normalizer of two p-Sylow intersection

Question

Let $N(P_1 \cap P_2)$ be the intersection of 2 p-Sylow, $P_1$ and $P_2$. I have 2 questions (which I put in a single question here because connected, and I tried to prove the last one).

First of all, given a group, is the intersection between p-Sylows always the same? (isomorphically) So if for instance I find two 2-Sylows of cardinality 8 whose intersection is a $\mathbb{Z}_2$, do I have that every intersection of every 2-Sylow is isomorphic to $\mathbb{Z}_2$? I had thought that if the action on the set of p-Sylow is double transitive then it's trivial, but is there some weaker criterion?

Then I was wondering if given the example above it is always true that $P_1<N(P_1 \cap P_2)$, because my teacher once used this fact, but I am not sure if it is a general property or it worked only in the specific case. I have thought that since the conjugate of $P_1$ by the action of $P_1$ is itself, then the elements of $P_1 \cap P_2$ are bound to go on $P_1$, and so $P_2$ is bound to go on a $P_k$ whose intersection with $P_1$ is again $P_1 \cap P_2$. Would this be enough to prove that we always have $P_1<N(P_1 \cap P_2)$?

Answer 1

The intersection of Sylow $p$-subgroups does not have to be the same. Let $F = \mathbb{Z}/p\mathbb{Z}$, and let $G = \textrm{GL}_3(F)$, the group of $3$ by $3$ matrices with entries in $F$ whose determinant is nonzero. The order of $G$ is $$p^3(p-1)(p^2-1)(p^3-1)$$

which shows you that the subgroup

$$P = \{ \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} : a, b ,c \in F\}$$

of order $p^3$ is a Sylow $p$-subgroup of $G$. Let

$$w_0 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}$$

$$w = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

Then let

$$P_1 := w_0Pw_0^{-1} = \{\begin{pmatrix} 1 & 0 & 0 \\ a & 1 & 0 \\ b & c & 1 \end{pmatrix} \}$$

$$P_2 = wPw^{-1} = \{ \begin{pmatrix} 1 & 0 & c \\ a & 1 & b \\ 0 & 0 & 1 \end{pmatrix}\}$$

So we have three Sylow $p$-subgroups $P, P_1 , P_2$, with $P \cap P_1$ trivial, but $P \cap P_2$ has order $p^2$. I'm not sure about your question with the normalizer, I'll have to think about it more.

Answer 2

It is not in general true that the every two Sylow subgroups are contained in the normalizer of their intersection. It is however true when these two subgroups are abelian. Indeed, use the fact that every two Sylow subgroups of a group are conjugate with each other and the commutativity of the subgroups(it is really straightforward). So, I guess that in the exercise your professor assigned you the subgroups were small and their commutativity should be apparent. Otherwise, it is a mistake to infer that.