Let \(P\in \operatorname{Syl}_p(G)\) where \(H\trianglelefteq G\) and show that \(P\cap H \in \operatorname{Syl}_p(H)\).
Solution
Let \(P\in\operatorname{Syl}_p(G)\) and \(H\trianglelefteq G\).
1. The subgroup \(P\cap H\) is a \(p\)-subgroup of \(H\).
It is a subgroup of the \(p\)-group \(P\), so its order is a power of \(p\).
2. Its index in \(H\) is prime to \(p\).
Since \(H\trianglelefteq G\), the product \(PH\) is a subgroup. The product formula gives \[ |PH|=\frac{|P||H|}{|P\cap H|}. \] Hence \[ [H:P\cap H] =\frac{|H|}{|P\cap H|} =\frac{|PH|}{|P|} =[PH:P]. \] Because \(P\le PH\le G\), \[ [G:P]=[G:PH][PH:P]. \] Thus \([H:P\cap H]=[PH:P]\) divides \([G:P]\). Since \(P\) is Sylow, \([G:P]\) is prime to \(p\), so \([H:P\cap H]\) is prime to \(p\).
A \(p\)-subgroup of \(H\) whose index is prime to \(p\) has the full \(p\)-part of \(|H|\). Therefore \[ \boxed{P\cap H\in\operatorname{Syl}_p(H).} \]