Can two cycles be the inverse of each other if they are disjoint? I am self-studying DF's Abstract Algebra and am currently working on the following exercise:
Prove that the order of an element in $S_{n}$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. 
I use the following lemma in my proof: If $|a|=m$ and $a^{n}=1$ then $m$ divides $n$.
I have already shown that if $\sigma\in S_{n}$ then if $l$ is the least common multiple of the lengths of the cycles in its cycle decomposition then $\sigma^{l}=1$. I am currently trying to show that $l$ is the least such positive integer such that $\sigma^{l}=1$.
To make things easy to see, I started to work with the case where $\sigma$ is the product of two cycles of lengths $m$ and $n$. Let $\sigma_{m}$ be the cycle of length $m$ and $\sigma_{n}$ be the cycle of length $n$. Then $\sigma=\sigma_{m}\sigma_{n}$. Suppose there exists an integer $l_{1}<l$ such that $\sigma^{l_{1}}=1$. It follows that $\sigma_{m}^{l_{1}}\sigma_{n}^{l_{1}}=1$. If $\sigma_{m}^{l_{1}}$ is not the inverse of $\sigma_{n}^{l_{1}}$ then by the lemma above, $m$ divides $l_{1}$ and $n$ divides $l_{1}$ so $l_{1}$ is a common multiple of $m$ and $n$ but $l$ is the least common multiple so $l<l_{1}$ which is a contradiction. 
My problem is what happens if $\sigma_{m}^{l_{1}}$ is the inverse of $\sigma_{n}^{l_{1}}$? Can that even happen since the two cycles are supposed to be disjoint?

Say the first cycle, $\sigma_1$ has that
$$ \sigma_1(a)= b\ne a$$
all cycles have at least two symbols (or we don't write them in a cycle decomposition) so this is a valid assumption. Then after applying the second cycle, $\sigma_2$, which does not have the symbols $a$ or $b$, by definition of disjoint. We then have
$$\sigma_1\sigma_2(a)=\sigma_1(\sigma_2(a))=\sigma_1(a)=b$$
still, hence it cannot be the identity of the group.
