Three solvable subgroups of pairwise coprime indexes implies solvable
Suppose is a finite group and are solvable subgroups of , such that the indices are pairwise relatively prime. Then, is a finite solvable group.
- Minimal normal implies elementary abelian in finite solvable
- Sylow subgroups exist
- Sylow implies order-dominating
- Solvability is subgroup-closed
- Solvability is extension-closed
Given: A finite group . Three solvable subgroups of such that are pairwise relatively prime.
To prove: is solvable.
Proof: We prove the claim by induction on the order of . The main part of the induction step is demonstrating the existence of a nontrivial solvable normal subgroup inside .
Note that for nontrivial, none of the three subgroups can be trivial, for that would contradict the assumption that the indices are relatively prime. Also, note that if any of the three subgroups equals the whole group, then the whole group is anyway solvable. So, we can restrict our attention to the case where all three subgroups are proper nontrivial solvable subgroups.
Existence of a nontrivial solvable normal subgroup
- has an elementary abelian nontrivial normal subgroup, say , a -group: Since is a finite group, it has a minimal normal subgroup. Fact (1) now yields that the minimal normal subgroup must be elementary abelian. Call this . Let be the prime for which is a power of .
- is contained in some conjugate of either or (without loss of generality, ): Since and are relatively prime, at least one of them must be relatively prime to . Suppose is relatively prime to . Then, by fact (2), has a -Sylow subgroup , which by order considerations is also -Sylow in . Further, fact (3) yields that is contained in some conjugate of . In particular, is contained in some conjugate of . Finally, since conjugate subgroups have the same index, we can replace with that conjugate to begin with, and thus assume without loss of generality that .
- : This follows from the assumption that the indices are relatively prime.
- The normal core of in contains : Since , the intersection of all conjugates , equals the intersection of all conjugates . However, since is normal in , we have that for all . In particular, . Thus, is contained in the intersection of all , so is contained in the normal core of in .
- The normal core of in is a nontrivial solvable normal subgroup of : By the previous step, , and itself is nontrivial, so is nontrivial. Since by definition, and is solvable, fact (4) yields that is solvable. Finally, the definition of normal core yields that is normal in .
The induction step
We have thus established that any finite group containing three solvable subgroups of pairwise coprime indices has a nontrivial solvable normal subgroup . Now, consider the quotient map . Under this map, are solvable subgroups of , and their indices continue to be pairwise relatively prime. Thus, is solvable by the induction hypothesis. We now use fact (5) to conclude that itself is solvable.