Group of p-power order isomorphism

[Locoism]

Homework Statement

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.

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...

 

[Kreizhn]

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 p^2 (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 \mathbb Z/p \times \mathbb Z/p 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?

 

[Deveno]

Homework Statement

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.

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...

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...