Let be an integer. Prove the following classification: every group of order is abelian if and only if , are distinct primes, , and does not divide for all and .

Suppose has the properties above; we proceed by induction on .

For the base case, let and let be a group of order . Then is either cyclic or elementary abelian, hence is abelian.

For the inductive step, suppose that every group of width at most which satisfies the given conditions for is abelian. Let be a group of width such that satisfies the given properties. Now every proper subgroup of has width at most , so that by the induction hypothesis, every proper subgroup of is abelian. By this previous exercise, is solvable.

By this previous exercise, there is a (maximal) subgroup such that is normal and has prime order; say . Let be a Sylow -subgroup of . Now by FTFGAG, is the direct product of its Sylow subgroups, and the Sylow subgroups of are normal in . In particular, is generated by elements of prime and prime squared order. Let have prime or prime squared order and let . If , then is cyclic, hence abelian. If is a proper subgroup of , then , so that . In either case, we have . Thus , and we have . Thus . Similarly, , so that . Thus , and is abelian.

Suppose every group of order is abelian. If (without loss of generality) , let be a nonabelian group of order . (We may take to be the direct product of a nonabelian group of order and a cyclic group.) Then is nonabelian of order . Thus for all . Now suppose (again without loss of generality) that divides . If , then we may construct a nonabelian group of order . Then is a nonabelian group of order . If , we may construct a nonabelian group of order , following the patterns here or here (depending on whether does not or does divide .) By taking the direct product with a cyclic group, there exists a nonabelian group of order .

Thus, if every group of order is abelian, then has the properties enumerated in the problem statement.

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How about this for right to left? Let Pi be the Sylow-p_i subgroup of G. Then for i=2,….k+1, |P1Pi|= (p_1^a1) * (p_i^a_i) so P1Pi is abelian by induction so P1 commutes with Pi for all these i so P1 commutes with = G. So P1 is in the center of G. Ditto with P2, P3, …, P(k+1). Since all these Sylow subgroups are in the center of G, by order considerations, the center of G is G so G is abelian. I’m not sure about this proof since I didn’t use the given divisibility condition anywhere.

If I understand correctly, you’re trying to show that must be abelian if we assume those conditions on the and . We have to use the divisibility condition somewhere, since otherwise we can construct nonabelian examples.

One problem I see is that it is not

a prioriobvious that is a subgroup, or even that is a proper subgroup. If this can be shown, then I think this argument can be made to work.