Let
$$1=G_0\leq G_1\leq … \leq G_{n-1} \leq G_n = G$$
be a central series of the group $G$. That is, $G_{i-1}/G_i\leq Z(G/G_i)$ for all $i$.
Let
$$1=Z_0(G)\leq Z_1(G)\leq … $$
$$… \leq \gamma_2(G)\leq \gamma_1(G)=G$$
be the upper central series and the lower central series of a group $G$, respectively. I have proved $$G_i\leq Z_i(G), \gamma_{i+1}(G)\leq G_{n-i}$$
I was wondering if there is any example of a group such that has a central series which is different from the upper and the lower central series.
Also I would like to prove that for a nilpotent group of nilpotency class $c$, $$\gamma_{c+1-i}(G)\leq Z_i(G)$$ but I do not see it. (For this, I have been thinking on using that a nilpotent group of class $c$ satisfies $$Z_c(G)=G,\gamma_{c+1}(G)=1$$ would that be useful?)
Any help?
Best Answer
Consider $G=V\times D_8$ (direct product of Klein four group with the group of the square) of order 32. The lower central series is $1<1\times Z(D_8)<G$. The upper series is $1<V\times Z(D_8)<G$. And three additional central series are given by $1<H\times Z(D_8)<G$ where $H$ is any of the three order 2 subgroups of $V$. Thus $V\times D_8$ is an example of what you are looking for.
Note that I have answered the strict version of this problem, where the additional central series have the same lengths as the upper and lower series. If they are allowed to be longer, the problem is even easier.
As for why the upper series' terms always contain the corresponding lower series, this is a standard result you can find in many textbooks. The essential idea is to start at the top, noting that $G'=\gamma_2(G)$ is the smallest normal subgroup with abelian quotient, so that $Z_{c-1}(G)\ge \gamma_2(G)$. And continue by an easy induction.