Finite Groups With a Certain Number of Cyclic Subgroups

In this short note we describe the finite groups G having |G| − 1 cyclic subgroups. This leads to a nice characterization of the symmetric group S3. In subgroup lattice theory, it is a usual technique to associate to a finite group G some posets of subgroups of G (see, e.g., [4]). One such poset is the poset of cyclic subgroups of G, usually denoted by C(G). Notice that there are few papers on the connections between |C(G)| and |G| ([2, 3] are examples). We also recall the following basic result of group theory. Theorem 1. A finite group G is an elementary Abelian 2-group if and only if |C(G)| = |G|. Inspired by Theorem 1, we study here the finite groups G for which (∗) |C(G)| = |G| − 1. First, we observe that certain finite groups of small orders, such as Z3, Z4, S3 and D8, have this property. Our main theorem proves that in fact these groups exhaust all finite groups G satisfying (∗). Theorem 2. Let G be a finite group. Then |C(G)| = |G| − 1 if and only if G is one of the following groups: Z3, Z4, S3 or D8. Proof. Assume that G satisfies (∗), let n = |G| and denote by d1 = 1, d2, . . . , dk the positive divisors of n. If ni = |{H ∈ C(G) | |H | = di }|, i = 1, 2, . . . , k, then k ∑ i=1 niφ(di) = n. Since |C(G)| = ∑k i=1 ni = n − 1, one obtains k ∑ i=1 ni (φ(di)− 1) = 1, which implies that there exists i0 ∈ {1, 2, . . . , k} such that ni0 = 1 and φ(di0) = 2 (i.e., di0 ∈ {3, 4, 6}); for an i = i0 we have either ni = 0 or φ(di) = 1 (i.e., di ∈ {1, 2}). http://dx.doi.org/10.4169/amer.math.monthly.122.03.275 MSC: Primary 20D60 March 2015] NOTES 275 We remark that di0 cannot be equal to 6 because in this case G would also have a cyclic subgroup of order 3, a contradiction. We infer that G contains a unique normal cyclic subgroup of order di0 , say H . Let X be a subgroup of G of prime order p. Then either p = 2 or p divides di0 . Hence by Cauchy’s theorem, either di0 = 4 and G is a 2-group, or di0 = 3 and G is a {2, 3}-group. If di0 = 3 it follows that H is the unique Sylow 3-subgroup of G, and consequently G = HK for some K ∈ Syl2(G). The theorem holds if G = H , so we may assume K = 1. As G has no cyclic subgroup of order 6, we have CK (H) = 1, so |K | = |Aut (H)| = 2. Therefore, G is the non-Abelian group S3 of order 6. Assume next that di0 = 4, so that G is a 2-group. If G = H , then the theorem holds, so we may assume that there is g ∈ G \ H . Then, as each member of G \ H is an involution, g is an involution inverting H via conjugation. Since |G : CG(H)| ≤ |Aut(H)| = 2, we conclude that G = H〈g〉 is the dihedral group of order 8. This completes the proof. The following corollary is an immediate consequence of Theorem 2. Corollary 3. S3 is the unique finite group G which is not a p-group and satisfies |C(G)| = |G| − 1. We also remark that a facile proof of Theorem 1 easily follows from the first part of the proof of Theorem 2. Finally, we indicate a natural open problem concerning the above study. Open problem. Describe the finite groups G satisfying |C(G)| = |G| − r , where 2 ≤ r ≤ |G| − 1. ACKNOWLEDGMENT. The author is grateful to the reviewer for his/her remarks which improved the pre-vious version of the paper. REFERENCES1. I. M. Isaacs, Finite Group Theory. American Mathematical Society, Providence, RI, 2008.2. G. A. Miller, On the number of cyclic subgroups of a group, Proc. Natl. Acad. Sci. USA 15 (1929) 728–731, http://dx.doi.org/10.1073/pnas.15.9.728.3. I. M. Richards, A remark on the number of cyclic subgroups of a finite group, Amer. Math. Monthly 91(1984) 571–572, http://dx.doi.org/10.2307/2323746.4. R. Schmidt, Subgroup Lattices of Groups. De Gruyter Expositions in Mathematics 14, De Gruyter, Berlin,1994, http://dx.doi.org/10.1515/9783110868647. Department of Mathematics, Alexandru loan Cuza University, Iaşi, [email protected] 276c© THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 122