When do distinct cycles (permutations) generate distinct cyclic groups? When do distinct cycles generate distinct cyclic groups?
Specifically, consider two cycles or order 5: $(a_1 \quad a_2 \quad a_3 \quad a_4 \quad a_5)$ and $(b_1 \quad b_2 \quad b_3 \quad b_4 \quad b_5)$.
Is it the case that these generate distinct cyclic groups if and only if they don't commute? I've been told that commutativity is relevant here, but I'm not sure why.
 A: Let $\sigma,\tau\in S_n$ be two cycles that generate the same group:
$$
\langle\sigma\rangle=\langle\tau\rangle
$$
Then $\tau\in\langle\sigma\rangle$, which implies $\tau=\sigma^k$ for some $k\in\mathbb N$. Therefore
$$
\sigma\circ\tau=\sigma\circ\sigma^k=\sigma^k\circ\sigma=\tau\circ\sigma
$$
so if $\sigma,\tau$ generate the same cyclic group, they also commute.

Suppose on the other hand that $\sigma,\tau\in S_n$ do commute. Let $\sigma\cap\tau$ be the set of elements from $\{1,...,n\}$ that are cycled by both $\sigma$ and $\tau$. Then I claim that for $i\in\sigma\cap\tau$ we have:
$$
\sigma(i),\tau(i)\in\sigma\cap\tau
$$
Assume WLOG that $\sigma(i)=j\notin\sigma\cap\tau$. Since $\sigma$ is a cycle, clearly $j$ is cycled by $\sigma$. Hence, this must mean that $j$ is not cycled by $\tau$:
$$
\tau\circ\sigma(i)=\tau(j)=j
$$
now consider $\tau(i)=k$:
$$
\sigma\circ\tau(i)=\sigma(k)=j
$$
where the last equality follows from commutativity of $\sigma,\tau$. But this implies $k=i$ by injectivity of $\sigma$ since $x=i$ is the unique solution to $\sigma(x)=j$. Thus $\tau(i)=i$ which contradicts $i\in\sigma\cap\tau$.
Inductively, this implies that $\sigma^k(i)\in\sigma\cap\tau$ and the same for $\tau^k(i)$, so that $\sigma\cap\tau$ must be the complete set of elements cycled by $\sigma$ and $\tau$ individually. Unless of course $\sigma\cap\tau=\emptyset$, in which case we have no element $i\in\sigma\cap\tau$ to lead us to the contradiction shown above.

Conclusion: Cycles $\sigma,\tau\in S_n$ generate the same cyclic group AND cycle the same set of elements iff they commute AND cycle at least one element in common. If they are disjoint cycles, naturally they do not.
