isomorphism of SL(2, Z2) and S3

Question

How can i prove that $SL_{2}(ℤ_{2})\cong S_3$?

It is easy to build explicit isomorphism but I think there is more beautiful solution.

Well, there is a unique (up to isomorphism) non abelian group of order $6$, since both $\mathrm{SL}_2(\mathbb{Z}_2)$ and $S_3$ are two non abelian groups of order $6$ they must be isomorphic. — Albert, Dec 14 '23 at 17:27
How many nonzero vectors are there in the vector space $Z_2^2$ ? — ahulpke, Dec 14 '23 at 17:39
@ahulpke: there are three, and elements of $SL_{2}(ℤ_{2})$ permute them; I think that is worthy of an answer, not just a comment — J. W. Tanner, Dec 14 '23 at 18:42
As you are looking for the enrichment your "bag of tools", have a look at my former answer here establishing a group isomorphism in a neighboring case : $PSL(2,\mathbb{Z_3}) \approx A_4$. — Jean Marie, Dec 14 '23 at 18:48

score 1 · Accepted Answer · answered Dec 14 '23 at 17:46

To elaborate on what Albert said :

First calculate the order of $SL(2,\mathbb Z_2),$ like here.
Find two matrices in $SL(2,\mathbb Z_2)$ which do not commute.
Use the fact that the only nonabelian group of order 6 is $S_3,$ and this is given here.

Here the crucial step is (3) and this technique is specific to lower order groups.

1 Answers1