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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?

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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. 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 $\langle\Bbb Z_m,+\rangle$.

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    @sizz: Two integers are relatively prime if their greatest common divisor is $1$, like the numerator and denominator of a fraction in lowest terms. $\langle(134)(25)\rangle$ is indeed isomorphic to $\langle\Bbb Z_6,+\rangle$.2012-11-20
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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.