1. ## Matrix Groups

Any help with the following problem is greatly appreciated:

For the part (a) of this problem, I must show that $M_{a,p}$ belongs to the general linear group of 2x2 matrices over p (a prime) iff $a \bmod\ p \neq p-1$.

I know from a definition that $GL(2,F)= \{ A\in Mat_{2} (F) | det(A) \neq 0 \in F \}$.

So here we must have $det \begin{bmatrix}1 & (a \bmod\ p)\\p-1 & 1\end{bmatrix}$ $= (a \bmod\ p) (p-1) \neq 0$

I'm kind of stuck here, what can I do next?

2. First, you have calculated the derivative incorrectly- you forgot the main diagonal.
$\left|\begin{array}{cc}1 & a (mod p) \\ p-1 & 1\end{array}\right|= 1- (p-1)[a (mod p)]$.

That will be 0 if and only if (p-1)[a (mod p)]= 1.

3. Thanks, that was a typo. So, we must have that $1- (p-1)(a \bmod\ p) \neq 0 \iff (p-1)(a \bmod\ p) \neq 1$. So I guess $(p-1)$ must be equal to $-1$ in $\mathbb{Z}_p$. Therefore $p-1 = -1 \neq a \bmod\ p$.

But how do I prove that (p-1)=-1?

And for part (b) I think they are asking for the inverse of

$M_{10,p} = \left|\begin{array}{cc}1 & (10 \bmod\ p) \\ p-1 & 1\end{array}\right|$

Then the inverse would be of the form:

$\frac{1}{det(M_{10,p})} \left|\begin{array}{cc}1 & -(10 \bmod\ p) \\ 1-p & 1\end{array}\right|$

But what values do I use for $det(M_{10,p})$ and 10 mod p?

4. Originally Posted by demode
Thanks, that was a typo. So, we must have that $1- (p-1)(a \bmod\ p) \neq 0 \iff (p-1)(a \bmod\ p) \neq 1$. So I guess $(p-1)$ must be equal to $-1$ in $\mathbb{Z}_p$. Therefore $p-1 = -1 \neq a \bmod\ p$.

But how do I prove that (p-1)=-1?

!!!... As $p=0\!\!\pmod p$ , obviously $p-1=0-1=-1\!\!\pmod p$ ...!

Tonio

And for part (b) I think they are asking for the inverse of

$M_{10,p} = \left|\begin{array}{cc}1 & (10 \bmod\ p) \\ p-1 & 1\end{array}\right|$

Then the inverse would be of the form:

$\frac{1}{det(M_{10,p})} \left|\begin{array}{cc}1 & -(10 \bmod\ p) \\ 1-p & 1\end{array}\right|$

But what values do I use for $det(M_{10,p})$ and 10 mod p?
.

5. Thank you. And for part (b) I know the inverse would be of the form

$\frac{1}{1-(p-1)(10 \bmod\ p)} \begin{bmatrix} 1&-(a \bmod\ p)\\ -(p-1)& 1 \end{bmatrix}$

Right? But what is $10 \bmod\ p$ when $p \in \{ 3,7 \}$? 7 or 3? Also I need to know (p-1). (I must know the value of the determinant in order to find its inverse).