Let $n \geq 3$. Then there exists an element $f \in S_n$ such that $f \neq g^3$ for any element $g \in S_n$, where $S_n$ denotes the symmetric group on $n$ letters. How to establish whether this assertion is true or false?
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Hints:
For any $\,n\ge 3\;$ there are elements of order $\;3\;$ in $\;S_3\;$
The map $\;\phi_3:S_n\to S_n\;,\;\;\phi_3(\sigma):=\sigma^3\;,\;\;n\ge3\;,\;$ cannot be injective...
DonAntonio
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Thanks, but I'm afraid I've not picked the hint. I guess we require the map to be not surjective. – Saaqib Mahmood Sep 15 '13 at 07:17

1Well, if $\;X\;$ is a finite set, then a map $\;f:X\to X\;$ is injective iff it is surjective... – DonAntonio Sep 15 '13 at 09:22
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Prove that $(123)\neq \sigma^3, \ \forall \sigma\in S_n$ where $(123)$ stands for the 3circle $1\mapsto2, \ 2\mapsto3, \ 3\mapsto1$. To prove it use that if $(123)= \sigma^3$, then the order of $\sigma$ must be $9$ and therefore $\sigma$ must be a (product of) $9$ circle(s). Then derive a contradiction.
In general for $n\geq m, \ (123\cdots m)\neq\sigma^m, \ \forall \sigma\in S_n$.
P..
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