Prove $GL_2(\mathbb{Z}/2\mathbb{Z})$ is isomorphic to $S_3$ 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
 A: 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.
A: You can do this in 2 steps :


*

*$GL_2(\mathbb{Z}/2\mathbb{Z})$ has order 6, and is non-abelian.

*Any non-abelian group of order 6 is isomorphic to $S_3$


To prove 2 : Let $G$ be any non-abelian group of order 6, then by Cauchy's theorem, there is a subgroup $H<G$ of order 2 and a subgroup $K$ of order 3. Since $[G:K] = 2$, $K$ is normal in $G$. If $H$ were normal in $G$, then
(a) $G = HK$
(b) $H\cap K = \{e\}$
(c) For all $h\in H, k\in K, hkh^{-1}k^{-1} \in H\cap K = \{e\}$
From (a), (b) and (c), you could conclude that $G$ is abelian, which it is not.
Hence, $H$ is not normal in $G$. Now let $G$ act on the set of left cosets of $H$ in $G$. This would give a homomorphism
$$
f: G\to S_3
$$
Now check that $\ker(f) < H$, and since $H$ is not normal in $G$, $\ker(f) \neq H$. The only possibility is that $\ker(f) = \{e\}$. Hence $f$ is injective. Since
$$
|G| = |S_3| = 6
$$
it follows that $f$ is surjective as well, and hence $f$ gives an isomorphism $G\cong S_3$
Have you seen either of these facts? If not, then say so in the comments and I can indicate the proofs.
