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Let $\sigma_1,\dots , \sigma_t$ be disjoint cycles in $S_n$ of lengths $l_1, \dots, l_t$ and $$ \sigma=\sigma_1 * \dots * \sigma_t.$$

Determine the order of $\sigma$ in terms of $l_1,\dots ,l_t$

I believe the answer is $ \mathrm{lcm} ( l_1,\dots ,l_t )$. Is this correct? How do I prove it?

Robin
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Tiger Blood
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  • Yes, that's correct. A simple way to note this is that each entry $m$ of a disjoint cycle $\sigma_k$ is mapped back to itself after $l_k$ times (and is independent of the other disjoint cycles) and since $\sigma$ is the composition of all the disjoint cycles, we conclude that $\sigma^n=\text{id}\iff l_k\mid n~\forall~k$ from which it follows that the least common multiple of all the $l_k$'s must be the order of $\sigma$ – Prasun Biswas Sep 27 '17 at 13:51

1 Answers1

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Yes and here is a hint as how to prove it:

Claim. For disjoint cicles $\sigma, \tau$ we have $$ \sigma \circ \tau = \tau \circ \sigma. $$

[Prove and apply this to get $$ \sigma^n = \sigma_1^n \circ \ldots \sigma_t^n $$ and use the disjointedness once more to conclude that $\sigma^n = \operatorname{id}$ iff $\forall i \le t \colon \sigma_i^n = \operatorname{id}$.]

Stefan Mesken
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