# Let $\gamma_1(G)=G$, $\gamma_i(G)=(\gamma_{i-1}(G),G)$. Show that $G$ is nilpotent iff $\gamma_m(G)=\langle e\rangle$ for some $m$.

I'm working on this exercise from Algebra by Hungerford:

Define a chain of subgroups $\gamma_i(G)$ of a group $G$ as follows: $\gamma_1(G)=G$, $\gamma_2(G)=(G,G)$, $\gamma_i(G)=(\gamma_{i-1}(G),G)$. Show that $G$ is nilpotent if and only $\gamma_m(G)=\langle e\rangle$ for some $m$.

The statement makes since intuitively, but I'm having trouble turning that into any sort of rigor. My method right now is using the fact that a group is nilpotent if and only if it is the direct product of its Sylow subgroups (I believe this boils down to showing the claim on an arbitrary Sylow subgroup, which seems simpler).

Although I don't know quite how to continue with method I've chosen, I was wondering if there was possibly a more straightforward way using the definition. I'm not sure how standard this definition is so I include it:

Let $G$ be a group. Define $C_1(G)=C(G)$ to be the center of $G$, $C_2(G)$ to be the inverse image of $C(G/C(G))$ under the canonical projection $G\mapsto G/C_1(G)$, and inductively define $C_i(G)$ to be the inverse image of $C(G/C_{i-1}(G))$ under the canonical projection $G\mapsto G/C_{i-1}(G)$. A group $G$ is nilpotent if $C_n(G)=G$ for some $n\in\mathbb{N}$.

While I understand the definition, I don't really see how this presents itself tangibly (if that makes sense); in other words, I don't understand what this means for an arbitrary element of a nilpotent group $G$.

Any help is appreciated. Thanks in advance.

• The two constructions are related by: $(C_i(G), G) \le C_{i-1}(G)$. This should help. – Ted Nov 8 '15 at 1:40

## 1 Answer

It is standard to extend the $C$'s down to 0 by setting $C_0(G) = 1$, so I will do that.

The two constructions are related by (*) below: $$(*): C_j(G) \ge \gamma_i(G) \iff C_{j+1}(G) \ge \gamma_{i-1}(G)$$

First we prove that if $G$ is nilpotent, then $\gamma_i(G)=1$ for some $i$.
If $G$ is nilpotent, then $C_n(G) = G$ and thus we have $$\gamma_1(G) = C_n(G).$$ Then by an easy induction using the $\Longleftarrow$ direction of (*), $$\gamma_i(G) \le C_{n-i+1}(G).$$ Letting $i=n+1$, we have $\gamma_{n+1}(G) \le C_0(G) = 1$.

The other direction is similar, starting with $C_0(G) = \gamma_m(G)$, using the $\Longrightarrow$ direction of (*).

• This is fantastic! Thank you! Do you have any hints on how I would go about showing (*)? – Blake Nov 8 '15 at 4:58
• $C_{j+1}$ is defined as all elements that commute with all elements of $G$ up to an element of $C_j$. Another way of saying that: $C_{j+1}$ is the largest subgroup of $G$ satisfying $(C_{j+1}, G) \le C_j$. Does that help? – Ted Nov 8 '15 at 5:43
• Yes that makes sense. I'll try and figure out how show $(*)$ from this now. Thank you for all of your help. – Blake Nov 8 '15 at 8:09
• Ok so the $\Rightarrow$ of $(*)$ is obvious (right?) and $\Leftarrow$ follows by $(C_{j+1},G)\leq C_j$. So I've think I've got it. I really appreciate you taking the time to help me on this. – Blake Nov 8 '15 at 8:16