Let $GL_2(Z_m)$ denote the multiplicative group of invertible $2 * 2$ matrices over the ring of integers modulo m. Find the order of $GL_2(Z_{p^{n}})$ for each prime ${p}$ and positive integer ${n}$.
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1Note that $\mathbb{Z}_{p^n}$ is not a field for $n>1$. It has zero divisors. For the case of $GL_2(F)$ of a finite field $F$, see here. – Dietrich Burde Jan 31 '15 at 19:29
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We reduce this problem to the one over $\mathbb{Z}_p$, and solve it for any dimension using the result mentioned in a comment.
Let $M = M_k(\mathbb{Z}_{p^{n}})$ be the set of all $k\times k$ matrices over $\mathbb{Z}_{p^{n}}$, and we need to find those for which $\det (A)$ is invertible, that is non-zero modulo $p$. Let $pM$ the submodule obtained by scalar multiplication by $p$. Then $M/pM \cong M_k(\mathbb{Z}_{p})$.
Moreover, $\det (A) \equiv \det (B) \mod p$ for $A \equiv B \mod pM$, for instance since $\det$ is a polynomial map in the entries of the matrix. Thus $\det(\cdot) \mod p$ is well-defined on $M/pM $, and it is preserved by the isomorphism.
Consequetly $$|GL_k(\mathbb{Z}_{p^n})| = |GL_k(\mathbb{Z}_{p})| \ |p M | $$ Since $pM = M_k (p \mathbb{Z}_{p^n}) \cong M_k ( \mathbb{Z}_{p^{n-1}})$ we have $|pM|= (p^{n-1})^{k^2}$, while $|GL_k(\mathbb{Z}_{p})|= \prod_{j=1}^k\left(p^k-p^{j-1}\right)$ (see for example Order of $\mathrm{GL}_n(\mathbb F_p)$ for $p$ prime as mentioned in a comment by Dietrich Burde).
Yielding $$|GL_k(\mathbb{Z}_{p^n})| = (p^{n-1})^{k^2}\prod_{j=1}^k\left(p^k-p^{j-1}\right).$$
For $k = 2$, this yields $p^{4n-4} (p^2 -1)(p^2 - p)$.
