Let \(GL_2(\mathbb Z_m)\) be the multiplicative group of invertible \(2\times2\) matrices over the ring of integers modulo \(m\). Find \[ |GL_2(\mathbb Z_{p^n})| \] for every prime \(p\) and positive integer \(n\).
Solution
Put \[ R_n\coloneqq\ZZ/p^n\ZZ, \qquad k\coloneqq\FF_p, \] and let \[ \rho:GL_2(R_n)\longrightarrow GL_2(k) \] be reduction modulo \(p\).
1. The reduction map \(\rho\) is surjective.
Let \(A\in GL_2(k)\) and lift its four entries arbitrarily to a matrix \(\widetilde A\in M_2(R_n)\). Since \[ \det(\widetilde A)\bmod p=\det(A)\neq0, \] the element \(\det(\widetilde A)\in R_n\) is not divisible by \(p\), hence is a unit of \(R_n\). The adjugate formula then shows that \(\widetilde A\) is invertible over \(R_n\). Thus \(\rho(\widetilde A)=A\).
2. The kernel of \(\rho\) has order\[ \lvert\ker\rho\rvert=p^{4(n-1)}. \]
An element of \(\ker\rho\) is exactly a matrix of the form \[ I+X, \qquad X\in M_2(pR_n). \] Conversely, every such matrix is invertible because \[ \det(I+X)\equiv1\pmod p, \] so its determinant is a unit of \(R_n\). Therefore the map \[ M_2(pR_n)\longrightarrow\ker\rho, \qquad X\longmapsto I+X, \] is a bijection.
The ideal \(pR_n\) has \(p^{n-1}\) elements, including the case \(n=1\), when \(pR_1=\{0\}\). Hence \[ \lvert\ker\rho\rvert =\lvert pR_n\rvert^4 =p^{4(n-1)}. \]
3. The group \(GL_2(k)\) has order\[ \lvert GL_2(k)\rvert=(p^2-1)(p^2-p). \]
The first column of an invertible \(2\times2\) matrix over \(k\) can be any nonzero vector in \(k^2\), giving \(p^2-1\) choices. Once the first column is chosen, the second column can be any vector outside its one-dimensional span, giving \(p^2-p\) choices. These choices are exactly the ordered bases of \(k^2\).
4. For every prime \(p\) and positive integer \(n\),\[ \boxed{ \lvert GL_2(\ZZ/p^n\ZZ)\rvert =p^{4(n-1)}(p^2-1)(p^2-p) }. \]
By steps 1 and 2, reduction modulo \(p\) gives a short exact sequence \[ 1\longrightarrow\ker\rho \longrightarrow GL_2(R_n) \overset{\rho}{\longrightarrow}GL_2(k) \longrightarrow1. \] Taking orders and applying steps 2 and 3 gives the displayed formula.
5. Q.E.D.
Step 4 gives the required order.