Center and abelianization of a finite p-group When working on a problem on characters of finite groups, I encounter the following. At the moment I don't have a good idea to solve it. Any help is appreciated.
Let $P$ be a $p$-group, where $p$ is a prime. Is there any relation between $\exp(Z(P))$ and $\exp(P/P')$, where $\exp$ is the exponent, $Z(P)$ is the center, and $P'$ is the commutator subgroup? Is it true that $\exp(P/P')\leq \exp(Z(P))$, or even that $\exp(P/P')= \exp(Z(P))$? Thank you in advance.
 A: You can have equality, or you can have either inequality.

*

*For $\exp(P/P')=\exp(Z(P))$: take your favorite $p$-group $Q$, of order $p^n$. Let $C_{p^m}$ be a cyclic group of order $p^m$, with $m\gt n$. Then $P=Q\times C_{p^m}$ has center of exponent $p^m$, and abelianization isomorphic to $(P/P')\times C_{p^m}$, which has exponent $p^m$ as well. Alternatively, any nonabelian group of order $p^3$ and exponent $p$ (for $p$ odd), or $D_8$ or $Q_8$ for $p=2$.


*For $\exp(P/P')\gt \exp(Z(P))$: let $p$ be an odd prime. Consider the set of all matrices of the form
$$\left(\begin{array}{ccc}
1 & \alpha & \gamma\\
0 & 1 & \beta\\
0 & 0 & 1
\end{array}\right)$$
where $\alpha$ and $\beta$ are taken in $\mathbb{Z}/p^n\mathbb{Z}$, and $\gamma$ is taken in $\mathbb{Z}/p^k\mathbb{Z}$, with $1\leq k\lt n$. Multiplication is then given by
$$\left(\begin{array}{ccc}
1 & a & c\\
0 & 1 & b\\
0 & 0 & 1\end{array}\right) \left(\begin{array}{ccc}
1 & r & t\\
0 & 1 & s\\
0 & 0 & 1
\end{array}\right) = \left(\begin{array}{ccc}
1 & a+r & c+t+a s\\
0 & 1 & b+s\\
0 & 0 & 1
\end{array}\right).$$
It is easy to verify that the abelianization is isomorphic to $\mathbb{Z}/p^n\mathbb{Z}\times\mathbb{Z}/p^n\mathbb{Z}$, hence is of exponent $p^n$. The center consists of the matrices in which both $\alpha$ and $\beta$ are multiples of $p^k$, so it is isomorphic to $\mathbb{Z}/p^{n-k}\mathbb{Z}\times \mathbb{Z}p^{n-k}\mathbb{Z}\times\mathbb{Z}/p^k\mathbb{Z}$. This has exponent $p^{\max(n-k,k)}$. Since $1\leq k\lt n$, then this is strictly smaller than $p^n=\exp(P/P')$.


*For $\exp(P/P')\lt \exp(Z(P))$, let $p$ be an odd prime, and
$$\begin{align*}
Q &= \langle x,y,z\mid x^p=y^p=z^p=1,\ xz=zx,\ yz=zy,\ yx=xyz\rangle\\
R &= \langle w\mid w^{p^2}=1\rangle\\
P &= (Q\times R)/\langle (z,w^{-p})\rangle.
\end{align*}$$
So $Q$ is the nonabelian group of order $p^3$ and exponent $p$, $R$ is cyclic of order $p^2$, and $P$ is the group obtained from $Q$ by adjoining a central $p$th root to $z$; call this root $r$ (that is, $r$ is the image of $(1,w)$ in the quotient). The center of $P$ is generated by $r$, hence has exponent $p^2$; the commutator subgroup of $P$ is generated by the image of $z$, so the abelianization is isomorphic to $(\mathbb{Z}/p\mathbb{Z})\times(\mathbb{Z}/p\mathbb{Z})\times(\mathbb{Z}/p\mathbb{Z})$, of exponent $p$.
