Are Intersections of Sylow p-Groups Always Trivial?

[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 p^k 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?

 

[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?

 

[Bachelier]

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?

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 .

 

[Deveno]

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 p^k 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?

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 sr^k = (r^k)^-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).

 

[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 .

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. p^k)