# Intersections of Sylow p-groups

1. Dec 7, 2011

### Bachelier

I want to understand this

If there exists more than one Sylow-p-subgroup of order p then for all these subgrps, their intersection is {e} the identity.

However if If there exists more than one Sylow-p-subgroup of order pk s.t. k>0, then their intersection is not necessarily the identity element.

Is this correct? Can someone provide a quick explanation and proof please?

Does it have to do with homomorphisms to permutation groups?

2. Dec 7, 2011

### mathwonk

the intersection of two subgroups is a subgroup of both. do you know the relation between the order of a group and the order of its subgroups?

3. Dec 7, 2011

### Bachelier

the order of the intersection grp must divide order of G, but it cannot be equal or larger than the order of the other intersecting p-subgroups .

Last edited: Dec 7, 2011
4. Dec 8, 2011

### Deveno

let's find an example of this, and then it will certainly show it is true, right?

so consider the dihedral group D6, of order 12. a sylow 2-subgroup of D6, would be of order 4. let's see if we can find 2 with non-trivial intersection.

let H = {1,r3, s, r3s}. since r3 is in the center, r3 and s commute, so this defines an abelian subgroup of order 4. now we need to find another one.

let K = {1,r3, rs, r4s}. to prove this is a group, we only need to show that r3 and rs commute.

r3(rs) = r4s (d'oh!)
(rs)r3 = (sr5)r3 = sr2 = r4s

(since srk = (rk)-1s).

note that H∩K = {1,r3}, which is non-trivial.

(the first half of your statement is obvious, any two groups of prime order must either conincide or intersect trivially, since the intersection would be a subgroup of both groups).

5. Dec 8, 2011

### Bachelier

Deveno, I haven't read your answer yet, but I was reviewing the Sylow chapter and recognized I made a mistake last night. The intersection must be a subgroup of each sylow p-group hence must divide the order of each sylow-p group. (i.e. pk)