TITLE: Prove $GL_2(\mathbb{Z}/2\mathbb{Z})$ is isomorphic to $S_3$
QUESTION [9 upvotes]: I'm asked to show that $G=GL_2(\Bbb Z/2\Bbb Z)$ is isomorphic to $S_3$. I have few ideas but I don't manage to put them all together in order to obtain a satisying answer.
I first tried using Cayley's theorem ($G$ is isomorphic to a subgroup of $S_6$), and I also noticed that $\operatorname{Card}(G)=\operatorname{Card}(S3)=6$ & that they're both non-abelian group.
Is this enough to say that considering $S_3$ is a subgroup of $S_6$ with the same cardinality than $G$, it has to be isomorphic to it ? 
Could anyone give me some elements to get a more rigorous proof or lead me to an other path to show this statement ? 
Thanks in advance

REPLY [15 votes]: If you don't know that the unique non-abelian group of order $6$ is $S_3$, you can go with a more explicit approach:
Find an isomorphism by showing that $GL_2(\mathbb Z/2\mathbb Z)$ is a group of permutations of a $3$-element set, the nonzero vectors in $\mathbb (\mathbb Z/2\mathbb Z)^2$. Number these vectors, then argue that every element of $GL_2$ induces a permutation of this set, which gives you a homomorphism between the two groups by sending each element of $GL_2$ to the corresponding permutation in $S_3$. Then show that this homomorphism is injective: two different elements of $GL_2$ permute the elements differently. Finally, count the number of elements in each group. Since the groups have the same order, your injective homomorphism is also surjective, and thus an isomorphism.