Special implies center is elementary abelian

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Statement

In a special group, the center is an elementary Abelian group

Proof

Let G be the group and Z = Z(G) = G' = \Phi(G) be the subgroup which is simultaneously the center, commutator subgroup and Frattini subgroup.

The Frattini part

The fact that Z is the Frattini subgroup is used up in the observation that G/Z is elementary Abelian.

The commutator subgroup part

The fact that Z is the commutator subgroup is used in the observation that commutators viz elements of the form [x,y] where x,y \in G, generate Z.

The center part

Since Z is the center, the map G \times G \to G given by (x,y) \mapsto [x,y] descends to a map from G/Z \times G/Z to G. As observed earlier, G/Z is elementary Abelian, so x^p \in Z for any x \in G.

Now since the center is the same as the commutator, the commutator map is a bihomomorphism, and we have:

[x,y]^p = [x^p,y] = 1

Since x^p \in Z by the above observation. Thus every commutator has order p or 1.

Putting the pieces together

We have the following facts:

  • Z is Abelian
  • Z is generated by elements having order p

Putting the pieces together, we see that every element in Z has order p so Z is elementary Abelian.