Characterize the natural numbers which are only the orders of abelian groups

Let n > 1 be an integer. Prove the following classification: every group of order n is abelian if and only if n = p_1^{a_1} \cdots p_k^{a_k}, p_i are distinct primes, a_i \in \{1,2\}, and p_i does not divide p_j^{a_j}-1 for all i and j.

(\Rightarrow) Suppose n has the properties above; we proceed by induction on k.

For the base case, let k = 1 and let G be a group of order n = p_1^{a_1}. Then G is either cyclic or elementary abelian, hence G is abelian.

For the inductive step, suppose that every group of width at most k which satisfies the given conditions for |G| is abelian. Let G be a group of width k+1 such that |G| satisfies the given properties. Now every proper subgroup of G has width at most k, so that by the induction hypothesis, every proper subgroup of G is abelian. By this previous exercise, G is solvable.

By this previous exercise, there is a (maximal) subgroup H \leq G such that H \leq G is normal and G/H has prime order; say G/H \cong Z_p. Let P be a Sylow p-subgroup of G. Now by FTFGAG, H is the direct product of its Sylow subgroups, and the Sylow subgroups of H are normal in H. In particular, H is generated by elements of prime and prime squared order. Let x \in H have prime or prime squared order and let y \in P. If \langle xy \rangle = G, then G is cyclic, hence abelian. If \langle xy \rangle is a proper subgroup of G, then xy = yx, so that \langle x \rangle \leq C_G(y). In either case, we have \langle x \rangle \leq C_G(P). Thus H \leq C_G(P), and we have HP = G \leq C_G(P). Thus P \leq Z(G). Similarly, HP = G \leq C_G(H), so that H \leq Z(G). Thus HP = G \leq Z(G), and G is abelian.

(\Leftarrow) Suppose every group of order n = p_1^{a_1} \cdots p_k^{a_k} is abelian. If (without loss of generality) a_1 \geq 3, let Q be a nonabelian group of order p_1^{a_1}. (We may take Q to be the direct product of a nonabelian group of order p^3 and a cyclic group.) Then Q \times Z_{p_2^{a_2} \cdots p_k^{a_k}} is nonabelian of order n. Thus a_i \in \{1,2\} for all i. Now suppose (again without loss of generality) that p_1 divides p_2^{a_2} - 1. If a_2 = 1, then we may construct a nonabelian group Q of order p_1p_2. Then Q \times Z_{p_3^{a_3} \cdots p_k^{a_k}} is a nonabelian group of order n. If a_2 = 2, we may construct a nonabelian group of order pq^2, following the patterns here or here (depending on whether p_1 does not or does divide p_2-1.) By taking the direct product with a cyclic group, there exists a nonabelian group of order n.

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

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  • John McPerson  On July 6, 2011 at 7:45 pm

    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.

    • nbloomf  On July 7, 2011 at 11:57 am

      If I understand correctly, you’re trying to show that G must be abelian if we assume those conditions on the p_i and a_i. 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 priori obvious that P_1P_i is a subgroup, or even that \langle P_1, P_i \rangle is a proper subgroup. If this can be shown, then I think this argument can be made to work.

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