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Let $\sigma \in S_n$ where $S_n$ is the symmetric group of degree $n$. Prove that if $\sigma$ is not the product of commuting $k$-cycles, then the cycle decomposition of $\sigma$ contains an $r$-cycle $\tau$ for some $r \neq k$.

We know that disjoint $k$-cycles commute, but not necessarily the other way around. So how do we use the fact that $\sigma$ is not the product of commuting $k$-cycles?

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    $\begingroup$ I'm confused. If the cycle decomposition of $\sigma$ does not contain an $r$-cycle for some $r\neq k$, then by definition, $\sigma$ is a product of disjoint (and hence commuting) $k$-cycles. You don't even need to assume anything about the order of $\sigma$. $\endgroup$ – tomasz Nov 4 '16 at 0:10
  • $\begingroup$ @tomasz Why are you assuming it doesn't contain an $r$-cycle? $\endgroup$ – user19405892 Nov 4 '16 at 0:13
  • $\begingroup$ I am arguing by contradiction. $\endgroup$ – tomasz Nov 4 '16 at 0:15
  • $\begingroup$ @tomasz If the cycle decomposition of $\sigma$ doesn't contain an $r$-cycle $\tau$ for some $r \neq k$, then it is the product of $k$-cycles. How do we know they are disjoint? $\endgroup$ – user19405892 Nov 4 '16 at 0:22
  • $\begingroup$ How do you define cycle decomposition? When we say "the cycle decomposition", we usually mean decomposition into disjoint cycles. They are disjoint by definition. $\endgroup$ – tomasz Nov 4 '16 at 0:28
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You are given a partition of $n$ and all given the information that it is not $n=k+k+\ldots +k$. Then it has one part not equal to $k$. It is not a consequence of the hypothesis. it is the hypothesis itself.

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