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If $H=\{\sigma\in S_n: \sigma (n)=n\}$, then H is a normal subgroup of $S_n$ for $n\geq3$.

How to solve this problem.If we have to disprove it then give an example.

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up vote 7 down vote accepted

Without (many) words:

$$n=4:\;\;\;\;(14)(123)(14)=(234)\notin H$$

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I want to ask you something about σ(n)=n. σ(n) is a permutation I want to know what is n? Is it a number? – user114873 Mar 31 '14 at 12:10
That means n is a permutation. – user114873 Mar 31 '14 at 12:12
@user114873, $\;S_n\;$ is the group of all bijections from the set $\;N:=\{1,2,...,n\}\;$ to itself (this is the easy-to-grasp definition. Symmetric groups are way more general but this definition will do for now), so $\;\sigma(n)=n\;$ means "a bijection $\;\sigma\;$ of $\;N\;$ leaving fixed the element $\;n\in N\;$ ". – DonAntonio Mar 31 '14 at 12:15
Very nice, Don Antonio! (+1) – amWhy Mar 31 '14 at 12:42
Thanks @amWhy :) – DonAntonio Mar 31 '14 at 12:42

A subgroup is normal iff it is invariant under conjugation.

Conjugation in $S_n$ is just renaming. More precisely, the permutation $\tau \sigma \tau^{-1}$ does exactly what $\sigma$ does, except that the numbers are renamed via $\tau$ (or $\tau^{-1}$, depending on which side you compose).

Your $H$ is not invariant under all possible renamings, only under those that fix $n$. Hence, $H$ is not normal. (Hence, also, the counterexample given by @DonAntonio.)

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