I know that $\mathbb{Z}_3$ is isomorphic to a subgroup $H\le S_3$ generated by the $3$-cycle $(123) \in S_3$.
So we should have $H = \{(123), (312), e\}$ where $e = (1)(2)(3)$.
What I don't understand is, which is respectively, the inverse of these elements?
Because if I do, for example
- $$(123)(312) = (312)$$
- $$(312)(123) = (312)$$
while I would expect to obtain the identity element $e$..
EDIT:
So $H$ is the wrong subgroup. But to show the isomorphism what I do is considering the injective map
\begin{align} \mathbb{Z}_3 &\rightarrow S_3\\ n&\mapsto (123)^n \end{align}
and so this would lead $e$ for $n=0$, $(123)$ for $n=1$, $(123)^2 = (312)$ for $n=2$.. Is this the wrong way to consider this identification?