Sylow p-subgroup of order p does not normalize any other Sylow p-subgroup

Question

Let $P_1,P_2$ be distinct Sylow p-subgroups of $G$ with order $p$.

Is it generally true that $P_1$ cannot normalize $P_2$?

I've seen algebra textbooks use this fact for $p=3,11$ and they quote 'a consequence of Sylow's Theorem' yet I cannot see how this can hold. Any help is appreciated.

Answer 1

In general it's true that no Sylow $p$-subgroup can normalize a different one (though particular elements can normalize both, not every element can). This is because a Sylow subgroup is normal in its normalizer and therefore is the only subgroup of that order in the normalizer (by the fact that all Sylow subgroups for the same prime are conjugate).

Specializing to your problem, if any nonidentity element in a Sylow $p$-subgroup of order $p$ normalized a different Sylow $p$-subgroup, then every element would because every nonidentity element is a generator. Therefore this can't happen.

Answer 2

A standard argument is:

Suppose $P$ is a $p$-Sylow group, and $Q$ is a group of order a $p$-power, normalizing $P$. Then $PQ$ is also group (because $Q$ normalizes $P$), which contains $P$, and has order a power of $p$. Hence, $PQ = P$, since $P$ is maximal with that property. Therefore $Q$ is a subgroup of $P$.

In particular, the only $p$-Sylow group to normalize $P$ is $P$ itself.