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Let $G$ be a group, if $x\in G$ has order $p$, for some prime $p$, and $A\space per\space G$ (that is, A is a permutable subgroup of $G$), then I want to show that $x$ normalizes $A$.
Any hints?

Robinson;
Definition (pag. 393): A subgroup H is said to be permutable in a group G if HK=KH whenever K≤G.
Exercise 6 (pag. 396): A permutable subgroup is normalized by every element of prime order.

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What is $A$ or «$A$ per $G$»? –  Mariano Suárez-Alvarez Oct 22 '12 at 19:23
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(How do you know it is very easy if you do not know where to start?! :-) ) –  Mariano Suárez-Alvarez Oct 22 '12 at 19:23
    
A is a permutable subgroup of G. The text says that is easy :D –  W4cc0 Oct 22 '12 at 19:25
    
Please edit the question and explain there all notation. –  Mariano Suárez-Alvarez Oct 22 '12 at 19:28
    
What is a permutable subgroup? –  Tobias Kildetoft Oct 22 '12 at 19:30

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up vote 3 down vote accepted

I do think you need to explain what a permutable subgroup is. It is a subgroup $A$ such that $AB = BA$ for every subgroup $B$? It does not suffice to assume that $A$ is permutable with $\langle x \rangle $ to obtain that $x$ normalizes $A$. If we take $G = A_{5},$ $A = A_{4}$ and $x$ of order $5$, then $A$ and $\langle x \rangle$ are permutable (with each other) but $x$ does not normalize $A.$

As mentioned in the linked Wikipedia article, in finite groups, all permutable subgroups are subnormal. That is the key point here, as hinted at in the Wikipedia aticle: $A$ is subnormal in the group $A\langle x \rangle,$ but is also a maximal subgroup of $A\langle x \rangle$, so must be normal in that group.

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Robinson; Definition (pag. 393): A subgroup $H$ is said to be permutable in a group $G$ if $HK=KH$ whenever $K\leq G$. Exercise 6 (pag. 396): A permutable subgroup is normalized by every element of prime order. –  W4cc0 Oct 22 '12 at 21:43
    
P.S. I believed that the notation was of common use, sorry for this. –  W4cc0 Oct 22 '12 at 21:46

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