An isomorphism from $GL_{2}(\mathbb{F_{2}})$ to $S_{3}$

Well I want to prove that $GL_{2}(\mathbb{F_{2}}) \simeq S_{3}$ then I need to find a bijection between them, so my attempt for the $\Leftarrow]$ part is to consider the permutation matrix, I will Ilustrate this with an example:

$\sigma = (1 \to3),(2\to1),(3\to2)$ and its permutation matrix is:

$$ \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right)$$

Then the pattern is to write a 1 in the place we are mapping our cycle.

But I dont know How to proceed with the $\Rightarrow]$ part, Can someone helpme to finish this bijection please? Thanks a lot in advance.

Note:

$\mathbb{F_{2}}$ is thefield of the integers mod two the consisting of {0,1} :)

THE ABOVE ATTEMPT IS WRONG THEN I NEED YOUR HELP IN BOTH SIDES OF THE ISOMORPHISM THANKS A LOT ;)

Solution 1:

Regard the group as $GL(V)$ where $V$ is a 2-dim'l vector space over $\mathbf{F}_2$. This vector space has 4 vectors, call them $\{0,u,v,w\}$. Any two non-zero vectors here are linearly independent. Check that $B_1=\{u,v\},\ B_2=\{u,w\},\ B_3=\{v,w\}$ are all the bases for $V$. A linear automorphsm is the same as one that carries one basis to another, so in this case it is the same as the permutation group on the collection of bases B${}=\{B_1,B_2,B_3\}$ and so it is $S_3$.

Solution 2:

Hints: Note that ${\rm GL}(2,\mathbb{F}_{2})$ has order $(2^{2}-1)(2^{2}-2) = 6.$ Note that the cyclic subgroup $\langle \left( \begin{array}{clcr} 0&1\\1&1 \end{array} \right) \rangle$ has order $3$, so has index $2$ and is normal. Note also that $\langle (123) \rangle$ is a normal subgroup of order $3$ in $S_{3}.$

Solution 3:

If you know that $|GL_2(\Bbb F_2)| = 6$, it suffices to show it is non-abelian, since there are precisely two isomorphism classes of groups of order $6$, groups isomorphic to $\Bbb Z_6$, and groups isomorphic to $S_3$ (It is instructive to prove this at least once in one's lifetime).

Alternatively, take any element of order $3$ in $GL_2(\Bbb F_2)$ and any element of order $2$, call them $A$ and $B$, respectively, and show $BA = A^2B$.

Note it is not sufficient to exhibit a subgroup of order $3$, you also need a non-normal subgroup of order $2$ (since $\Bbb Z_6$ has a normal subgroup of order $3$, namely $\{0,2,4\}$).

Finally, we can have: $\begin{bmatrix}1&1\\1&0\end{bmatrix} \mapsto (1\ 2\ 3)$, $\begin{bmatrix}0&1\\1&0\end{bmatrix} \mapsto (1\ 2),\ \begin{bmatrix}0&1\\1&1\end{bmatrix} \mapsto (1\ 3\ 2)$,

$\begin{bmatrix}1&1\\0&1\end{bmatrix} \mapsto (1\ 3), \begin{bmatrix}1&0\\1&1\end{bmatrix} \mapsto (2\ 3)$, $I \mapsto e$

Showing this is indeed a homomorphism is tedious, however.