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let $G$ be a group of order divisible by $p,q$. for any $p$-sylow subgroup $P$ and $q$-sylow subgroup $Q$. I know the intersection $P\cap Q = \{e\}$ because the orders are different. what about two $p$-sylow subgroups, can I say their intersection is only $e$?

thanks. benny.

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See this question. –  Mikko Korhonen Feb 21 '12 at 17:25
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As a simple counterexample, consider the subgroups $<r^3,s>$, $<r^3,rs>$ and $<r^3,r^2s>$ of $D_{12}$. –  Carl Feb 21 '12 at 17:38
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If P is the direct product of two cyclic groups of order 4, then either P is the only Sylow 2-subgroup of G, or there is another Sylow 2-subgroup of G intersecting P in more than the identity. This basically follows from Sylow's theorem and is a nice example of a group that controls its embeddings very well. Brauer even classified groups with P as a Sylow 2-subgroup (using more than Sylow theorems). –  Jack Schmidt Feb 21 '12 at 18:49
    
It is not enough for the orders of two subgroups to be "different" in order to conclude that their intersection is trivial. You need to use the fact that their orders are relatively prime. –  Arturo Magidin Feb 21 '12 at 20:25
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1 Answer

up vote 6 down vote accepted

Firstly, having different orders does not guarantee trivial intersection. For instance in a cyclic group of order $4$, we subgroup of order $2$ and a subgroup of order $4$ will intersect in a subgroup of $2$. What guarantees trivial intersection is coprime orders.

Will two Sylow $p$ subgroups intersect trivially always?

Definitely not. There are plenty of examples, where intersection of two Sylow $p$-Subgroups is not identity.

Firstly, two Sylow $p$-Subgroups intesect trivially if the order of the Sylow is prime. That is, the maximal exponent of $p$ in $|G|$ is $1$. The proof of this fact follows from Lagrange's Theorem. However, I give no guarantee about the converse.

Another related question on when two Sylow Subgroups have trivial intersection is here.

However, you can still say something about the index of intersection of Sylow Subgroups. For instance, look here. Related is a fact called Poincare's Theorem which is stated in many forms. One of them is:

Let $G$ be a finite simple group. Let $H$ be a proper subgroup of $G$. Then the order of $G$ divides $[G:H]!$

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thanks. great answer!! –  benjamin Feb 21 '12 at 17:52
    
@benjamin I am glad to have been of some help. :-) –  user21436 Feb 21 '12 at 17:53
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