- #1

fishturtle1

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- Homework Statement
- Let ##G## be a group. We define the lower central series of ##G## as

$$G = \gamma_1 \ge \gamma_2 \ge \gamma_3 \ge\dots$$

where ##\gamma_1 = G## and ##\gamma_{i+1} = [\gamma_i, G]##. ##\textbf{My question is:}## Why is ##\gamma_{i+1} \rhd \gamma_i## for all ##i##?

- Relevant Equations
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Let ##G, H## be groups. We define ##[G,H] = \langle \lbrace xyx^{-1}y^{-1} : x \in G, y \in H \rbrace \rangle##.

Lemma 5.30.: i) If ##K \lhd G## and ##K \le H \le G## then ##[H, G] \le K## if and only if ##H/K \le Z(G/K)##.

ii) If ##H, K \le G## and ##f : G \rightarrow L## is a homomorphism, then ##f([H,K]) = [f(H), f(K)]##.

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My attempt: If ##i = 1##, then ##\gamma_1 = G \rhd G' = \gamma_2##. We proceed by induction on ##i##. Consider an element ##xyx^{-1}y^{-1}## where ##x \in \gamma_i## and ##y \in G##. Since ##\gamma_i \rhd G##, we have ##yx^{-1}y^{-1} = x_0 \in \gamma_i##. So, ##xyx^{-1}y^{-1} = xx_0 \in \gamma_i \le G##. It follows that ##\gamma_{i+1} = [\gamma_i, G] \le \gamma_i \le G##.

Can I have a hint how to proceed to show ##\gamma_{i+1} \lhd G##, please?

Edit: The textbook says 'It is easy to check ##\gamma_{i+1}(G) \le \gamma_i(G)##. Moreover, Lemma 5.30i) shows that ##[\gamma_i(G), G] = \gamma_{i+1}(G)## gives ##\gamma_i(G) / \gamma_{i+1}(G) \le Z(G/\gamma_{y+1}(G))##'. I think I've been able to show ##\gamma_{i+1}(G) \le \gamma(G)##. But I think I need to show ##\gamma_i \lhd G## for all ##i## before I can get the second line of the textbook.

Can I have a hint how to proceed to show ##\gamma_{i+1} \lhd G##, please?

Edit: The textbook says 'It is easy to check ##\gamma_{i+1}(G) \le \gamma_i(G)##. Moreover, Lemma 5.30i) shows that ##[\gamma_i(G), G] = \gamma_{i+1}(G)## gives ##\gamma_i(G) / \gamma_{i+1}(G) \le Z(G/\gamma_{y+1}(G))##'. I think I've been able to show ##\gamma_{i+1}(G) \le \gamma(G)##. But I think I need to show ##\gamma_i \lhd G## for all ##i## before I can get the second line of the textbook.

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