# Understanding the normalizer of a Sylow $p$-subgroup

The proofs I have seen of the second and third Sylow theorems use the following result (which my professor stated is the key as to why the Sylow theorems work):

Lemma. Let $$P$$ be a Sylow $$p$$-subgroup of a finite group $$G$$. If $$Q$$ is a $$p$$-group contained in $$N_G(P)$$, then $$Q$$ must be contained in $$P$$.

Proof. One can show that because $$Q \le N_G(P)$$, the product $$PQ$$ is a subgroup of $$G$$, and furthermore that

$$|PQ| = \frac{|P||Q|}{|P \cap Q|}.$$

Note that $$|P|$$, $$|Q|$$, and $$|P \cap Q|$$ are all powers of $$p$$, so it follows that $$|PQ|$$ is a $$p$$-subgroup. Clearly, we have $$|P| \leq |PQ|$$, and because $$P$$ is Sylow, we also have $$|PQ| \leq |P|$$. Hence, $$PQ = P$$, so $$Q$$ must be contained in $$P$$.

I can follow the proof easily, but even then, it is not clear to me as to why this lemma should be true. For instance, here are some things that seem rather arbitrary to me:

• Why does $$Q$$ have to be contained in the normalizer of $$P$$? Is there something special about the normalizer of a Sylow $$p$$-subgroup (or subgroups in general)? Is any of this related to normal subgroups?
• Why do we consider the product $$PQ$$? That is, when one is trying to prove the above result, what intuition(s) might lead to the above proof?
• What is the importance of $$p$$ being a prime and $$P$$ being a Sylow $$p$$-subgroup? Are these, perhaps, the strongest conditions one can find for this sort of behavior to be true?

Admittedly, these questions probably stem from my inexperience with the group theoretic concepts being used. That being said, does anyone have some insight into this lemma, beyond what is stated in the above proof?

1. $$P$$ is normal in its normalizer, normalizer is the biggest subgroup with this property, and it contains $$Q$$ by assumption.
2. The product $$PQ$$ turns out to be a $$p$$-subgroup containing $$P$$, so it is equal to $$P$$, and it contains $$Q$$, so $$Q$$ is in $$P$$.
3. Without $$p$$ being a prime item $$2$$ is wrong. It is also wrong if $$P$$ is not Sylow.