0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ What you seem to have writte is not the lower but rather the upper central series. $\endgroup$ – DonAntonio Mar 23 '16 at 15:37
  • $\begingroup$ yes! thank you! $\endgroup$ – BetaY Mar 23 '16 at 15:42
  • $\begingroup$ In the dihedral group of order $16$, $Z(G)$ has order $2$ but $G/Z(G)$ is dihedral of order $8$ and has exponent $4$. $\endgroup$ – Derek Holt Mar 23 '16 at 15:49
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.