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I am solving some previous year's question paper of our college and found the following problem:

Let $p$ be a prime number. Let $G$ be a group of all $2 \times 2$ matrices over $Z_p$ with determinant $1$ under matrix multiplication. To find the order of $G$.

I am struck with the problem. Please Help!

User8976
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    What you are looking for is the order of the group $SL_2(\mathbb{Z}/p\mathbb{Z})$. This link gives a pretty good explanation. –  Aug 13 '15 at 18:41

3 Answers3

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Hint: look at two cases, according to whether the top left element is $0$ or not.

Robert Israel
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First select the elements of the main diagonal.

There $p-1$ ways to select them so that their product is $1$.

If the product is $1$ the product of the elements of the other diagonal must be $0$. There are $2p-1$ ways to select them so at least one of them is a multiple of $p$.

This gives us $(p-1)(2p-1)$ matrices.

There are $p^2-p+1$ selections in which the product of the diagonal is not $1$.

No matter what the product of the diagonal is, the product for the other diagonal shall be fixed, and will be non-zero. There are $p-1$ ways to choose them so that they give the selected product.

So we have $(p-1)(2p-1)+(p^2-p+1)(p-1)=(p-1)(p^2+p)=p^3-p$ such matrices.

Asinomás
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  • Did you somehow include the case where the main diagonal product is $0$ and the other diagonal product is $p-1$? I didn't see it but maybe it's folded into the second part of the argument. – user2566092 Aug 13 '15 at 18:47
  • yes, as long as the product of the main diagonal is not $1$ we will be forced to make the product of the other diagonal $a$, where $a$ will not be $0$(when the product of the main diagonal is $0$ we have $a= p-1$). There will always be $p-1$ ways to make the product of the other diagonal $a$. The case you are reffering to is when $a=p-1$. – Asinomás Aug 13 '15 at 18:51
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A general matrix here is $\begin{bmatrix} a&b \\ c&d \end{bmatrix}$ with $ad-bc=1$. If $a=0$ then $d$ can be anything and for any non-zero $b$, $c$ is $\frac{-1}{b}$. Hence there are $p$ choices for $d$, $p-1$ choices for $b$ and one choice for each $a$ and $c$, giving us $p(p-1)$ matrices. If $a\neq0$ then $d=\frac{1+bc}{a}$ and $b$ and $c$ can be anything giving us $(p-1)p^2$ such matrices for a total of $p(p-1)+(p-1)p^2=p^3-p$ matrices.

ZSMJ
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