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Let $G = GL_2(\mathbb{Z}/p^2\mathbb{Z})$. Find $|G|$.

$G$ is the group of invertible $2 \times 2$ matrices with entries in the ring $\mathbb{Z}/p^2\mathbb{Z}$. In this case, I know invertibility is equivalent to the condition that the determinant of the chosen matrix is a unit of the ring $\mathbb{Z}/p^2\mathbb{Z}$. Now, $(\mathbb{Z}/p^2\mathbb{Z})^{\times} = \{\bar{a} : (a, p^2) = 1 \}.$ So, we know $a$ cannot be a multiple of $p$. My idea is to count the number of elements that do give a determinant a multiple of $p$, and then subtract this quantity from the total number of possible matrices, $p^8$. So we are looking for solutions of the equation $ad-bc \equiv 0(mod p),$ or $ad \equiv bc (mod p)$. I know for the field case, a common trick is to fix the first column and find conditions on the second column that give the desired result.

So fix $a = a_0$ and $c = c_0$ and consider $a_0d \equiv bc_0 (mod p).$ A class of solutions that jumps out at me is $b \equiv a_0 (mod p)$ and $d \equiv c_0 (mod p).$ But are these the only solutions? Is this even the right way to go about this problem?

Any help would be appreciated!

L. Tim
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  • Use $x\equiv y\pmod{p}$ for $x\equiv y\pmod{p}$. – Shaun Feb 20 '23 at 20:13
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    $|GL_2(\Bbb{F}_p)|= (p^2-1)(p^2-p)$ (choose a non-zero vector then another one not in the subgroup it generates) and $GL_2(\Bbb{Z}/(p^2))\to GL_2(\Bbb{F}_p)$ is surjective with kernel the matrices $\equiv I\bmod p$ so the kernel has $p^4$ elements. – reuns Feb 20 '23 at 20:34
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    Hint: [1] Reduce all the entries $\pmod p$ and you get a matrix in $M_2(\mathbb Z/p\mathbb Z)$. [2] Prove that the reduced matrix $\pmod p$ is invertible if and only if the original matrix is invertible. [3] How many elements are there in $GL_2(\mathbb Z/p\mathbb Z)$? [4] "Lift" that number to the number of invertible matrices in $GL_2(\mathbb Z/p^2\mathbb Z)$ –  Feb 20 '23 at 20:34
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    See also this post and this post. – Dietrich Burde Feb 20 '23 at 20:35
  • @reuns What's the surjective homomorphism $GL_2(\Bbb{Z}/(p^2))\to GL_2(\Bbb{F}_p)$? Is it like this? $$\pmatrix{a+\Bbb{Z}/(p^2) & b+\Bbb{Z}/(p^2)\ c+\Bbb{Z}/(p^2)&d+\Bbb{Z}/(p^2)}\mapsto \pmatrix{a+\Bbb{Z}/(p) & b+\Bbb{Z}/(p)\ c+\Bbb{Z}/(p)&d+\Bbb{Z}/(p)}$$ – Nothing special Oct 08 '24 at 19:47

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