classify all groups of order P2 where p : prime ?

Question

note here P2 is p square .

Answer 1

Call your group G, with |G| = p² (for some prime p). Cauchy's theorem tells us that some element g in G whose order is p.

is a subgroup of G of order p. By the Sylow theorems, it's the only such subgroup, in fact.

Because || is half |G|, we know it is a subgroup (normal even). Importantly, it's a maximal subgroup - no maximal subgroup can be the center of the group, and no group divisible by a single prime can have a trivial center.

The center Z(G) of this group must be a subgroup, but we know that Z(G) has order 1, p, or p². Since it's not trivial, it's not order 1. Since it can't be , it's not order p. Thus the center is order p².

From this, we can determine that G is abelian - the center of Gis the portion of G that is commutative.

From that, we can conclude that G looks like Zp² or Zp×Zp. Those are the only two abelian groups of order p² (due to Krönecker decompositions).

Answer 2

I believe the only two answers are the cyclic group of order p^2 and the product of the cyclic group of order p with itself.

To attempt a proof:

Clearly, all elements are of order 1, p, or p^2. And so there are only two possibilities:

A. There's an element of order p^2, in which case we're done.

B. All non-identity elements have order p.

I don't immediately see the proof for case B, but usually conjugacy classes play a role somewhere.

Answer 3

just to make sparkling: the case |Z(G)| = p, does no longer ensue, because of the fact it finally finally ends up in a contradiction. so G/Z(G) cyclic, mutually as |G| = p^2, forces us to have G/Z(G) = {Z(G)} the id subgroup of G/Z(G), so Z(G) = G.