Is this property Dedekind modular?

Question

Let $G$ be a finite group with $H,K\trianglelefteq G$ and $P\in Syl_p(G)$. I want to show that \begin{align} HK\cap P&=(H\cap P)(K\cap P) \newline HP\cap KP&=(H\cap K)P \end{align} I know the Dedekind Modular Law says that if $A,B,C\leq G$ and $B\leq C$ then $$AB\cap C=(A\cap C)B $$ but I have no clue how to apply it in my problem. Could someone give me some help ? Thank you

Answer 1

$\textbf{Theorem 1}$ Let $H,K \unlhd G$ and $P \in Syl_p(G)$. Then $(P \cap H)(P \cap K)=P \cap HK$.

$\textbf{Proof}$ 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 $(P \cap H)(P \cap K)=P \cap HK$. $\square$

$\textbf{Theorem 2}$ Let $H,K \unlhd G$ and $P \in Syl_p(G)$. Then $PH \cap PK=P(H \cap K)$.

$\textbf{Proof}$ 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 $PH \cap PK$: $$|PH \cap PK|=\frac{|PH||PK|}{|PHK|}=\frac{|P||H||P||K|}{|P \cap H||P\cap K||PHK|}=\frac{|P||H||P||K|}{|P \cap H||P \cap K|} \cdot \frac{|P \cap HK|}{|P||HK|}$$

$$=\frac{|P||H||P||K|}{|P \cap H||P \cap K|} \cdot \frac{|H \cap K||P \cap HK|}{|P||H||K|}=\frac{|P||H \cap K||P \cap HK|}{|P \cap H||P \cap K|}$$

$$= \text { (using Theorem 1) } \frac{|P||H \cap K||P \cap H||P \cap K|}{|P \cap H||P \cap K||P \cap H \cap K|}=\frac{|P||H \cap K|}{|P \cap H \cap K|}=|P(H \cap K)|.$$ Since of course $P(H \cap K) \subseteq PH \cap PK$, the two subgroups must be equal. $\square$

Note The following is also true but somewhat harder to prove.

$\textbf{Theorem 3}$ Let $G=HK$, $H$,$K$ subgroups of $G$ and let $p$ a prime dividing the order of $G$. Then there exists a $P \in Syl_p(G)$ such that $P=(P\cap H)(P \cap K)$, with $P \cap H \in Syl_p(H)$ and $P \cap K \in Syl_p(K)$.