# Show that $\operatorname{Aff}(3)$ is isomorphic to $S_3$, the symmetric group of all permutations of 3 objects.

Show that $$\operatorname{Aff}(3)$$ is isomorphic to $$S_3$$, the symmetric group of all permutations of 3 objects.

Where

$$\operatorname{Aff}(3):={\{( \begin{array}{cc} a & b \\ 0 & 1 \end{array}): a,b\in\mathbb{Z}_3}, a\neq0\}$$, $$\mathbb{Z}_3$$ are the integers module 3.

Idea: I know that $$S_3$$ has 6 elements, and the $$\operatorname{Aff}(3)$$ matrices are: $$( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array})$$, $$( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array})$$ $$( \begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array})$$, $$( \begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array})$$ $$( \begin{array}{cc} 2 & 1 \\ 0 & 1 \end{array})$$, $$( \begin{array}{cc} 2 & 2\\ 0 & 1 \end{array})$$

But how do I find an isomorphism? Can you help me please

If $$a\mapsto \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$$ and $$b\mapsto \begin{pmatrix} 2 & 0 \\ 0 & 1\end{pmatrix}$$, then $$\operatorname{Aff}(3)$$ has as a presentation $$\langle a, b\mid a^3, b^2, bab=a^{-1}\rangle,$$ which is, in turn, a presentation for $$S_3$$. Hence $$\operatorname{Aff}(3)\cong S_3.$$