Application of a property about nilpotent group

There is a quite fundamental theorem about the nilpotent groups that if $Z_1 \leq Z_2 \leq \cdots \leq Z_m=G$ is the upper central series of $G$ and suppose that $Z(G)$ has finite exponent dividing $e$, then so does $Z_i/Z_{i-1}$.

Does it follows from here that $G/Z(G)$ also has finite exponent dividing $e$.

I am thinking a potential argument using $G/Z_{i} \cong (G/Z_{i-1})/ (Z_i/Z_{i-1})$ inductively. Is this feasible?

• What you seem to have writte is not the lower but rather the upper central series. – DonAntonio Mar 23 '16 at 15:37
• yes! thank you! – BetaY Mar 23 '16 at 15:42
• In the dihedral group of order $16$, $Z(G)$ has order $2$ but $G/Z(G)$ is dihedral of order $8$ and has exponent $4$. – Derek Holt Mar 23 '16 at 15:49

Consider the following group: $$G= \begin{Bmatrix} \begin{bmatrix} 1 & * & * & \cdots & *\\ & 1 & * &\cdots & *\\ & & \ddots &\cdots & *\\ & & & & 1 \end{bmatrix}_{n\times n}\colon *\in\mathbb{Z}_p \end{Bmatrix}.$$ The upper central series is interesting for this group: $$1 \leq \begin{Bmatrix} \begin{bmatrix} 1 & 0 & \cdots & 0 & *\\ & 1 & \cdots & 0 & 0\\ & & \ddots &\cdots & 0\\ & & & & 1 \end{bmatrix} \end{Bmatrix} \leq \begin{Bmatrix} \begin{bmatrix} 1 & \cdots & 0 & * & *\\ & 1 & \cdots & 0 & *\\ & & \ddots &\cdots & 0\\ & & & & 1 \end{bmatrix} \end{Bmatrix} \leq \cdots$$ [fill up anti-diagonals from right corner successively].
Each section $Z_i/Z_{i-1}$ is elementary abelian $p$-group, but you can arbitrarily increase the exponent by increasing the size of matrices.