Research Article  Open Access
Joshua Harrington, Lenny Jones, Alicia Lamarche, "Characterizing Finite Groups Using the Sum of the Orders of the Elements", International Journal of Combinatorics, vol. 2014, Article ID 835125, 8 pages, 2014. https://doi.org/10.1155/2014/835125
Characterizing Finite Groups Using the Sum of the Orders of the Elements
Abstract
We give characterizations of various infinite sets of finite groups under the assumption that and the subgroups of satisfy certain properties involving the sum of the orders of the elements of and . Additionally, we investigate the possible values for the sum of the orders of the elements of .
1. Introduction
For any nonempty subset of a finite group , we let denote the sum of the orders of all elements of (counting multiplicities). Amiri et al. [1] gave a characterization of cyclic groups using . They showed that if is a cyclic group and is a noncyclic group of the same order, then . In 2011, H. Amiri and S. M. J. Amiri [2] turned their attention to the minimum value of , and they showed that, for nilpotent groups of a fixed order, the minimum value is attained when each Sylow subgroup has prime exponent. In addition, they showed that if there exists a nonnilpotent group of fixed order , then the minimum value of , as ranges over all groups of order , will be attained for a nonnilpotent group. Moreover, this minimum value of on such a nonnilpotent group is strictly smaller than the value of on any nilpotent group of order . In the same article, H. Amiri and S. M. J. Amiri conjectured that if is a simple group and is a nonsimple group of order , then . However, this conjecture has recently been shown to be false [3]. In 2012, H. Amiri and S. M. J. Amiri [4] showed that for all proper subgroups of , where is the symmetric group and is the alternating group.
In this paper, we give characterizations of various infinite sets of finite groups , given certain restrictions on or (or both), where is a proper subgroup of . More precisely, we prove the following.
Theorem 1. Let be a finite abelian group. Then divides for every subgroup of if and only if is cyclic of squarefree order.
Theorem 2. Let be a finite abelian group. Then for every proper subgroup of if and only if for any prime and any positive integer .
Theorem 3. For any fixed , there exist infinitely many finite groups , both abelian and nonabelian that satisfy the conditions in each of the following situations: (1),(2).
Theorem 4. Let be a positive integer. Then Moreover, the number is the smallest possible constant such that for all .
2. Preliminaries and Notation
We assume that all groups in this paper are finite. For an element in a group , we let denote the order of in . For a group , a prime that divides , and a positive integer , we define We let and denote, respectively, the symmetric group and the alternating group on the set . We let denote Euler’s totient function.
We now state, without proof, two useful results from the literature, the first of which was mentioned previously in Section 1.
Theorem 5 (see [1]). If is a cyclic group and is a noncyclic group of the same order, then .
Lemma 6 (see [2]). If and are finite groups, then , with equality if and only if .
The following lemma, which we also state without proof, is a special case of Theorem A in [2].
Lemma 7. Let be a prime and let be a positive integer. If is an abelian group with , then
Lemma 7 shows that the minimum value of , as ranges over all abelian groups of fixed order is achieved on the elementary abelian group of that order. Lemma 7 will be useful in proving Theorem 2.
While not difficult to prove, the next proposition does not currently appear in the literature and will be useful throughout this paper. This result determines formulas for in specials cases.
Proposition 8. Let and be positive integers, and let be a prime.
Proof. Let be a positive integer. Observe that . Hence, the number of elements of order in is and therefore Thus,
3. The Proof of Theorem 1
For the sake of convenience, we say that is a divisible group if divides for all subgroups of . We can deduce the following result concerning divisible groups from Lemma 6.
Lemma 9. Let and be finite groups with , and let . Then is a divisible group if and only if both and are divisible groups.
Proof. Suppose first that is a divisible group. Assume, by way of contradiction and without loss of generality, that is not divisible. Then there is some subgroup of such that does not divide . Consider the subgroup of . Using Lemma 6, since is a divisible group, we have that
for some positive integer . But then , which contradicts our assumption that does not divide .
