- Let \(P\in \syl_p(G)\) and let \(H\normal G\). Show that \(P\intersect H \in \syl_p(H)\).
Solution
Let \[ |P|=p^a, \qquad |H|=p^b m, \qquad p\nmid m. \] Since \(P\) is Sylow in \(G\), \(p^a\) is the largest power of \(p\) dividing \(|G|\).
1. The product \(PH\) is a subgroup of \(G\).
Because \(H\trianglelefteq G\), the product of the subgroup \(P\) with the normal subgroup \(H\) is a subgroup.
2. Write \(|P\cap H|=p^c\). Then \(c\ge b\).
By the product formula, \[ |PH|=\frac{|P||H|}{|P\cap H|} =p^{a+b-c}m. \] Since \(PH\le G\), the \(p\)-part of \(|PH|\) cannot exceed the \(p\)-part \(p^a\) of \(|G|\). Therefore \[ a+b-c\le a, \] so \(c\ge b\).
3. One also has \(c\le b\).
The intersection \(P\cap H\) is a \(p\)-subgroup of \(H\). Since \(p^b\) is the largest power of \(p\) dividing \(|H|\), Lagrange’s theorem gives \[ |P\cap H|=p^c\le p^b. \] Thus \(c\le b\).
4. Hence \(P\cap H\) is a Sylow \(p\)-subgroup of \(H\).
By 2 and 3, \(c=b\). Therefore \[ |P\cap H|=p^b, \] which is the full \(p\)-part of \(|H|\). Hence \(P\cap H\in\operatorname{Syl}_p(H)\).