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Group of p-power order isomorphism

  1. Oct 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Let G be a group of order p2, where p is a positive prime.

    Show that G is isomorphic to either Z/p2 or Z/p × Z/p.

    3. The attempt at a solution

    Am I completely wrong here or is this just the definition of a p-Sylow subgroup? what I mean is that if g is of order p2 then there is a subgroup of order p and of order p2, which are isomorphic to Zp and Zp2 (respectively).
    Also, if the Sylow group is isomorphic to Zp, it is abelian, would that consequently make G abelian? Not too sure how to put all this into mathematical form...
  2. jcsd
  3. Oct 28, 2011 #2
    Well, by definition G is a p-group and the only non-trivial Sylow p-subgroup of a p-group is the whole group itself.

    There is clearly a subgroup of order [itex] p^2 [/itex] (G itself), but a priori there is no reason to assume it's cyclic. I assume you're using Cauchy's theorem to say there is a subgroup of order p, and certainly it must be cyclic. You could use this to argue that the group is isomorphic to [itex] \mathbb Z/p \times \mathbb Z/p [/itex] but there is an easier way.

    Do you know about the class equation? Can you show that the center of a p-group is always non-trivial?
  4. Oct 28, 2011 #3


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    since G is a p-group, it IS it's own p-Sylow subgroup. so the sylow theorems won't help you here.

    try this: either G has an element of order p2, or it doesn't, so...
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