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Let $\sigma = (1 2 \dots 9) \in S_{10}$.

a) Calculate the size of the normalizer $N_{S_{10}}(<\sigma >)$.

b) Describe exactly the elements in $N_{S_{10}}(<\sigma >)$.

I am not sure how to approach this. I understand that we look for permutations that fixes $10$, and yet I can't see what to do further...

I know that $\tau \sigma \tau^{-1} =(\tau(1) \dots \tau(9))$ by definition, and also that for $\tau$ to be in $N_{S_{10}}(<\sigma >)$ than it is required that $\tau \sigma \tau^{-1} =(\tau(1) \dots \tau(9)) = \sigma^i$ for some $i$.

How can I continue from here?

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  • $\begingroup$ Let $g \in N = N_{S_{10}}(\langle \sigma \rangle)$. Since $\sigma \in N$, by multiplying $g$ by a power of $\sigma$ you can assume $g(1)=1$. Now $g^i(1)=i+1$, so $g\sigma g^{-1} = g^i \Leftrightarrow g(2)=i+1$, so $g(2)=2,3,5,6,8$ or $9$. $\endgroup$ – Derek Holt Jul 25 '18 at 7:48
  • $\begingroup$ @Derek Holt It is really difficult for me to follow all of the assumptions you made... Could you please elaborate? $\endgroup$ – ChikChak Jul 28 '18 at 21:05
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I'll try to continue from where you stopped...

It has to be $o(g\sigma g^{-1})=o(\sigma^i)=9$ so $gcd(i,9)=1\Rightarrow i=1,2,4,5,7,8$

The $g\in S_{10}$ with the above property (for fixed $i$) form a coset of $C_{S_{10}}(<\sigma>)=<\sigma>$ and the number of such cosets is $6$ (because we have $6$ $i$'s)

For example for the $g\in S_{10}$ with $g\sigma g^{-1}=\sigma^2$ we see that all $t\in S_{10}:t\sigma t^{-1}=\sigma^2$ are $gC(<\sigma>)=\{t: t=gu \text{ with } u\sigma u^{-1}=\sigma\}$

And since $|<\sigma >|=9$, $|N_{S_{10}}(<\sigma>)|=9\cdot 6=54$

For (b) the elements of $N_{S_{10}}(<\sigma>)$ are these of the form $g\sigma^l$ with $\sigma^l\in<\sigma> $ and $g\in S_{10}$ with the property described in the first line.

I hope it is correct!

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