problem in understanding $|\frac{\Omega_1(Z_3(G))}{Z(G)}|=p^3$

Question

in some paper for certain finite p-group $G$ (p is prime number greeter than 3) we have $|\frac{\Omega_1(Z_3(G))}{Z(G)}|=p^3$ but in my understanding of $\Omega_1(A)=\langle g \in A|o(g)=p\rangle $ for every $g\in \Omega_1(Z_3(G))$ we should have $g^p=1$ then how it is possible that $g^{p^3}\in Z(G)$ (knowing that it is possible that $g^{p^3} \not =1$)
if it's possible please help me
$Z_i(G)$ is i th member of central series

Answer 1

In general $g\in \Omega_1(X)$ does not mean $g^p=1$, it just means $g=g_1\cdots g_m$ where for each $i$, $g_i^p=1$. If you're in a class 3 group and $p>3$ you can think of a Lie ring instead. Mod out the center to a get a class 2 Lie ring, which in your case is now generated by elements of order $p$. There you probably can fashion such an argument and certainly examples of such groups should exist -- think of an immediate descendant of the extra-special group of order $p^3$ and exponent $p$.