Let $\mathbb D_6=\{e,a,a^2,b,ab,a^2b : a^3=b^2=e\,\, \text{and}\,\, ba=a^2b\}$ and $GL =\{A \in{\rm Mat}_2(\mathbb{Z}_2) : A\text{ is invertible} \}$
Prove that $\mathbb D_6$ and $GL$ are isomorphic (group).
My attempt: To prove that $\mathbb D_6$ and $GL$ are isomorphic we need to find a function $f$ such that is a bijection and $f(ab)=f(a)f(b)$ for all $a,b\in GL$
Note that $GL = \{ \begin{bmatrix}1&0\\0&1\end{bmatrix}, \begin{bmatrix}0&1\\1&0\end{bmatrix}, \begin{bmatrix}1&1\\1&0\end{bmatrix}, \begin{bmatrix}1&1\\0&1\end{bmatrix}, \begin{bmatrix}1&0\\1&1\end{bmatrix}, \begin{bmatrix}0&1\\1&1\end{bmatrix}\}$
Let $f:GL\rightarrow\mathbb{D}_6$ such that
$\begin{bmatrix}1&0\\0&1\end{bmatrix} \longrightarrow e$
$\begin{bmatrix}1&1\\1&0\end{bmatrix} \longrightarrow a$
$\begin{bmatrix}0&1\\1&1\end{bmatrix} \longrightarrow a^2$
$\begin{bmatrix}0&1\\1&0\end{bmatrix} \longrightarrow b$
$\begin{bmatrix}1&1\\0&1\end{bmatrix} \longrightarrow ab$
$\begin{bmatrix}1&0\\1&1\end{bmatrix} \longrightarrow a^2b$
Note that by construction, $f$ is bijective. How can i prove that $f(ab)=f(a)f(b)$? Not in the manual way, as I do not want to make $6!$ combinations.