Let \(G=GL_2(\FF_2)\), the group of invertible \(2\times2\) matrices over the field with two elements. Prove that \[ G\cong S_3. \]
Solution
Let \[ V\coloneqq\FF_2^2\sm\{0\}. \] Then \(V\) has exactly three elements.
1. The natural action of \(G\) on \(V\) defines an injective homomorphism\[ \rho:G\longrightarrow\operatorname{Sym}(V)\cong S_3. \]
Every element of \(G\) is an invertible linear map on \(\FF_2^2\), so it sends nonzero vectors to nonzero vectors and therefore permutes \(V\). This gives a homomorphism \[ \rho:G\to\operatorname{Sym}(V). \] If \(T\in\ker\rho\), then \(T\) fixes every nonzero vector. In particular, for the standard basis \(e_1,e_2\) one has \[ Te_1=e_1, \qquad Te_2=e_2. \] Hence \(T=I_2\), so \(\ker\rho=\{I_2\}\) and \(\rho\) is injective.
2. \(\abs G=6\).
The first column of an invertible \(2\times2\) matrix over \(\FF_2\) can be any nonzero vector, giving \(4-1=3\) choices. Once the first column is chosen, the second column must lie outside its one-dimensional span, which contains two vectors, giving \(4-2=2\) choices. Thus \[ \abs G=(4-1)(4-2)=6. \]
3. \(G\cong\boxed{S_3}\).
By step 1, \(G\) embeds in \(S_3\). Step 2 gives \[ \abs G=6=\abs{S_3}. \] Therefore the injective homomorphism \(\rho:G\to S_3\) is surjective and hence an isomorphism.
4. Q.E.D.
Step 3 proves the required isomorphism.