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.