Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have problems proving this:

"If G is a solvable group, then a maximal normal subgroup has prime index".

I see that $\frac{G}{M}$ has to be simple. If I had that it has to be abelian too, I would have that $\frac{G}{M}$ would be cyclic, and then it will be done. But I'm stuck at this.

Any ideas?

Thanks a lot!

share|improve this question

1 Answer 1

up vote 4 down vote accepted

If $G$ is solvable then any image of $G$ is solvable. If $G/M$ is not abelian, it's a non-abelian simple group. These groups are trivially not solvable. So $G/M$ must be abelian.

share|improve this answer

Your Answer

 
discard

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.