Is it always true that intersection of two Sylow p-subgroups is trivial?

Question

I would like to investigate a few properties of Sylow p-subgroup, especially, when $p^n \;|\; |G|$ for $n > 1$.

If $n = 1$ and $P_1, P_2$ are Sylow $p$-subgroups, then if $id \neq g \in P_1 \cap P_2$, then $\{g, \cdots, g^p\} = P_1 = P_2$, so the proof is done. I would like to know what happens if $n \geq 2$; I cannot use the same trick as before since there's is no guarantee that a Sylow p-subgroup $P \cong \mathbb Z/p^n$

(I am aware of the result that if $P_1, P_2$ are Sylow $p$-subgroup, then $P_1$ and $P_2$ are conjugate to each other. If I were to use it, then it is sufficient to show that if $H_1, H_2 < G$ such that $H_1, H_2$ are conjugates, then $H_1 \cap H_2 = \{id\}$ or $H_1 = H_2$. But how, and is it even true? So really, I am asking two separate questions here.)

@ancientmathematician $P_8 \subset S_4$, so possible candidate of $P_8$ is $D_4$ ($Q_8$ is not a subgroup of $S_4$.) — James C, Nov 23 '20 at 14:54
True. But why not look and see which it is? If the example is too big try $S_3\times S_2$ with $p=2$. — ancient mathematician, Nov 23 '20 at 14:55
@ancientmathematician Brilliant. $\Big({id, (13)} \times {id, (12)}\Big) \cap \Big({id, (23)} \times {id, (12)}\Big) = {id \times id, id \times (12)}$, and both are Sylow 2-subgroup. So that automatically implies that the second statement is false. — James C, Nov 23 '20 at 15:02
Not only is it not always true, you can say more about the number of $p$-Sylow subgroups when you know the intersections are not trivial. See here. — Arturo Magidin, Nov 23 '20 at 21:09

Qiaochu Yuan · Accepted Answer · 2020-11-23T20:43:23.133

3

No, there's no reason this should be true. A simple conceptual reason is that the center $Z(G)$ may have nontrivial elements of order $p$, in which case the Sylow $p$-subgroup of $Z(G)$ is a subgroup of every Sylow $p$-subgroup. So we can consider, for example, the Sylow $2$-subgroups of $C_2 \times S_3$, which has center $C_2$, as ancientmathematician says in the comments. More generally the same argument works if $G$ has a normal $p$-subgroup.

edited Nov 23 '20 at 20:43

answered Nov 23 '20 at 19:44

Qiaochu Yuan

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@QiachuYuan I will call $Z_p$, the Sylow $p$-subgroup of $Z(G)$. What's the reason that every Sylow p-subgroup must contain $Z_p$? – James C Nov 24 '20 at 01:52
@James: Sylow $p$-subgroups are maximal with respect to inclusion (this is either a definition or part of the Sylow theorems) so every $p$-subgroup is contained in a Sylow. Conjugating, every normal $p$-subgroup is contained in every Sylow. – Qiaochu Yuan Nov 24 '20 at 02:03
Got it. Thanks. – James C Nov 24 '20 at 02:05

Is it always true that intersection of two Sylow p-subgroups is trivial?

1 Answers1