Isomorphism between integers and symmetry group I am a little slow on this this.
Consider the subgroup $<\sigma>$ where $\sigma = (1 3 746)$ of $S_7$. Why is $<\sigma>$ isomorphic to $<\mathbb{Z_5},+>$?
I know that two elements of $\sigma$ are fixed, which explains the '5' case, but doesn' explain how they are isomorphic to each other?
 A: Calculate the successive powers of $\sigma$: 
$$\begin{align*}
\sigma&=(13746)\\
\sigma^2&=(17634)\\
\sigma^3&=(14367)\\
\sigma^4&=(16473)\\
\sigma^5&=\text{identity permutation}
\end{align*}$$
In other words, $\langle\sigma\rangle$ is a cyclic group of order $5$. All cyclic groups of order $5$ are isomorphic, so it’s isomorphic to $\langle\Bbb Z_5,+\rangle$. One isomorphism is given by $\sigma^k\mapsto k$ for $k=0,1,2,3,4,5$.
With a little more experience you’ll realize that you don’t need to do the actual calculations: an $m$-cycle in a permutation group always has order $m$. Thus, if $\sigma$ is an $m$-cycle, $\langle\sigma\rangle$ is always isomorphic to $\langle\Bbb Z_m,+\rangle$, and one possible isomorphism is the map $\sigma^k\mapsto k$ for $0\le k<m$. There are others. Exercise: If $a$ is relatively prime to $m$, then the map $\sigma^k\mapsto ak\bmod m$ is an isomorphism of $\langle\sigma\rangle$ to $\langle\Bbb Z_m,+\rangle$.
A: Note that $\sigma$ has order $5$. Now if the group $G=\langle g \rangle$  and $g$ has order $n$ then $G$ is isomorphic to $(\mathbb{Z}_n,+)$. 
The function $f:G \to (\mathbb{Z}_n,+) , g^k \mapsto k$ is an isomorphism.
