# 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.