if $Q$ and $P$ are distinct $p$-Sylow subgroups then $Q\not\subseteq N_G(P)$.

Question

I have been told to use the following to prove another claim, but I would like to prove this anyway for myself. However I can't tell why it's true. I think it's true, but can't see why! Here it is:

If $P$ and $Q$ are distinct $p$-Sylow subgroups in a group $G$ then $Q\not\subseteq N_G(P)$.

Could I use the theorem below? If so, how? I mean $Q$ isn't a $p$-subgroup, which is why I'm unsure.

If $G$ is a group, $P \leq G$ is a $p$-Sylow subgroup, for some prime $p$, then for each $H \leq G$ a $p$-subgroup such that $H \leq N_G(P)$ then $H \leq P$.

Thanks!

Answer 1

Of course you can apply this theorem. Any $p$-Sylow group is a $p$-subgroup per definition.

Here is a simple proof of your claim: If $Q \leq N_G(P)$ then its quotient $QP/P$ is a $p$-subgroup of $N_G(P)/P$. Since $|N_G(P):P|$ divides $|G:P|$ and $P$ is a $p$-Sylow group, $p$ does not divide the group order, so there is no nontrivial $p$-subgroup of $N_G(P)/P$. Hence $QP/P$ is trivial which means $Q \leq P$.

Answer 2

Distinct Sylow-p groups are conjugate. If two subgroups are conjugate one cannot be contained in the normalizer of the other.

Answer 3

There is still another approach. If $Q$ would normalize $P$, then $PQ$ is a subgroup of $N_G(P)$, since of course $P \subseteq N_G(P)$. But $|PQ|=|P||Q|/|P \cap Q|$, which shows that $PQ$ is a $p$-subgroup of $N_G(P)$, so it must lie in some Sylow $p$-subgroup of $N_G(P)$. Since $P$ is normal in $N_G(P)$, it is the only Sylow $p$-subgroup of $N_G(P)$, so $PQ \subseteq P$, whence $PQ=P$, which is equivalent to $Q \subseteq P$. But $|P|=|Q|$, and we get $P=Q$, contradicting $P$ and $Q$ being distinct.