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
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.
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.