Let $S$ be a normal p-subgroup of a finite group $G$. Prove that $S \subseteq P$ for every Sylow p-subgroup $P$ of $G$.
Now, I know that this involves the Sylow Theorems, of course. This is very new to me but my thinking so far is this.
S is a normal p-subgroup of a finite group G, so we can say that S has some order of a power of p, $|S|=p^k$ where p is prime and S is a subgroup of some Sylow p-subgroup $K$ (by the second Sylow Th). If I let $P$ be any other Sylow p-subroup, we must show $S \subseteq P.$
Using the third Sylow Theorem, we can state that $K$ and $P$ are conjugates. Thus, there exists $x \in G$ s.t. $xKx^{-1}=P$ Thus, $xSx^{-1} \subseteq P$. But, since S is normal in G, we have $xSx^{-1}=S$.
Please let me know if this is sound theory and if not, please offer corrections.
Best Answer
Yes, exactly that's the way here to use the Sylow theorems.