Lower Central Series - Understanding the Induction Process

[fishturtle1]

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.

 

[fresh_42]

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.

 

[fishturtle1]

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?

 

[fresh_42]

Your notation is a bit confusing. We usually write it as $G=G^0 \rhd G^1\rhd G^2\rhd \ldots$
$$
gG^{n+1}g^{-1}=g[G^n,G]g^{-1}=[gG^ng^{-1},gGg^{-1}]\subseteq [G^n,G]=G^{n+1}
$$

 

[fishturtle1]

Thank you! I think I finally understand.