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Is the symmetric group S3 cyclic?

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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