Suppose now that both and are divisible, and let be a subgroup of . Since , it follows that , where is a subgroup of , and is a subgroup of . Then, by Lemma 6, we have that
which implies that is a divisible group.
In light of Lemma 9, it is enough to focus our attention on groups, where is a prime. Recall for a given positive integer and a prime that divides that is the subset of containing all elements of order in . Note that when . Two obvious, but useful, consequences of this observation are given below.
Lemma 10. Let be a prime and let be a group. Then .
Proof. Let be the exponent of . Then
Corollary 11. Let be a prime and let be a group. If is a divisible group and is any proper subgroup of , then
Lemma 12. Let and be positive integers, and let be a prime. Suppose that . Then is a divisible group if and only if .
Proof. Clearly, is a divisible group if , so assume that . First suppose that , so that . Then, using Proposition 8 with , we have
Let be a subgroup of such that . Then by Proposition 8, and thus
since . Hence, is not a divisible group when .
Now assume that , and let be a subgroup of such that . Again, using Proposition 8, we can write
where
Since , it follows that , and hence . Thus, is not a divisible group and the proof is complete.
Lemma 13. Let be a prime and suppose that , where is nontrivial, with . Then is not a divisible group.
Proof. Let , and define Then and Thus since
Lemma 14. Let where are not all equal. Then
Proof. Note that and since are not all equal. Suppose that and define Let and observe that Thus, we may assume without loss of generality that . Let be a subgroup of such that Since , we have that from which the lemma follows.
We are now in a position to establish Theorem 1. For the convenience of the reader, we restate Theorem 1 below as Theorem 15 using the divisible terminology.
Theorem 15. Let be an abelian group. Then is a divisible group if and only if is cyclic of squarefree order.
Proof. If is cyclic of squarefree order, then the fact that is a divisible group follows immediately from Lemma 6. So, now assume conversely that is a divisible group. To establish the theorem, it is enough, by Lemmas 6, 12, and 13, to show for a prime that where with not all equal, is not a divisible group. Let be as in (29), and assume by way of contradiction that is a divisible group. Let be a subgroup of with Note that . Using this observation, Lemma 14, and Corollary 11, it follows that Using the fact that to solve the inequality we get that , which is impossible, and the proof is complete.
The following is immediate from Theorem 15.
Corollary 16. If is an abelian divisible group, then every proper subgroup of is a divisible group.
Remark 17. The converse of Corollary 16 is false. For example, let , where is a prime. Then every proper subgroup of is a divisible group, but is not a divisible group by Lemma 12.
3.1. A Remark on Nonabelian Divisible Groups
The nonabelian situation is much less clear than the abelian case. In fact, we have been unable to find a single nonabelian divisible group, although we are somewhat hesitant to conjecture their nonexistence. On the other hand, using Lemma 9, we are able to construct infinitely many nonabelian groups that are not divisible. Since , we see that is not divisible. Then, by Lemma 9, it follows that is not divisible for any prime .
4. The Proof of Theorem 2
We say a group is a group if for all proper subgroups of . For the convenience of the reader, we use this new terminology to restate Theorem 2 below as Theorem 18.
Theorem 18. Let be a finite abelian group. Then is a group if and only if for any prime and any positive integer .
Proof. Using Proposition 8, it is easy to show that if
for any prime and any positive integer , then is a group.
So, assume conversely that is an abelian group, and write
where and are primes. Assume first that , and let be a subgroup of such that
where . We claim that . To see this, we have
which shows that is not a group if . Hence, assume that so that is a group for some prime . Suppose that and write
where
Assume first that . Let , where is trivial if , so that . Let be a subgroup of such that
Then
Thus and , where .
Now suppose that so that . Let be the subgroup of with . Then, using Proposition 8, we have
which implies that . Hence, and or , which completes the proof of the theorem.
