I was studying the group theory book by John S. Rose and this is the problem 257, which is driving me nuts. It looks so easy, but i can't solve it.
Let $G$ be a finite group, $H$, $K$ be normal subgroups and $P$ a sylow p-subgroup . Then $PH \cap PK=P\left( H\cap K\right)$.
Of course, one incluision is trivial: if $x\in P\left( H\cap K\right)$ there exists $p\in P$ and $y\in H\cap K$ such that $x=py$, but $py\in PH$ and $py\in PK$, so $x\in PH \cap PK$.
But i don't know how to proceed with the other inclusion. This is my work so far:
Since we are working with a sylow subgroup, the only information we have is related with the order of the group, so maybe we can prove that both sides have the same cardinality and we will be done.
Let $\lvert G \rvert= p^nr$ with $p\nmid r$, $\lvert H \rvert= p^hs$ with $p\nmid s$ and $\lvert K \rvert= p^kl$ with $p\nmid l$. By definition of $P$ we have that $\lvert P \rvert= p^n$ and since $H$ and $K$ are normal subgroups, $H\cap P$ and $K\cap P$ are sylow p-subgroups of $H$ and $K$ respectively, so $\lvert H\cap P \rvert= p^h$ and $\lvert K\cap P \rvert= p^k$.
Now, I don't know what to do, I tried to work with the formula for the order of the product of subgroups and with the index, but I got no results.
I'll appreciate your help, many thanks.
Best Answer
Lemma Let $H,K \unlhd G$ and $P \in Syl_p(G)$. Then $(P \cap H)(P \cap K)=P \cap HK$.
Proof We use a counting argument to show that indeed $(P \cap H)(P \cap K)=P \cap HK$. Observe that $|(P\cap H)(P\cap K)|=\frac{|P\cap H| \cdot|P\cap K|}{|P\cap H\cap K|}=\frac{|H|_p \cdot |K|_p}{|P \cap H \cap K|}$, where the index $p$ denotes the largest power of $p$, dividing the order of the group between its $|\cdot|$. Now, $|P \cap H \cap K| = |H \cap K|_p$, since $P \cap H \cap K$ is a Sylow $p$-subgroup of the normal subgroup $H \cap K$.
Hence, $$|(P\cap H)(P\cap K)| = \frac{|H|_p \cdot |K|_p}{|H \cap K|_p}=\left(\frac{|H| \cdot |K|}{|H \cap K|}\right)_p=|HK|_p.$$
Since the $p$-subgroup $(P \cap H)(P \cap K) \subseteq P \cap HK \in Syl_p(HK)$ (note $HK$ is also normal in $G$), this forces $(P \cap H)(P \cap K)=P \cap HK$. $\square$
Proposition Let $H,K \unlhd G$ and $P \in Syl_p(G)$. Then $PH \cap PK=P(H \cap K)$.
Proof Note that the sets $PH, PK, PHK$ are in fact all subgroups by the normality of $H$ and $K$. Let us calculate the order of $PH \cap PK$: $$|PH \cap PK|=\frac{|PH||PK|}{|PHK|}=\frac{|P||H||P||K|}{|P \cap H||P\cap K||PHK|}=\frac{|P||H||P||K|}{|P \cap H||P \cap K|} \cdot \frac{|P \cap HK|}{|P||HK|}\\=\frac{|P||H||P||K|}{|P \cap H||P \cap K|} \cdot \frac{|H \cap K||P \cap HK|}{|P||H||K|}=\frac{|P||H \cap K||P \cap HK|}{|P \cap H||P \cap K|}= \text { (using the Lemma) }\\ \frac{|P||H \cap K||P \cap H||P \cap K|}{|P \cap H||P \cap K||P \cap H \cap K|}=\frac{|P||H \cap K|}{|P \cap H \cap K|}=|P(H \cap K)|.$$ Since of course $P(H \cap K) \subseteq PH \cap PK$, the two subgroups must be equal. $\square$