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.

Suppose A, B are normal abelian subgroups of some finite group G. Let [G:A]=m, [G:B]=n, where gcd(m,n)=1.

Can G be non-abelian?

--

I've been attempting to show that G must be abelian, but I'm starting to think that this isn't necessarily true.

I can show that |G|=|AB|=mnq, |A∩B|=q where q is some positive integer. I can also show that there are cases in which q does not equal 1 (Let G=Z2xZ3xZ4) (if q=1, AxB would then be isomorphic to AB, and so G would be abelian.).

share|improve this question
    
I can't parse the "Can it be shown that G is not necessarily abelian?"...did you mean to ask whether it is possible to prove that G necessarily is not abelian? –  DonAntonio Dec 7 '12 at 4:09
    
@DonAntonio How about parse it this way: "does there exist a finite group G with two normal abelian subgroups of coprime indices?" An affirmative answer to this would demonstrate that, upon the hypothesis Ida poses (finite G having such A,B), G is not necessarily abelian. –  anon Dec 7 '12 at 4:21
    
Then you meant to ask "does there exists a finite non-abelian group with...", right? Thanks –  DonAntonio Dec 7 '12 at 4:27
    
...Right. ${}{}{}$ –  anon Dec 7 '12 at 4:27
5  
So $A \cap B < Z(G)$, which makes $G$ nilpotent. But any prime $p$ dividing $|G|$ must either not divide $m$ or not divide $n$, so a Sylow $p$-subgroup is contained in either $A$ or $B$, and hence is abelian. –  Derek Holt Dec 7 '12 at 8:44
show 1 more comment

1 Answer

up vote 2 down vote accepted

This elaborates upon Derek's solution in the comments. If $[G:A]$ and $[G:B]$ are relatively prime, then $[G:AB]$ is $1$ since it divides both of these indices, hence $G=AB$. Furthermore, commutators of normal subgroups are distributive, so we compute the derived subgroup of $G$ as

$$G'=[G,G]=[AB,AB]=[A,A][A,B][B,A][B,B]=[A,B].$$

Since $A,B\triangleleft G$ are normal, $[A,B]\le A\cap B$. Since $A,B$ are both abelian and $G=AB$, we can tell that the intersection $A\cap B\le Z(G)$ is central. Hence $[G',G]=1$ and $G$ is nilpotent (of class $\le2$).

A finite group is nilpotent if and only if it is a direct product of its Sylow $p$-subgroups. Each Sylow subgroup must be contained in one of $A$ or $B$ due to their coprime indices, in which case every Sylow subgroup is abelian. Therefore $G$ is abelian since it is ($\cong$) a direct product of abelian groups.

share|improve this answer
add comment

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.