Is the symmetric group S3 cyclic?

The Mathematician — Wed, 26 Oct 2022 13:00:30 +0000

The symmetric group S3 is not cyclic because it is not abelian. We could prove this in a different way by checking all elements one by one.

Proof. Recall that

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

As each exponent on the identity element is an identity element, we also need to check 5 elements: $(12)$

\begin{align*} (12) &= (12) \\ (12)(12) &= e \\ \end{align*}

$(13)$

\begin{align*} (13) &= (13) \\ (13)(13) &= e \\ \end{align*}

$(23)$

\begin{align*} (23) &= (23) \\ (23)(23) &= e \\ \end{align*}

$(123)$

\begin{align*} (123) &= (123) \\ (123)(123) &= (132) \\ (123)(123)(123) &= e \\ \end{align*}

$(123)$

\begin{align*} (132) &= (132) \\ (132)(132) &= (123) \\ (132)(132)(132) &= e \\ \end{align*}

No single element of

S_3

can generate the symmetric group

S_3

. So the symmetric group S3 is not cyclic.

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Is the symmetric group S3 cyclic?