= Solution
Invertible <matrices> are closed under multiplication, contain $I$, have associative multiplication, and have inverses by the adjugate formula over the field $\mathbb F_p$. The first column can be any nonzero <vector> and the second any <vector> outside its span, so
$$
|GL_2(\mathbb F_p)|=(p^2-1)(p^2-p).
$$
For $p=2$, the <matrix> $\begin{pmatrix}0&1\\1&1\end{pmatrix}$ has order three. There is no element of order six: $GL_2(\mathbb F_2)\cong S_3$, whose element orders are $1,2,3$.
For $p>2$, $SL_2(\mathbb F_p)=\ker\det$ is a proper normal <subgroup>. For $p=2$, the order-three <subgroup> in the copy of $S_3$ is proper and normal.
In $GL_2(\mathbb F_{11})$, take
$$
H=\left\{\begin{pmatrix}a&b\\0&1\end{pmatrix}:a\in A, b\in\mathbb F_{11}\right\},
$$
where $A\le\mathbb F_{11}^{\times}$ is the <subgroup> of order five. This <semidirect product> has order $55$ and is nonabelian because a nontrivial diagonal element does not commute with translations.
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