On CLT and non-CLT groups

Q: What is a CLT group and how is it related to the Converse of Lagrange’s Theorem?

A finite group is CLT if it has subgroups of every possible order dividing its size, meaning it satisfies the Converse of Lagrange’s Theorem. Otherwise it is non-CLT.

Q: For which integers d does the paper construct a solvable group G with no subgroup of order d?

For every integer d≥2 that is not a prime power, the paper shows there exists a finite solvable group G such that d divides |G|, the prime divisors match, and G has no subgroup of order d.

Q: What is the CLT-degree dCLT(G) and what key property does it have?

The CLT-degree dCLT(G) measures the proportion of divisors of |G| that occur as subgroup orders. It satisfies 0<dCLT(G)≤1, is multiplicative for coprime-order products, and equals 1 exactly when G is CLT.

Sep 15, 2026

Uploaded by Franzy

This note studies finite groups in the context of the Converse of Lagrange’s Theorem, distinguishing CLT groups from non-CLT groups. For every integer d≥2 that is not a prime power, it constructs a finite solvable group G such that d divides |G|, the prime divisors of |G| and d match, and G has no subgroup of order d. It also defines the CLT-degree, proves density results, and derives corollaries answering related questions.

LanguageEnglish

CategoryResearch & Report

FormatPDF

Pages8

Reading Time20 min

AccessFree to download

Introduction

Proof of Theorem 1.1

Auxiliary results

Proof of Theorem 1.1 (structure)

arXiv :2403 .05774v1 [math .GR] 9 Mar 2024 On CLT and non-CLT groups Marius T􀀕arn􀀕auceanu March 9, 2024 Abstract In this note, we prove that for every integer d ≥ 2 which is not a prime power, there exists a 􀀌nite solvable group G such that d | |G| ,􀀙 (G) = 􀀙 (d) and G has no subgroup of order d. We also introduce the CLT-degree of a 􀀌nite group and answer two questions about it. MSC2020 : Primary 20D60; Secondary 20D99, 20E99 . Key words : CLT group, solvable group, Frobenius group. 1 Introduction All groups considered in this note are 􀀌nite. A group is said to be CLT if it possesses subgroups of every possible order (that is, it satis􀀌es the Converse of Lagrange’s Theorem) and non-CLT otherwise. It is well-known that CLT groups are solvable (see [11]) and that supersolvable groups are CLT (see [10]) . Moreover, the inclusion between the classes of CLT groups and solvable groups, as well as the inclusion between the classes of supersolvable groups and CLT groups are proper (see, for example, [3]) . Recall also the papers [1, 2, 13] which construct non-CLT groups of order p􀀋 q􀀌 with p, q primes and 􀀋,􀀌 ∈ N∗ . The 􀀌rst starting point for our work is given by the smallest examples of non-CLT groups, namely A4 and SL(2 , 3) . It is well-known that they have no subgroup of order 6 and 12, respectively. This leads to the following natural question: Given an integer d ≥ 2, is it possible to construct a group whose order is divisible by d, but having no subgroup of order d? Obviously, the answer to this question is ”NO” if d is a prime power. So, in what follows we will assume that d is a composite number and we will denote by 􀀙(d) the set of primes dividing d. Also, for a 􀀌nite group G we will denote 􀀙 (G) = 􀀙(|G|) . Our 􀀌rst result shows that the answer to the above question is ”YES” for all composite numbers d. Theorem 1.1 . For every integer d ≥ 2 which is not a prime power, there exists a 􀀌nite solvable group G such that d | |G| , 􀀙 (G) = 􀀙(d) and G has no subgroup of order d. The second starting point for our work is given by a question of Martino Garonzi on MathOver􀀍ow [4]: Is there a constant c > 0 such that D􀀜(~~ ~~|(GG|)) > c for any 􀀌nite group G? Here D (G) denotes the number of divisors d of |G| for which there exists a subgroup of G of order d and 􀀜 (|G|) denotes the number of all divisors of |G| . Note that this question remained unanswered for more than ten years. It suggests to consider the function D (G) dCLT (G) = , 􀀜 (|G|) which will be called the CLT-degree of G. Clearly, dCLT (G) measures the probability of a 􀀌nite group G to be CLT. It is easy to see that this new function satis􀀌es the following properties: - 0 < dCLT (G) ≤ 1, for any 􀀌nite group G. Moreover, dCLT (G) = 1 if and only if G is CLT. - dCLT is multiplicative, that is dCLT (G1 ×G2 ) = dCLT (G1 )dCLT (G2 ), for any 􀀌nite groups G 1 , G2 of coprime orders. - If N is a normal subgroup of G, then 􀀜 (|G|) dCLT (G/N) ≤ dCLT (G) . 􀀜 (|G|/|N|) - If |G| = pn11 · · ·pnkk with k ≥ 2, then dCLT (G) ≥ PQki~~1~~(nnii~~ ~~21) . Moreover, we have equality if and only if G has all proper subgroups of prime power order. Such a group is either cyclic of order pq, where p and q are distinct primes, or a semidirect product Cnp ⋊ Cq , where Cq acts irreducibly on Cnp . Thus n + 3 dCLT (Cnp ⋊ Cq ) = 2n + 2 and, in particular, dCLT (A4 ) = ~~5~~6 . Let G be the class of all 􀀌nite groups. Our second result is stated as follows. Theorem 1 .2. The set Im(dCLT ) = {dCLT (G) | G ∈ G} is dense in [0 , 1] . Obviously, Theorem 1 .2 gives a negative answer to Garonzi’s question. Corollary 1.3 . There is no constant c > 0 such that D􀀜(~~ ~~|(GG|)) > c for any 􀀌nite group G. We also infer that: Corollary 1.4 . There is no constant c < 1 such that dCLT (G) ≤ c for any 􀀌nite non-CLT group G. Note that an example of a family of 􀀌nite non-CLT groups (Gn )n≥1 with limn→∞ dCLT (Gn ) = 1 will be given at the end

FAQs

What is a CLT group and how is it related to the Converse of Lagrange’s Theorem?

A finite group is CLT if it has subgroups of every possible order dividing its size, meaning it satisfies the Converse of Lagrange’s Theorem. Otherwise it is non-CLT.

For which integers d does the paper construct a solvable group G with no subgroup of order d?

For every integer d≥2 that is not a prime power, the paper shows there exists a finite solvable group G such that d divides |G|, the prime divisors match, and G has no subgroup of order d.

What is the CLT-degree dCLT(G) and what key property does it have?

The CLT-degree dCLT(G) measures the proportion of divisors of |G| that occur as subgroup orders. It satisfies 0<dCLT(G)≤1, is multiplicative for coprime-order products, and equals 1 exactly when G is CLT.

Sep 15, 2026

Uploaded by Franzy