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I would please like some help on the exercise 257 of the Rose Group Theory. Let $G$ be a finite group and $H$ and $K$ be normal subgroups of $G$, and $P$ a Sylow p-subgroup of $G$. Then $(PH)\cap (PK)=P(H\cap K)$.

One inclusion is straightforward: $P(H\cap K)\subset (PH)\cap (PK)$ since $H\cap K \subset H,K$. Let $g\in G$ such that $g=p_1h=p_2k$ where $p_i \in P, h\in H,k\in K$. I want to show $p_1=p_2$. The normality of H and K tells me that PH and PK are subgroups of G. By order considerations, P is a p-Sylow of PH, PK and $P(H\cap K$). By normality, $H \triangleleft PH, K \triangleleft PK,H\cap K \triangleleft P(H\cap K)$. I thought about the Frattini's argument but nothing there.

Nicolas
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Notice that $H\cap K$ is also a normal subgroup of $G$. Let $\bar G =G/H\cap K$. Notice also that $\bar P$ is also a sylow $p$ subgroup of $\bar G$.

$\bar P \bar H \cap \bar P \bar K=\bar P (\bar H\cap \bar P \bar K)$ by Dedekind rule.

Let $h\in \bar H\cap \bar P \bar K$ then $h=ak$ for $a\in \bar P,k\in \bar K$. Note that the elements $h,k$ commutes with each other as $\bar H\cap \bar K=1$. Hence

$$(hk^{-1})^{|a|}=a^{|a|}=e$$ $$h^{|a|}=k^{|a|}\implies h^{|a|}=e$$

Hence $h$ is element of prime power order. Thus, $ \bar H\cap \bar P \bar K$ is an $p$ group $\implies \bar P (\bar H\cap \bar P \bar K)$ is an $p$ group containing $\bar P$. By the maximality of $\bar P$, we have

$$\bar P \bar H \cap \bar P \bar K=\bar P$$ Hence $$PH\cap PK=P(H\cap K)$$

Note For Dedekind Rule, check here.

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mesel
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    Thanks very much Mesel. That is a nice proof actually. – Nicolas Mar 21 '15 at 18:30
  • @Nicolas: You are welcome. – mesel Mar 21 '15 at 18:34
  • I just add a few notes for myself mostly. $\overline{P}$ is a Sylow of $\overline{G}$ as it is the image by the canonical surjection $G: \rightarrow G/{H\cap K}$ and this map is surjective. Then H and K normal in G implies that their bar images are also normal in $\overline{G}$. Hence $[\overline{H},\overline{K}] \subset \bar{H} \cap \bar{K}=1$. Last, $\overline{PH \cap PK} \subset \overline{PH}\cap \overline{PK} = \overline{P}$ therefore $PH \cap PK \subset P(H\cap K)$. If I am not making any mistake. – Nicolas Mar 21 '15 at 18:37
  • @Nicolas: yes your arguments are true. – mesel Mar 21 '15 at 18:44
  • @Nicolas: If you think that the given answer is true, you should mark it as an accepted answer. – mesel Mar 22 '15 at 10:43
  • Dear mesel, I have happily accepted your answer (sorry i just found out that I can accept answers). Thanks again for your answer, a brilliant application of the Dedekind Rule. – Nicolas Mar 22 '15 at 11:25
  • @Nicolas: I suggest you to have look at http://math.stackexchange.com/tour . – mesel Mar 22 '15 at 22:53