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Let $G$ be a finite group. Let $H \leq K\unlhd G$. If for each $P$ Sylow subgroup of G there exists $x \in G$ such that $HP^{x}=P^{x}H$ then for each $P \cap K$ of $K$ there exists $y \in K$ such that $H(P\cap K)^{y}=(P\cap K)^{y}H$.

I know that $H(P\cap K)^{x}=(P\cap K)^{x}H$, but $x$ may not be an element of $K$. How may I prove this?

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up vote 1 down vote accepted

Let $Q$ be a $p$-Sylow subgroup of $K$. Then $Q$ extends to a $p$-Sylow subgroup of $G$, $P$, and $P\cap K = Q$. If $P'$ is a $p$-Sylow subgroup of $G$, then there exists $z$ such that $(P')^z = P$, hence $(P'\cap K)^z = P'^z\cap K^z = P'^z\cap K = P\cap K = Q$, so $P'\cap K$ is a $p$-Sylow subgroup of $K$.

Thus, for every Sylow subgroup $P$ of $G$, $P\cap K$ is a Sylow subgroup of $K$.

So let $P$ be a Sylow subgroup of $G$. We know there exists $x$ such that $(P\cap K)^x H = H(P\cap K)^x$. Since $(P\cap K)^x=P^x\cap K$ is a Sylow subgroup of $K$, there exists $y\in K$ such that $(P\cap K)^y = (P\cap K)^x$.

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Thank you very very much. – user28083 Apr 16 '12 at 20:03
Assume, in the above question, that $x \in T \unlhd G$. Does this imply that $y \in T \cap K$? – user28083 Apr 18 '12 at 12:28

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