# Commutator subgroup and a center

• Dec 2nd 2008, 07:59 PM
aliceinwonderland
Commutator subgroup and a center
Hi, all

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

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

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

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

I will appreciate if someone gives me an intuitive enlightment of the above one.
• Dec 3rd 2008, 05:26 AM
ThePerfectHacker
Quote:

Originally Posted by aliceinwonderland
Hi, all

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

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

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

What bothering me is that why $\displaystyle [C_{i}(G), C_{i}(G)]$ is a proper subgroup of $\displaystyle C_{i-1}(G)$ 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 $\displaystyle \leq$ rather than $\displaystyle <$.
Consider, $\displaystyle G=S_3$ since $\displaystyle Z(G) = \{ (1)\}$.
Then central series just consists of trivial subgroups.