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Let $G$ be the group of all $2\times 2$ matrices \begin{bmatrix}a&b\\c&d\end{bmatrix} where $a,b,c,d$ are integers modulo $p$ for $p$ prime such that $ad-bc\not =0$.$G$ forms a group relative to multiplication. Then how do I find the order of $G$?

I really don't know how to proceed. All I could try is look at $p=3$. So, I tried to count the number of solutions to $ad-bc=0 \bmod p$ manually by squaring and adding the number of solutions to $ad=0 \bmod p$, $ad=1 \bmod p$, $ad=2 \bmod p$.Let that sum be $k$. Then the order of $G$ is $3^4-k=81-k$. The same method however does not work with a general prime $p$. Would someone give me a hint to solve this problem? Thanks.

Batominovski
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  • The two rows have to be linearly independent, so the first row must be nonzero and you have $p^2-1$ choices for $(a,b)$ and then $(c,d)$ must not be a multiple of $(a,b)$ so you have $p^2-p$ choices. – Derek Holt Jul 28 '15 at 15:46
  • http://math.stackexchange.com/questions/1200622/group-of-all-2-times2-matrices-where-a-b-c-and-d-are-integers-modul?rq=1 http://math.stackexchange.com/questions/901654/order-of-group-gl-2-left-mathbbf-p-right http://math.stackexchange.com/questions/296047/order-of-general-linear-group-of-2-times-2-matrices-over-mathbbz-3?rq=1 ... – Matthew Towers Jul 28 '15 at 15:46

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Combinatorics Solution: It might be easier to find the number of solutions where $ad - bc = 0$. There are two cases to consider: $a = 0$ and $a \neq 0$. In the second case, you can pick $b,c$ arbitrarily, and then we are forced to have $d = bc/a$ (remember nonzero elements have inverses mod $p$).

Linear Algebra Solution: Think of the columns of the matrix as a basis of $\mathbb{F}_p^2$, ie as a basis of the vector space of 2-component column vectors whose entries are integers mod p. The first column can be chosen as any nonzero vector. Then the second column must be a vector which is not a scalar multiple of the first column. How many distinct scalar multiples of the first column are there?

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You said hint so neither of these are complete, but let me know if you need more details.

Alex Zorn
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