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Intersections of Sylow p-groups

  1. Dec 7, 2011 #1
    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. jcsd
  3. Dec 7, 2011 #2


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    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?
  4. Dec 7, 2011 #3
    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
  5. Dec 8, 2011 #4


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    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).
  6. Dec 8, 2011 #5
    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)
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