Special implies center is elementary abelian

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.