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We begin by the following definition.

Definition: Let $d \in \mathbb{N}$. The $(2d + 1)$-dimensional Heisenberg group over the ring of $p$-adic integers $\mathbb{Z}_p$, denoted by $\mathbb{H}_d(\mathbb{Z}_p)$, is given by $$\mathbb{H}_d(\mathbb{Z}_p) = \left\{ \begin{bmatrix} 1 & x^t & z \\ 0 & I_d & y \\ 0 & 0 & 1 \end{bmatrix} \in \mathrm{GL}_{d+2}(\mathbb{Z}_p) : x, y \in \mathbb{Z}_p^d, \, z \in \mathbb{Z}_p \right\},$$ where $I_d$ is the $d \times d$ identity matrix, and $x^t$ denotes the transpose of the vector $x$.

For each $n \in \mathbb{N}$, we define the subgroup $\mathbb{H}_d(p^n\mathbb{Z}_p)$ of $\mathbb{H}_d(\mathbb{Z}_p)$ by $$\mathbb{H}_d(p^n\mathbb{Z}_p) = \left\{ \begin{bmatrix} 1 & x^t & z \\ 0 & I_d & y \\ 0 & 0 & 1 \end{bmatrix} \in \mathrm{GL}_{d+2}(\mathbb{Z}_p) : x, y \in (p^n \mathbb{Z}_p)^d, \, z \in p^n \mathbb{Z}_p \right\}.$$

Then my question is the following:

Question: How can one show that $\mathbb{H}_d(\mathbb{Z}_p)/{\mathbb{H}_d({p^n\mathbb{Z}_p})}$ is isomorphic to $\mathbb{H}_d(\mathbb{F}_{p^n})$ as groups, where $\mathbb{F}_{p^n}$ is the finite field of order $p^n$?

The above isomorphism mentioned in the question is mentioned without any proof in a recent preprint by J.P. Velasquez-Rodriguez (section 3.1 on page 9). To me it seems like the natural projection map $\pi:\mathbb{H}_d(\mathbb{Z}_p)\rightarrow \mathbb{H}_d\left(\mathbb{Z}_p/{p^n\mathbb{Z}_p}\right) = \mathbb{H}_d\left(\mathbb{Z}/{p^n\mathbb{Z}}\right)$ induces an isomorphism between $\mathbb{H}_d(\mathbb{Z}_p)/{\mathbb{H}_d({p^n\mathbb{Z}_p})}$ and $\mathbb{H}_d\left(\mathbb{Z}/{p^n\mathbb{Z}}\right)$, but I am not sure whether I am wrong or how to settle the above-mentioned problem. Thanks in advance for any relevant help or suggestion in this regard.

Martin Brandenburg
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Surajit
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1 Answers1

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This is obviously a mistake due to confusing the rings $\mathbf Z/p^n\mathbf Z$ and $\mathbf F_{p^n}$, which are not isomorphic when $n \geq 2$.

To "settle" the issue, just look at the explicit case $d = 1$, where $$ H_1(R) = \left\{\begin{pmatrix}1&x&z\\0&1&y\\0&0&1\end{pmatrix} : x, y, z \in R\right\} $$ for any commutative ring $R$. Since $$ \begin{pmatrix}1&x&z\\0&1&y\\0&0&1\end{pmatrix}^n = \begin{pmatrix}1&nx&\binom{n}{2}xy + nz\\0&1&ny\\0&0&1\end{pmatrix} $$ for all integers $n$, when $R$ has prime characteristic $p$ where $p$ is odd we have $$ \begin{pmatrix}1&x&z\\0&1&y\\0&0&1\end{pmatrix}^p = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix} $$ since $\binom{p}{2} \equiv 0 \bmod p$ when $p \not= 2$. Thus all nontrivial elements of $H_1(\mathbf F_{p^n})$ have order $p$. On the other hand, most nontrivial elements in $H_1(\mathbf Z/p^n\mathbf Z)$ when $n \geq 2$ do not have order $p$, e.g.,
$$ \begin{pmatrix}1&1&1\\0&1&1\\0&0&1\end{pmatrix}^p = \begin{pmatrix}1&p&\binom{p}{2} + p\\0&1&p\\0&0&1\end{pmatrix}, $$ which is not the identity in $H_1(\mathbf Z/p^n\mathbf Z)$ when $n \geq 2$.

When $p = 2$ and $n \geq 2$, instead of using the $p$th power use the $p^2$-power: all elements of $H_1(\mathbf F_{2^n})$ have order dividing $4$, while there are nontrivial elements of $H_1(\mathbf Z/2^n\mathbf Z)$ that don't have order dividing $4$, e.g., $$ \begin{pmatrix}1&1&1\\0&1&1\\0&0&1\end{pmatrix}^4 = \begin{pmatrix}1&4&10\\0&1&4\\0&0&1\end{pmatrix} $$

KCd
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