PGL(2,2) is isomorphic to S3

We will prove in this article that PSL(2,2) is isomorphic to S3, that GL(2,2) is isomorphic to S3, and GL(F2) is isomorphic to S3. By definition,

PSL(V)

is equal to the quotient

GL(V)/Z(V)

where

Z(V)

is the centralizer of

GL(V)

. In this case, the centralizer is just the identity element itself. So we have that

GL(2,2) = PGL(2,2)

. Further, recall that

\begin{equation*} S_3 = \{e,(12),(13),(23),(123),(132)\}. \end{equation*}

Proof. Define the mapping

\rho: S_3 \longrightarrow PGL(2,2)

with the next elements

\begin{align*} (e) \longmapsto \mathbb{I}_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \ (123) \longmapsto \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}, \ (132) \longmapsto \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\end{align*}

\begin{align*} (12) \longmapsto \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},, \ (12)(123) \longmapsto \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \ (12)(132) \longmapsto \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\end{align*}

Note that

(12)(132) = (13)

(12)(123) = (23)

and

(123)^2 = (132)

. The map is surjective as there exists an element

S_3

for any element of

PGL(2,2)

. That the mapping is a group homomorphism is easy to see and we will give one example, and the rest is for the reader to verify:

\begin{align*} \rho(123)\rho(123) = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \rho(132) \end{align*}

Repeat this process for any element. What is left is to prove injectivity, but we do see already in the mapping above that

\ker(\rho) = \{\mathbb{I}_2\}

. So we get

S_3 \cong PGL(2,2)