58 Chapter 1. Groups I Conversely, define an equivalence relation on a group G by x ≡ y if x = y that is, x and y are equivalent if they generate the same cyclic subgroup. Denote the equivalence class containing an element x by gen(C), where C = x thus, gen(C) consists of all the generators of C. As usual, equivalence classes form a partition, and so G is the disjoint union: G = C gen(C), where C ranges over all cyclic subgroups of G. In Theorem 1.39(ii), we proved that |gen(C)| = φ(|C|), and so |G| = ∑ C φ(|C|). By hypothesis, for any divisor d of n, the group G has at most one cyclic subgroup of order d. Therefore, n = C |gen(C)| = C φ(|C|) ≤ d|n φ(d) = n, the last equality being Proposition 2.71(iii). Hence, for every divisor d of n, we must have φ(d) arising as |gen(C)| for some cyclic subgroup C of G of order d. In particular, φ(n) arises there is a cyclic subgroup of order n, and so G is cyclic. • Here is a variation of Theorem 1.93 (shown to me by D. Leep) which constrains the number of cyclic subgroups of prime order in a finite abelian group G. We remark that we must assume that G is abelian, for the group Q of quaternions is a nonabelian group of order 8 having exactly one (cyclic) subgroup of order 2. Theorem 1.94. If G is an abelian group of order n having at most one cyclic subgroup of order p for each prime divisor p of n, then G is cyclic. Proof. The proof is by induction on n = |G|, with the base step n = 1 obviously true. For the inductive step, note that the hypothesis is inherited by subgroups of G. We claim that there is some element x in G whose order is a prime divisor p of |G|. Choose y ∈ G with y = 1 its order k is a divisor of |G|, by Lagrange’s Theorem, and so k = pm for some prime p. By Exercise 1.32 on page 27, the element x = ym has order p. Define θ : G → G by θ : g → gp (θ is a homomorphism because G is abelian). Now x ∈ ker θ, so that | ker θ| ≥ p. If | ker θ| p, then there would be more than p elements g ∈ G satisfying gp = 1, and this would force more than one subgroup of order p in G. Therefore, | ker θ| = p. By the First Isomorphism Theorem, G/ ker θ ∼ = im θ ⊆ G. Thus, im θ is a subgroup of G of order n/p satisfying the inductive hypothesis, so there is an element z ∈ im θ with im θ = z . Moreover, since z ∈ im θ, there is b ∈ G with z = bp. There are now two cases. If p n/p, then xz has order p · n/p = n, by Proposition 1.89, and so G = xz . If p | n/p, then Exercise 1.33 on page 27 shows that b has order n, and G = b . •

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