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Question In my group theory course, I am asked to show for $\sigma \in S_n$ that if $\sigma (1 \cdots n) = (1 \cdots n) \sigma$ then $\sigma = (1 \cdots n)^i$ for certain $i$.

My answer Let $\sigma \in S_n$. If $$\sigma (1 \cdots n) = (1 \cdots n) \sigma$$ then $$\sigma (1 \cdots n) \sigma^{-1} = (1 \cdots n).$$ This implies $$( \sigma(1) \cdots \sigma(n)) = (1 \cdots n).$$ We conclude $\sigma = ()$, which is equal to $(1\cdots n)^i$ for $i$ that are multiples of $n$.

Problem This just seems really silly. Why would the question mention "certain $i$"? There are many ways to write $(1)$...

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  • $\begingroup$ I think by $\sigma=(1)$ you mean $\sigma$ is the identity permutation, which you'd denote as $()$ (compactly) or $(1)(2)\cdots(n)$ (full cycle notation). It wouldn't make sense to include the cycle $(1)$ but not the cycles $(2),(3),\cdots,(n)$ when writing the identity permutation. One could also write $\sigma=1$, since sometimes $1$ is the identity element of a multiplicatively written group, but that could be confusing (since $1$ also denotes an element of a set that permutations act on) so $\sigma=e$ would be better. $\endgroup$ – arctic tern Jul 26 '16 at 22:43
  • $\begingroup$ @arctictern It is a notation my course materials use. I agree it is a bit weird. Will edit it to your suggestion (). $\endgroup$ – Timon van der Berg Jul 26 '16 at 22:45
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Say $n=3$ for example. Then $(\sigma(1)\sigma(2)\sigma(3))=(123)$. But that's not enough to conclude the equality $\sigma(k)=k$ for $k=1,2,3$. Why? Because $(123)$ can also be written $(231)$ and $(312)$, so we could also have other posisble $\sigma$s as well.

Hint: suppose $\sigma(1)=i$, then try to prove $\sigma=(12\cdots n)^i$.

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  • $\begingroup$ Ah, I see. If $\sigma(1) = i$ then $\sigma = (i, i+1,\cdots,i+n)$. Also, $(1,\cdots,n)^i = (i, i+1, \cdots, i+n)$. In the same way we can construct a $\sigma$ for every $i \in \{ 1, \cdots n\}$. --- something like this? $\endgroup$ – Timon van der Berg Jul 26 '16 at 22:54
  • $\begingroup$ @TimonvanderBerg Yes (as long as you interpret those numbers mod $n$). $\endgroup$ – arctic tern Jul 26 '16 at 22:57

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