Let $ p $ be prime and let $ P_n(p) $ denote the Pauli group on $ n $ qudits each of size $ p $. Then $ P_n(p) $ and $ \text{Heis}_{2n+1}(\mathbb{F}_p) $ are both extraspecial $ p $ groups of order $ p^{2n+1} $. Are they isomorphic?
I think the answer is obviously yes using the symplectic representation of the Pauli group, but I don't see this fact explicitly stated anywhere so I was wondering if I'm missing something. Basically an arbitrary element of $ \text{Heis}_{2n+1}(\mathbb{F}_p) $ is given by $ (v,\zeta^i,w) $ where $ v,w \in \mathbb{F}_p^n $ and $ \zeta $ is a central element of order $ p $ (in the standard representation $ \zeta $ is just a primitive $ p $th root of unity). Then the bijection $ \text{Heis}_{2n+1}(\mathbb{F}_p) \to P_n(p) $ $$ (v,\zeta^i,w) \mapsto \zeta^i X^vZ^w $$ should be an isomorphism. Where by $ X^v $ we mean for example $ X^{(1,0,1)}=X_1 \otimes I \otimes X_3 $.
Note that here I make the somewhat unusual choice of excluding $ i $ from the qubit Pauli group (in other words generating it purely from $ X,Z $ type operators). See Definition of the Pauli group and the Clifford group