I have recently been doing many exercises on nilpotent groups and I'm having a bit of a problem using the upper central series to prove theorems. I can prove them using the lower one just fine (mainly because of it's nice relation with the canonical projection), but I'd really like to explore alternatives.
I will show what I mean with the following example:
If $H \leq Z(G)$ and $G/H$ is nilpotent, then $G$ is nilpotent
My original hunch is to use the upper central series, because it explicitly mentions the center of the group. In fact, this is the main suggestion I have seen on this site (namely, here and here). So let's give it a try:
If $\{1\} \unlhd N_1/H \unlhd N_2/H \unlhd$ … $\unlhd N_k/H \unlhd G/H$ is the upper central series for $G/H$, then $$\frac{N_i/H}{N_{i-1}/H} = Z\left(\frac{G/H}{N_{i-1}/H}\right)$$
Now we define a series for $G$ as $\{1\} \unlhd H \unlhd N_1 \unlhd$ … $\unlhd N_k \unlhd G$. Since $H \leq Z(G)$, we already have the first step in showing this is a central series.
From here, the only way I can see to proceed uses the Third Isomorphism Theorem: $$\frac{N_i}{N_{i-1}}\simeq \frac{N_i/H}{N_{i-1}/H} = Z\left(\frac{G/H}{N_{i-1}/H}\right) \simeq Z\left(\frac{G}{N_{i-1}}\right)$$
But this gives us an isomorphism, not an equality, and is thus not really what we need.
At this point, I get stuck, and end up just falling back on the lower central series.
Could anyone please explain a bit how to proceed, or how to use an argument that doesn't rely on the LCS?
Thanks in advance!
Best Answer
Hint: First try to prove that for any normal series $$1 = G_0 \triangleleft G_1 \triangleleft \dots \triangleleft G_n = G$$ we have $G_{i+1}/G_i \leq Z(G/G_i)$ if and only if $[G_{i+1}, G] \leq G_i$.
Applying the above in your case you have to show that $[N_{i+1}, G] \leq N_i$ for $i = 1,2,\dots, k-1$.