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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.

user114873
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2 Answers2

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Without (many) words:

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

DonAntonio
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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
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    @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
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    Very nice, Don Antonio! (+1) – amWhy Mar 31 '14 at 12:42
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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.)

lhf
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