5. Investigating the Range of
It is easy to get bounds on as the next proposition shows.
Proposition 19. Let be a finite group. Then with equality holding in both inequalities in (43) if and only if is trivial.
Proof. For any , we have that . Thus, the proposition follows from the definition of .
A natural question to ask is whether there exist nontrivial groups with arbitrarily close to the bounds given in (43). Theorem 3 provides an affirmative answer to this question.
5.1. The Proof of Theorem 3
Proof of Theorem 3. To prove part (1) of Theorem 3, first let , where is a prime and is a positive integer. Then, using Lemma 6, and Proposition 8 with , we have that
Solving the equation
for gives
Since
it follows that . Then, since , we have
A similar computation gives the same conclusion when , and the proof of part (1) of Theorem 3 is complete.
To prove part (2) of Theorem 3, first let , where is prime and is a positive integer. Then, using Lemma 6, and Proposition 8 with , we have that
Solving the equation
for gives
Since
it follows that . Also, since , we see that
Thus,
Therefore,
A similar computation yields the same conclusion when , which completes the proof of Theorem 3.
5.2. The Proof of Theorem 4
It is clear that estimating is, up to scaling, equivalent to estimating the average order of an element in . This problem was first raised by Erdös and Turán [5] in 1968 when they conjectured that Conjecture (56) was proven by Schmutz [6] in 1989 when he showed that It follows from (57) that where The proof of (57) is nontrivial and requires some very technical results concerning partitions. In 1991, Goh and Schmutz [7] sharpened (57) by proving that where The proof of (60) also requires some partition theory.
In the spirit of Theorem 3, it is our goal here to determine, if possible, the smallest possible constant such that in a nontrivial manner. In other words, does there exist a smallest real number for which (62) is true, and can we determine it? Theorem 4 provides an affirmative answer to these questions. There are two advantages to the bound given for in Theorem 4 over the results in [6, 7]. First of all, Theorem 4 is proven for all , not just for sufficiently large values of . The second advantage is that the proof of Theorem 4 that we present here does not rely on any results from partition theory. We use the following result, due to Massias [8] in 1984.
Theorem 20. Let denote the largest order of an element in . Then
In 1903, Landau [9] proved that , but Theorem 20 represents the first explicit upper bound for . Although somewhat technical, Massias’ proof of Theorem 20 uses elementary analytic techniques requiring little more than calculus.
We also need the following lemmas.
Lemma 21. Let be an integer and let , with . If fixes either or , then .
Proof. Observe that if fixes both and , then So, suppose that fixes , but not . Then we may write for some with , where and are disjoint, and for all . Thus, which completes the proof.
Lemma 22. If , then
Proof. If , then Thus, from which the lemma follows.
Lemma 23. If is an integer, then
Proof. The lemma is easily verified for . We proceed by induction on . Then since when , and the proof is complete.
Lemma 24. Let . Then for all .
Proof. The proof is straightforward using calculus.
Lemma 25. If is an integer, then
Proof. Let Since , we have
Proof of Theorem 4. The proof is by induction on . It is easy to verify computationally that the theorem is true for all , with . For , let and for each with . Then is a complete set of left coset representatives for in . Thus,
In Table 1, we compare the rounded values of , , , and the bound for given in (60), for . The bound given in (58) is much worse for these small values of than any bound given in Table 1, so we omit it.

Remark 26. The sequence is listed in the OnLine Encyclopedia of Integer Sequences and can be found at http://oeis.org/A060014.
We end with the following conjecture, which is supported by numerical evidence.
Conjecture 27. Let be a positive integer. Then Moreover, the number is the smallest possible constant such that for all .
Remark 28. The sequence is listed in the OnLine Encyclopedia of Integer Sequences and can be found at http://oeis.org/A060015.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The authors thank the referees for the valuable comments.
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Copyright
Copyright © 2014 Joshua Harrington et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.