Let $G_r=\mathrm{GL}_n(\mathbb{Z}/p^r)$. For $x\in G_r$ the centraliser $C_{G_{r}}(x)$ is abelian iff $x$ is regular iff the reduction mod $p$ of $x$ is regular. This is due to G. Hill, Regular elements and regular characters of $\mathrm{GL}_n(\mathcal{O})$, J. Algebra 174 (1995), no. 2, 610–635. The case $r=1$ is easy and was known earlier.

Let $G_r=\mathrm{GL}_n(\mathbb{Z}/p^r)$. For $x\in G_r$ the centraliser $C_{G_{r}}(x)$ is abelian iff $x$ is regular iff the reduction mod $p$ of $x$ is regular. This is due to G. Hill, Regular elements and regular characters of $\mathrm{GL}_n(\mathcal{O})$, J. Algebra 174 (1995), no. 2, 610–635. The case $r=1$ is easy and was known earlier.

To add some details, note that Hill's Theorem 3.6 holds when the residue field is $\overline{\mathbb{F}}_p$, but I think his proof goes through for any algebraically closed residue field $\overline{k}$. If $k$ is any field and $C_{\mathrm{GL}_n(k)}(A)$ is abelian for $A\in \mathrm{GL}_n(k)$, then using for example the rational canonical form one can see that $A$ must be $\mathrm{GL}_n(k)$-conjugate to a companion matrix, and so $k^n$ is a cyclic $k[A]$-module. Extending scalars to $\overline{k}$ we get that $A$ is regular as an element of $\mathrm{GL}_n(\overline{k})$.

Let $G_r=\mathrm{GL}_n(\mathbb{Z}/p^r)$. For $x\in G_r$ the centraliser $C_{G_{r}}(x)$ is abelian iff x is regular iff the reduction mod $p$ of x is regular. This is due to G. Hill, Regular elements and regular characters of $\mathrm{GL}_n(\mathcal{O})$, J. Algebra 174 (1995), no. 2, 610–635. The case $r=1$ is easy and was known earlier.

Let $G_r=\mathrm{GL}_n(\mathbb{Z}/p^r)$. For $x\in G_r$ the centraliser $C_{G_{r}}(x)$ is abelian iff $x$ is regular iff the reduction mod $p$ of $x$ is regular. This is due to G. Hill, Regular elements and regular characters of $\mathrm{GL}_n(\mathcal{O})$, J. Algebra 174 (1995), no. 2, 610–635. The case $r=1$ is easy and was known earlier.