Let $M$ belongs to $GL(2,p)$ where $p$ is a prime number, and $\det M$ generate $GL(1,p)$, so I want to prove that the order of $M$ is coprime to $p$.
I think if $M^{np}=I_2$ that means $M^n=I_2$ but how to do next?
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1@user Not at all ; for example we might have ${\sf det}(M)=(-1)$. – Ewan Delanoy Nov 10 '13 at 08:33
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2$p=2$, $M=\pmatrix{0&1\cr1&0\cr}$ seems to be a counterexample. – Gerry Myerson Nov 10 '13 at 08:43
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2It seems that the result is true when $p$ is odd. – Ewan Delanoy Nov 10 '13 at 08:52
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If you look at the answer to this question can see that the only case to study is $M=\left(\begin{array}{cc} a & 1 \\ 0 & a \end{array}\right).$ (In the other two cases the order of matrix divides $p-1$, respectively $p^2-1$, and therefore is coprime with $p$.) In this case $a^2$ generates $\mathbb F_p^{\times}$, that is, the order of $a^2$ in $\mathbb F_p^{\times}$ is $p-1$. But this is not possible for $p\ge 3$: if the order of $a$ is $m$, then $m\mid p-1$ and the order of $a^2$ is $m/\gcd(2,m)$ which is less than $p-1$.
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@GerryMyerson In my answer it's tacitly assumed that $p\ge 3$ as long as you settled the question into the negative for $p=2$. (However, I can't see which part of Arturo Magidin argument can fail for $p=2$.) – Nov 10 '13 at 14:16
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My mistake, I think. The matrix $M$ in my comment on the question is in fact of the repeated-eigenvalue, non-diagonalizable form, so the parenthetical statement does not apply to it. – Gerry Myerson Nov 10 '13 at 22:43