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Let be a prime $p$ and $H = \{X \in \mathrm{GL}_2(\mathbb{Z}_p) \mid \det(X) = 1\}$. I know that the order of a element $g \in \mathrm{GL}_2(\mathbb{Z}_p)$ is the less $k$ such that $g^k = e$, but I don't understand how to use this to find the order of H.

Wellington Silva
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    That's two different meanings of the term "order." This is asking for $|H|$. Apply the first isomorphism theorem to the determinant map. – Randall Dec 17 '23 at 22:18

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The first isomorphism theorem states that if you have a group morphism $f:G\longrightarrow G'$ then it induces an isomorphism $G/{\rm Ker}(f)\overset{\simeq}{\longrightarrow}{\rm Im}(f)$. Applying this to $\det:{\rm GL}_2(\mathbf{F}_p)\longrightarrow\mathbf{F}_p^*$ gives you an isomorphism ${\rm GL}_2(\mathbf{F}_p)/H\overset{\simeq}{\longrightarrow}\mathbf{F}_p^*$ because $H={\rm Ker}(\det:{\rm GL}_2(\mathbf{F}_p)\longrightarrow\mathbf{F}_p^*)$ and $\det$ is surjective. This means that $$ p-1=|\mathbf{F}_p^*|=|{\rm GL}_2(\mathbf{F}_p)/H|=\frac{|{\rm GL}_2(\mathbf{F}_p)|}{|H|}=\frac{(p^2-1)(p^2-p)}{|H|} $$ therefore $|H|=p(p^2-1)$.

Tuvasbien
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  • Why is $|\text{GL}_2(\mathbf F_p)|=(p^2-1)(p^2-p)$? – Nothing special Dec 17 '23 at 23:31
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    @Nothingspecial There are $p^2-1$ possibilities for the first row (you can chose any non-zero vector in $\mathbf{F}_p^2$) and there are $p^2-p$ possibilities for the second row because you have to chose it in $\mathbf{F}_p^2\setminus\mathbf{F}_pv$ where $v$ is the first row. – Tuvasbien Dec 17 '23 at 23:33
  • Makes sense..! However, is there an explicit formula for $|\text{GL}_n(\mathbf F_p)|$ in general? – Nothing special Dec 17 '23 at 23:35
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    Yes, the same reasoning gives $$ |{\rm GL}n(\mathbf{F}_p)|=\prod{k=0}^{n-1}(p^n-p^k)=(p^n-1)(p^n-p)\cdots (p^n-p^{n-1}). $$ – Tuvasbien Dec 17 '23 at 23:36
  • I learned from this post that this formula works for any finite field... (Just we replace $p$ by $p^i$...) I have never had so much realization from a simple comment I made... Thank you for helping me with my curiosity! – Nothing special Dec 20 '23 at 22:50