Find subgroup of $GL(n,\mathbb{Z}_{19})$ with index 3
My try:
I thought about looking at $\det:GL(n,\mathbb{Z}_{19}) \to\mathbb{Z}_{19}^*$ and finding a subgroup of index 3 of $\mathbb{Z}_{19}^*$.
I know that $[\mathbb{Z}_{19}^*:H]=[GL(n,\mathbb{Z}_{19}):\det^{-1}(H)]$ given that $H$ is a normal subgroup.
So, I need to find a normal $H<\mathbb{Z}_{19}^*$ such that $[\mathbb{Z}_{19}^*:H] = 3$
How can I do so?
It is a well known fact, that for a prime $p$, $\mathbb{Z}_p^*$ is cyclic of order $p-1$. In your case it is $\langle 2 \rangle_{18}$. Thus, the subgroup of $\mathbb{Z}^*$ that you need, is generated by $2^3 = 8$. And thus, the subgroup, you were initially looking for is $\{A \in GL(n, 19)| \exists n \in \mathbb{N}, det(A) = 8^n \}$