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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

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    The $2$-Sylow subgroups of the symmetric group $S_4$ embedded in $S_5$ are also $2$-Sylow subgroups of the $S_5$. Now take two different embeddings of the $S_4$ (one fixing $5$, the other $4$). Their intersection is $S_3$ (fixing $4$ and $5$) having $2$-Sylow subgroups generated by a $2$-cycle. Now choose in both $S_4$'s a $2$-Sylow subgroup containing let's say the $2$-cycle $(1 2)$. Do you see why none of the $2$-Sylow subgroups normalizes (=centralizes, as we speak about $Z_2$) this intersection? – 2017-02-10
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    So, if I have understood, you take the intersection of two different $D_4$ embedded in $S_5$ and if you take one fixing $5$ and the other $4$, then the intersection is a 2-group in $S_3$ which has to be a transposition, $(12)$ for instance. Then if you act coniugating, you can send (can you really do this? not so sure aboout that) one of the two groups in a way that now the intersection isn't $(12)$ but the other one in $D_4$. So (if it exists a coniugation that does this) it is not true that the intersection of the two 2-Sylows contains $D_4$. But where's the error in the reasoning above? – 2017-02-10
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    Some elements of $D_4$ will map $(1 2)$ to different $2$-cycle, let's say $(3 4)$, so the intersection with $P_k$ won't be $(1 2)$ but $(3 4)$. – 2017-02-10
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    Yeah indeed, I don't know what I was thinking, also the reasoning at the bottom of my question is wrong. The coniugate of $P_1 \cap P_2$ by $P_1$ is in $P_1$, but you can't say it is exactly $P_1 \cap P_2$, it could be something isomorphic (as you showed using the transpositions in $D_4$). – 2017-02-10

2 Answers 2

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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.

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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.