The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$

Group Theory Problems and Solutions in Mathematics

Problem 124

Let $p$ be a prime number.
Let $G$ be a non-abelian $p$-group.
Show that the index of the center of $G$ is divisible by $p^2$.
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Proof.

Suppose the order of the group $G$ is $p^a$, for some $a \in \Z$.
Let $Z(G)$ be the center of $G$. Since $Z(G)$ is a subgroup of $G$, the order of the center is also a power of $p$, that is, $|Z(G)|=p^b$, for some $b \in \Z$.
Then we have the index $[G: Z(G)]=p^{a-b}$.

If $a-b=0$, then we have $G=Z(G)$ and $G$ is an abelian group. This contradicts with the assumption that $G$ is non-abelian. So $a-b \neq 0$.

If $a-b=1$, then the order of the quotient $|G/Z(G)|=[G:Z(G)]=p$ is a prime, thus $G/Z(G)$ is a cyclic group.
Recall that if the quotient by the center is cyclic, then the group is abelian.
Thus the group $G$ is abelian, which again a contradiction.

Therefore, we must have $a-b \geq 2$, hence $p^2$ divides the index $[G: Z(G)]=p^{a-b}$.
This concludes the proof.


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