Lower central series

  • #1
334
44
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:
---------------------------
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)]##.
-------------------------
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.
 
Last edited:

Answers and Replies

  • #2
15,355
13,385
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##?
It isn't. ##\gamma_{i+1}\lhd \gamma_i##
Relevant Equations:: ---------------------------
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)]##.
-------------------------

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.
You are thinking far too complicated. It is more or less trivial. E.g.
$$
\underbrace{h\underbrace{gh^{-1}g^{-1}}_{\in \gamma_n\lhd G}}_{\in \gamma_n}\; , \;h\in \gamma_n\, , \,g\in G \Longrightarrow \gamma_{n+1}=[\gamma_n,G]\subseteq \gamma_{n}
$$
and the same for the invariance of the subgroup.

A nice example of a double induction by the way.
 
  • Informative
Likes fishturtle1
  • #3
334
44
Thanks for the reply!

So we're trying to prove ##(\gamma_{n} \lhd G) \Rightarrow (\gamma_{n+1} \le \gamma_n) \Rightarrow (\gamma_{n+1} \lhd G)##.

For the second implication, we want to show ##\gamma_{n+1} \lhd G##. Let ##g \in G## and ##x \in \gamma_{n+1}##. Then ##gxg^{-1} \in \gamma_n##. ...

Would it be enough to show ##g(xhx^{-1}h^{-1})g^{-1} \in \gamma_{n+1}## for all ##x \in \gamma_n, h, g \in G##? We'd have
$$g(xhx^{-1}h^{-1})g^{-1} = g(xy)g^{-1} = z$$
for some ##y, z \in \gamma_n## but i'm not sure if this is the right way to go.

Sorry but am I on the right track here?
 
  • #4
15,355
13,385
Your notation is a bit confusing. We usually write it as ##G=G^0 \rhd G^1\rhd G^2\rhd \ldots##
$$
gG^{n+1}g^{-1}=g[G^n,G]g^{-1}=[gG^ng^{-1},gGg^{-1}]\subseteq [G^n,G]=G^{n+1}
$$
 
  • Informative
Likes fishturtle1
  • #5
334
44
Thank you! I think I finally understand.
 

Related Threads on Lower central series

Replies
5
Views
627
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
6K
  • Last Post
Replies
2
Views
2K
Top