# Commutator subgroup and a center

• Dec 2nd 2008, 08:59 PM
aliceinwonderland
Commutator subgroup and a center
Hi, all

$C_{i}(G)$ denotes a center in central series and [G, G] denotes a commutator subgroup of G.

My textbook says $[C_{i}(G), C_{i}(G)] < C_{i-1}(G)$.

Since $C_{i-1}(G)$ is a normal subgroup of $C_{i}(G)$ and $C_{i}(G) / C_{i-1}(G)$ is abelian, it is plausible that $C_{i-1}(G)$ contains $[C_{i}(G), C_{i}(G)]$.

What bothering me is that why $[C_{i}(G), C_{i}(G)]$ is a proper subgroup of $C_{i-1}(G)$ rather than being just a subgroup?

I will appreciate if someone gives me an intuitive enlightment of the above one.
• Dec 3rd 2008, 06:26 AM
ThePerfectHacker
Quote:

Originally Posted by aliceinwonderland
Hi, all

$C_{i}(G)$ denotes a center in central series and [G, G] denotes a commutator subgroup of G.

My textbook says $[C_{i}(G), C_{i}(G)] < C_{i-1}(G)$.

Since $C_{i-1}(G)$ is a normal subgroup of $C_{i}(G)$ and $C_{i}(G) / C_{i-1}(G)$ is abelian, it is plausible that $C_{i-1}(G)$ contains $[C_{i}(G), C_{i}(G)]$.

What bothering me is that why $[C_{i}(G), C_{i}(G)]$ is a proper subgroup of $C_{i-1}(G)$ rather than being just a subgroup?

I will appreciate if someone gives me an intuitive enlightment of the above one.

I think your book might have a mistake and that it should be $\leq$ rather than $<$.
Consider, $G=S_3$ since $Z(G) = \{ (1)\}$.
Then central series just consists of trivial subgroups.