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.

  • $\begingroup$ Thanks a lot. Can you please flesh out how the first case $\iff G$ Abelian? And can you unfold the intuition please? $\endgroup$ – Nai Feb 24 '14 at 7:27
  • $\begingroup$ $G'=\{1\} \Leftrightarrow \forall x,y \in G: x^{-1}y^{-1}xy=1 \Leftrightarrow \forall x,y \in G: xy=yx \Leftrightarrow G$ is abelian. The intuition? If $G$ is simple try to to find "standard" normal subgroups ($Z(G), G', ...)$ which have to be trivial or the whole group. $\endgroup$ – Nicky Hekster Feb 25 '14 at 5:02
  • $\begingroup$ Thanks a lot. Upvoted. Can you please flesh out the first $\iff$: $G'=\{1\} \Leftrightarrow \forall x,y \in G: x^{-1}y^{-1}xy=1$? And can you please answer in your answer and move comments into your answer? $\endgroup$ – Nai Feb 25 '14 at 9:43
  • $\begingroup$ (3.) I'm still perplexed. Your last sentence in your comment overhead looks just like the definition of a simple group. What's different? What's the intuition? $\endgroup$ – Nai Feb 25 '14 at 9:44

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