Permutable Subgroups 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. 
 A: 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. 
A: 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.
A: Here one just needs to recall that a permutable subgroup is ascendant. Then if $x$ is not contained in $A$, it follows that $A<N_{A\langle x\rangle}(A)$, 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.
