True or False: (19h.) The commutator subgroup of a simple group G must be G itself.
Answer: http://www.auburn.edu/~huanghu/math5310/alg-hw-ans-
i think 3.pdf
False. G must be nonabelian.
I wrote the relevant definitions here. Because $G$ is simple hence $N = \{id\}$ or $G$.
The hinge looks like Theorem 15.20 hence I want to use it.
Possibility 1: If $N = \{id\}$, then $G/N = G/\{id\} \quad \simeq G$. How to sally forth?
Possibility 2: If $N = G$, then $G/G = \{G\} \quad \simeq \{id\}$. How to sally forth?
(3.) What's the intuition?
You do not have to go through factor groups. $G'$ is a normal subgroup of $G$, so there are two possibilities: $G'=\{1\}$ or $G=G'$. The first case is being equivalent to $G$ being abelian (note that every commutator is the identitiy element!). Abelian and simple means isomorphic to a cyclic group of prime order. So your statement is false, unless you require that $G$ is non-abelian.