3

How many invertible $2\times2$ matrices over $\mathbb{Z}_{26}:=\mathbb{Z}/26\mathbb{Z}$ are there?

Writing $\begin{pmatrix}a & b \\ c & d\end{pmatrix},$ then both of the relations $2\nmid ad-bc$ and $13\nmid ad-bc$ must hold. It yields that $a$ and $b$ cannot be simultaneously divisible by $2$ nor by $13$ – we get that there are $26^2-13^2-2^2+1$ possible pairs $(a,b).$

But what about the bottom row? I know there are $(13^2-13)(2^2-2)$ possible pairs, but I cannot find any combinatorial reasoning (of course the number may be written in other forms as well, like $26^2-2\cdot13^2-13\cdot2^2+26$ or $26^2-26\cdot13-26\cdot2+26$ etc., but none of them helped me to get "a combinatorial idea behind").

Remark: I know about this topic, from it I would get the number of the matrices immediately, but it requires some uneasy theory. On the other hand, using PIE I can get one factor/part of the number immediately. That is why I believe there may be "easier", more combinatorial, solution.

Olexandr Konovalov
  • 7,002
  • 2
  • 34
  • 72
byk7
  • 865
  • 1
    By the Chinese remainder theorem we have $GL_2(\mathbb{Z}{26}) \cong GL_2(\mathbb{Z}_2) \times GL_2(\mathbb{Z}{13})$. Do you know how to compute the size of $GL_2(\mathbb{Z}_p)$ for $p$ a prime? – Qiaochu Yuan Oct 10 '18 at 01:29
  • @QiaochuYuan I know, it is $(p^2-1)(p^2-p)$. But how exactly we the isomorphism from the CRT? – byk7 Oct 10 '18 at 01:37
  • CRT gives an isomorphism $\mathbb{Z}{pq} \cong \mathbb{Z}_p \times \mathbb{Z}_q$ of rings, for $p, q$ distinct primes (of course the argument is more general than this). This extends to an isomorphism $M_2(\mathbb{Z}{pq}) \cong M_2(\mathbb{Z}_p) \times M_2(\mathbb{Z}_q)$ of rings, where $M_2$ means the ring of $2 \times 2$ matrices. Now take the group of units of both sides. – Qiaochu Yuan Oct 10 '18 at 02:52
  • I see. But the thing that I haven't still got is how is the isomorphism extended... – byk7 Oct 10 '18 at 09:34
  • You just need to check that $M_2(R \times S) \cong M_2(R) \times M_2(S)$ for $R, S$ two rings, in addition to using CRT. – Qiaochu Yuan Oct 10 '18 at 17:28
  • Consider $\varphi:M_2(R\times S)\to M_2(R)\times M_2(S),\ (a_{ij})\mapsto(\pi_R(a_{ij}),\pi_S(a_{ij}))=\bigl((r_{ij}),(s_{ij})\bigr),$ where $a_{ij}=(r_{ij},s_{ij}).$ Clearly it is a surjective homomorphism and $\ker\varphi=\pi^{-1}R(\boldsymbol{0}_R)\cap\pi^{-1}_S(\boldsymbol{0}_S)={\boldsymbol{0}{R\times S}},$ where $\boldsymbol{0}$'s are the corresponding zero matrices. So $\varphi$ is injective. Correct? – byk7 Oct 10 '18 at 21:51

0 Answers0