulo any integer, it seems a little strange that the Wikipedia article insists on $p$ being an odd prime.)
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The Heisenberg group can be defined over any commutative unital ring. It is the group of all uni-upper- triangular matrices over that ring. If the ring is the ring of integers modulo some $n$ you get your group. Traditionally, though, people assume $n=p$ to be prime. Then the group is one of two nonabelian groups of order $p^3$.
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does it satisfy the same generator/relation presentation though when $p$ is not prime? – hunter Jul 20 '20 at 00:41
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Yes, if rhe ring is the ring of integers mod $n$. For an arbitrary ring it may be not finitetely generated. Usual cases are ring of all integers or a field. If it is the field of real numbers you get the Lie group Nil. – markvs Jul 20 '20 at 00:44