# The normalizer of permutation

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?

• 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 $9$. – Derek Holt Jul 25 '18 at 7:48
• @Derek Holt It is really difficult for me to follow all of the assumptions you made... Could you please elaborate? – ChikChak Jul 28 '18 at 21:05

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!