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Find isomorphism between $S_3$ and $GL_2(F_2)$.

proof: Let $A = \begin{pmatrix} a& b\\ c & d \end{pmatrix}$. Where $\det (A) \neq 0$.

And recall $S_3$ is the permutation group with {1,2,3}.

Then there are $6$ different matrices Let $ \begin{pmatrix} 1& 1\\ 1 & 0 \end{pmatrix}$ , $ \begin{pmatrix} 1& 1\\ 0 & 1 \end{pmatrix}$ , $ \begin{pmatrix} 0& 1\\ 1 & 1 \end{pmatrix}$ , $ \begin{pmatrix} 1& 0\\ 1 & 1 \end{pmatrix}$ , $ \begin{pmatrix} 1& 0\\ 0 & 1 \end{pmatrix}$ , $ \begin{pmatrix} 0& 1\\ 1 & 0 \end{pmatrix}$ . And $|GL_2(F_2)| = 6$. And $|S_3| = 3! = 6$ both have orders $6$.

I was trying to construct a map by trying to compare their Cayley tables, but in general matrix multiplication does not commute. Can someone please help me? Maybe we can construct a map, but I am not sure. I am stuck. Thank you for any help.

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    $\begingroup$ Note that there is only one non-abelian group of order 6 (which is $S_3$). You can also see the answer here $\endgroup$
    – GAVD
    Sep 3, 2015 at 4:59
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    $\begingroup$ How many elements are there in $(F_2)^2\setminus \{0\}$? $\endgroup$
    – user24142
    Sep 3, 2015 at 5:00
  • $\begingroup$ Following @user24142 comment. You know that $\mathbb{F}_2^2$ is a vector space and $(0,0)$ is fixed by any isomorphism (which, in turn is permuting the other elements..). $\endgroup$
    – rafforaffo
    Sep 3, 2015 at 15:50

1 Answer 1

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Observe that every element of $S_3$ can be written in terms of $\alpha=(1 \, 2)$ and $\beta=(1 \, 2\, 3)$. Also observe that $|\alpha|=2$ and $|\beta|=3$. Now look for matrices in $GL(\mathbb{Z}_2)$ which have orders $2$ and $3$ respectively. Suppose they are $A$ and $B$. Then consider the function $f(\alpha)=A$ and $f(\beta)=B$. Extend it (using the homomorphic property) to create an isomorphism.

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