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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 $G$»?2012-10-22
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    (How do you know it is very easy if you do not know where to start?! :-) )2012-10-22
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    A is a permutable subgroup of G. The text says that is easy :D2012-10-22
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    Please edit the question and explain *there* all notation.2012-10-22
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    What is a permutable subgroup?2012-10-22
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    [Wikipedia](http://en.wikipedia.org/wiki/Quasinormal_subgroup)2012-10-22
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    What are you asking?2012-10-22
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    I have edited the question. I hope now is more clear...2012-10-22
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    @W4cc0 May I ask that after such a long time, have you solved the question…?2018-10-09
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    @SteveJacob, yes, I think I managed to solve the question.2018-10-12
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    @W4cc0 May I ask for a solution? Or shall I ask it again in StackExchange? Thanks.2018-10-22
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    @SteveJacob I've posted the answer here.2018-10-31

3 Answers 3

1

Here one just needs to recall that a permutable subgroup is ascendant. Then if $x$ is not contained in $A$, it follows that $A, and so that $x$ normalize $A$.

Note that a subgroup $H$ of a group $G$ is said to be ascendant if there is an ascending series of subgroups starting from $H$ and ending at $G$, such that every term in the series is a normal subgroup of its successor.

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    But now you have to tell us what ascendant means.2018-11-03
4

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](http://www.scribd.com/doc/89421590/A-Course-in-the-Theory-of-Groups-2ed-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.2012-10-22
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    P.S. I believed that the notation was of common use, sorry for this.2012-10-22
2

First, $xAx^{-1}$ is permutable. Then $AxAx^{-1}$ is a subgroup of A{x} which contains A.

Therefore, if $xAx^{-1}\neq A$, we can derive that $AxAx^{-1}=A{x}$.

$\Longrightarrow$ $x^{-1}=a_1xa_2x^{-1}$

$\Longrightarrow$ $e=a_1xa_2$

$\Longrightarrow$ $x\in A$ , a contradiction.