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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?

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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.

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    $\begingroup$ Pity about the restriction to length $\ge 2$, for cycles of length $1$ could be called unicycles. $\endgroup$ – André Nicolas Jul 31 '14 at 23:53
  • $\begingroup$ @Andre I quite agree, I am still displeased with the killjoy that jumped in and named 2-cycles transpositions instead of bicycles. $\endgroup$ – Adam Hughes Jul 31 '14 at 23:55
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let $\sigma$ and $\kappa$ be the cycles suppose $\sigma(x)\neq x$ , then $\kappa(x)=x$ so $\sigma(\kappa(x))\neq x$ so they are not inverses.

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