The order of $H = \{X \in \mathrm{GL}_2(\mathbb{Z}_p) \mid \det(X) = 1\}$ with the definition

Question

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.

Answer 1

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 $|H|=p(p^2-1)$.