Nilpotent Group and Normal Series

Question

I tried some questions in abstract algebra (Groups, Rings, Modules, Fields) in January and February and I was unable to solve some. I am asking those now because I was critically ill then. I have done a graduate level course in group theory.

Prove that a group $G$ is nilpotent if and only if there is a normal series $$ G = G_0 > G_1 > · · · G_n = \{ e\}$$ such that $G_i/G_{i+ 1} < C(G/ G_{i+1})$ for every $i$.

Group is nilpotent implies there exists a n such that $C_n (G) =G$,where $C_n(G) $ is inverse image of $C(G/ C_{n-1}(G))$. But i was unable to relate it to normal series.

Conversely,if a normal series is given such that $G_i/ G_{i+ 1} < C(G/ G_{i+1})$ I am still unable to relate it with $C_n (G)$.

I think the problem arises because I don't think i know of a result which correlates $C_n (G) $ with normal series.

The problem is that nilpotent groups is one of my weaknesses, although I understand results and theorems completely but such is not the case with the problems.

Answer 1

I will use the usual notation: $Z(K)$ is the center of the group $K$. $Z_0(G)=\{e\}$, and $Z_{k+1}(G)$ is the subgroup such that $Z_{k+1}(G)/Z_k(G) = Z(G/Z_k(G))$.

Your definition is:

A group $G$ is nilpotent if and only if there exists $n\geq 0$ such that $Z_n(G)=G$.

You are trying to show:

A group $G$ is nilpotent if and only if there exists a normal series $$G=G_0 \gt G_1 \gt\cdots \gt G_m=\{e\}$$ such that $G_i/G_{i+1}\leq Z(G/G_{i+1})$ for each $i$.

For the "if" part: assume there is such a series. We want to show that there is a $n$ such that $Z_n(G)=G$.

First, note that $G_{m-1}\leq Z_1(G)=Z(G)$. Indeed, we know that $G_{m-1}/G_m=G_{m-1}$ is contained in $Z(G/G_m)= Z(G)$.

I claim that $G_{m-2}\leq Z_2(G)$. Indeed, we know that $G_{m-2}/G_{m-1}$ is contained in $Z(G/G_{m-1})$. Since $G_{m-1}\leq Z(G)$, then $G/G_{m-1}$ surjects onto $G/Z(G)$; and if $f\colon H\to K$ is a surjective group homomorphism, then $f(Z(H))\subseteq Z(K)$.

Thus, $$\frac{G_{m-2}}{Z(G)}\cong \frac{G_{m-2}/G_{m-1}}{G_{m-1}/Z(G)} \leq Z\left(\frac{G}{Z(G)}\right).$$ Hence, $G_{m-2}\leq Z_2(G)$.

This should give you the idea. Now proceed by induction and show that $G_{m-i}\leq Z_i(G)$, and thus conclude that $Z_m(G)=G$.

(This is why the series with the centers $Z_i(G)$ is called the "upper central series": it lies "above" any central series.)

For the "only if" clause, note that the series $$G = Z_n(G)\gt Z_{n-1}(G)\gt\cdots\gt Z_1(G)\gt Z_0(G)=\{e\}$$ is a normal series, and that by definition $Z_i(G)/Z_{i+1}(G)\leq Z(G/G_{i+1})$, so the centers directly give you the groups $G_i$ that you need.