Equivalent characterizations of nilpotent and cyclic groups

If G is finite, prove that (1) G is nilpotent if and only if it has a normal subgroup of each order dividing |G|, and (2) is cyclic if and only if it has a unique subgroup of each order dividing |G|.


We begin with some lemmas.

Lemma 1: Let G be a group, and let H_1, \ldots, H_k be characteristic subgroups of G. Then H = H_1\cdots H_k is characteristic in G. Proof: H is a subgroup because each H_i is normal. If \alpha is an automorphism of G, then \alpha[H] = \alpha[H_1] \cdots \alpha[H_k] = H_1 \cdots H_k = H. \square

Lemma 2: Let G be a finite group. If P \leq G is a normal Sylow subgroup, then P is characteristic. Proof: P is the unique subgroup of order |P|. \square

Now to the main result; first we prove (1).

(\Rightarrow) We proceed by induction on the breadth (number of prime divisors with multiplicity) of G.

For the base case, if G is a finite nilpotent group of breadth 1, then G \cong Z_p is an abelian simple group. Thus G has a normal subgroup of order 1 and p, the only divisors of |G|.

For the inductive step, suppose that every finite nilpotent group of breadth k \geq 1 has a normal subgroup of each order dividing |G|, and let G be a finite nilpotent group of breadth k+1. Now Z(G) is nontrivial, so there exists an element x \in Z(G) of prime order, and \langle x \rangle G is normal. We showed in a lemma to the previous exercise that G/\langle x \rangle is nilpotent, and has breadth k. By the inductive hypothesis, G/\langle x \rangle has a normal subgroup of each order d dividing |G/\langle x \rangle|. By the Lattice Isomorphism Theorem, a subgroup \overline{H} \leq G/\langle x \rangle is normal if and only if \overline{H} = H/\langle x \rangle for some normal subgroup H \leq G. Thus we see that G has a normal subgroup of order d for all d dividing |G| such that p|d. Now let d divide |G|, with p not dividing d, and let H \leq G be a normal subgroup of order pd. Since H \leq G, H is a finite nilpotent group. By Theorem 3, H is the direct product of its Sylow subgroups. Write H = P \times Q, where P is the Sylow p subgroup of H and Q the direct product of the remaining Sylow subgroups. By Lemma 2, each factor of Q is characteristic in H, so that Q is characteristic in H. Note that |Q| = d, thus Q \leq G is normal and has order d.

(\Leftarrow) Suppose G has a normal subgroup of each order d dividing |G|. In particular, G has a normal (thus unique) Sylow p-subgroup for each prime p. By Theorem 6.3, G is nilpotent.

Now we prove (2).

The (\Rightarrow) direction was proved in Theorem 2.7.

(\Leftarrow) Suppose G is finite and has a unique subgroup of order d for each d properly dividing |G|. In particular, all the Sylow subgroups of G are unique, hence normal; say there are k of these. Suppose some Sylow p-subgroup P of G is not cyclic; then by this previous exercise, P has a subgroup congruent to Z_p \times Z_p, and thus G has distinct subgroups of order p, a contradiction. Thus each Sylow subgroup of G is cyclic; say P_i = \langle x_i \rangle. Then |x_1 \cdots x_k| = |G|, so that G = \langle x_1 \cdots x_k \rangle is cyclic.

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