MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to know a proof of this fact: "every simple group has a minimal simple group as a subquotient." (If $H$ and $K$ are two subgroups of $G$ s.t. $H\lhd K$, then $\frac{K}{H}$ is called a subquotient of $G$) many thanks.

share|cite|improve this question
up vote 3 down vote accepted

If every proper subgroup of G is soluble, then it is done. So there is a non-soluble subgroup H with every subgroup soluble. Let K be maximal normal subgroup of H, then H/K is a minimal simple group.

share|cite|improve this answer
So do not need $G$ itself to be simple. It is enough that $G$ is not solvable (allowing the possibility $H = G$). – Geoff Robinson Jan 2 '13 at 8:01
Thanks a lot for the answer and correction. – sebastian Jan 2 '13 at 9:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.