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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

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It is obvious that Aff(3) is a group consisting of 6 elements cause a can get 2 values and b can get 3 values.

There are only 2 groups of order 6. One is S3 and the second one is Z6. Z6 is commutative, while S3 not. Find two matrices in Aff(3), which are not commutative, therefore You prove that Aff(3)!=Z6, therefore must be S3.

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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 $\operatorname{Aff}(3)\cong S_3.$

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