Is Z3 a Simple and Solvable groups?

Question

The group $C_3$ is cyclic of order $3$, and is simple. I've learned both that all abelian groups are solvable and that no simple group is solvable, which is a contradiction. What is the truth of the matter?

Answer 1

Non-abelian simple groups are not solvable. This is because simple groups don't have proper normal subgroups so the chain of normal subgroups is always trivial $\{e\}<G$. Hence if $G/\{e\}\simeq G$ is nonabelian then $G$ is not solvable.

Every abelian group is solvable on the other hand, becuase the same chain $\{e\}< G$ provides the answer.

In particular the cyclic group $C_n$ is always solvable even for prime $n$ (for which it is additionaly simple).

All in all: the only solvable and simple groups are $C_p$ for prime $p$.

Answer 2

All non-abealian simple groups are not solvable, but the abealian simple groups are solvable.

Proof: Since $G$ is simple, the only possible subnormal chain goes $\{1\}<G$. For $G$ to be normal, we need $G/\{1\}$ to be abealian. But $G/\{1\}=G$ so this is true precisely when $G$ is abelian.

Answer 3

There is something else that is good to know here. The following two statements are equivalent (and this is in fact easy to prove).

A non-abelian simple group has even order.

A group of odd order is solvable.

However, the proof of one of the statements, being an old conjecture of William Burnside, is highly non-trivial and is the famous Theorem of Feit and Thompson (1963). At the time it required a paper of $2^8-1$ pages, see http://projecteuclid.org/euclid.pjm/1103053941. It was the culmination of a long spell of research in finite group theory and group representation theory.
Almost 50 years later, in September 2012, a fully formal proof, checked with the Coq proof assistant, was announced by Georges Gonthier and fellow researchers at Microsoft Research and INRIA.