Symmetric Group on 3 Letters is Isomorphic to Dihedral Group D3

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Theorem

Let $S_3$ denote the Symmetric Group on 3 Letters.

Let $D_3$ denote the dihedral group $D_3$.


Then $S_3$ is isomorphic to $D_3$.


Proof

Consider $S_3$ as presented by its Cayley table:

$\begin{array}{c|cccccc}

\circ & e & (123) & (132) & (23) & (13) & (12) \\ \hline e & e & (123) & (132) & (23) & (13) & (12) \\ (123) & (123) & (132) & e & (13) & (12) & (23) \\ (132) & (132) & e & (123) & (12) & (23) & (13) \\ (23) & (23) & (12) & (13) & e & (132) & (123) \\ (13) & (13) & (23) & (12) & (123) & e & (132) \\ (12) & (12) & (13) & (23) & (132) & (123) & e \\ \end{array}$


Consider $D_3$ as presented by its group presentation:

$D_3 = \gen {a, b: a^3 = b^2 = e, a b = b a^{-1} }$


and its Cayley table:

$\begin{array}{c|cccccc}
     & e     & a     & a^2   & b     & a b   & a^2 b \\

\hline e & e & a & a^2 & b & a b & a^2 b \\ a & a & a^2 & e & a b & a^2 b & b \\ a^2 & a^2 & e & a & a^2 b & b & a b \\ b & b & a^2 b & a b & e & a^2 & a \\ a b & a b & b & a^2 b & a & e & a^2 \\ a^2 b & a^2 b & a b & b & a^2 & a & e \\ \end{array}$


Let $\phi: S_3 \to D_3$ be specified as:

\(\ds \map \phi {1 2 3}\) \(=\) \(\ds a\)
\(\ds \map \phi {2 3}\) \(=\) \(\ds b\)


Then by inspection, we see:

\(\ds \map \phi {1 3 2}\) \(=\) \(\ds a^2\)
\(\ds \map \phi {1 3}\) \(=\) \(\ds a b\)
\(\ds \map \phi {1 2}\) \(=\) \(\ds a^2 b\)

and the result follows.

$\blacksquare$


Sources

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