How to prove that a permutation must be a product of $3$-cycles if $ \rho^3=I$？

Question

I was trying to solve the following problem:

Suppose that the permutation $\rho$ of $\{1,...,n\}$ satisfies $\rho^3=I$. Show that $\rho$ is a product of 3-cycles, and deduce that if $n$ is not divisible by 3 then $\rho$ fixes some $k$ in $\{1,...,n\}$.

I was able to show the sufficiency of $\rho$ being a product of 3-cycles, but I cannot show the necessity of it. How should I do it?

Answer 1

Let $\rho=\sigma_1\sigma_2\dots\sigma_r$ be the cyclic decomposition of $\rho$ in $S_n$. The $(\sigma_i)_{1\le i\le r}$ are cycles and are pairwise disjoint. Thus they commute with each others. Thus $\rho^3=\sigma_1^3\sigma_2^3\dots\sigma_r^3=I$. Since the cycles are disjoints, what can you say about $\sigma_1^3,\sigma_2^3,\dots,\sigma_r^3$? Then you should be able to see why $(\sigma_i)_{1\le i\le r}$ are $3$-cycles.

Answer 2

The permutation $\rho$ has a decomposition as a product of disjoint, hence commuting, (non-trivial) cycles: $\;\rho=\gamma_1\dotsm\gamma_r$. The order of $\rho$ is the l.c.m. of the orders of the cycles, so each $\gamma_i$ has order $3$. As the order of a cycle is its length, this means each $\gamma_i$ is a $3$-cycle.