# Normal Abelian Subgroups with Relatively Prime Indices

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?

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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.).

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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
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

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.