Complete classification of the groups for which converse of Lagrange's Theorem holds

It is known that the converse of Lagrange's Theorem isn't true in general. More precisely it is known that the following proposition:

If $$G$$ is a finite group of order $$n$$ and $$m\mid n$$ then there exists a subgroup $$H$$ of $$G$$ such that $$\operatorname{order}(H)=m$$.

isn't true for all finite groups $$G$$.

My questions are:

• For which groups $$G$$ does the converse of Lagrange's Theorem (as stated above) hold? More precisely, if $$G$$ is a group for which the converse of Lagrange's Theorem as I mentioned above holds then what properties must $$G$$ satisfy?

• If there is no complete classification of such $$G$$s then can someone give me references to works by other mathematicians where they try to give at least a partial classification of these $$G$$s?

Please note that I am not interested in knowing a complete classification of the groups for which a partial converse holds (Sylow's Theorems does the job in some sense). I want to know a complete classification of the groups for which the converse of Lagrange's Theorem as I mentioned above holds.

• In any finite Abelian group, the converse will hold!! – Chinnapparaj R Feb 14 '17 at 14:31
• @R.Chinnapparaj: But if the converse of Lagrange's Theorem hold for a finite group $G$, must it be Abelian? Please read my question again. – user170039 Feb 14 '17 at 14:32
• As a first step, the group will be solvable by Hall's theorem. – Tobias Kildetoft Feb 14 '17 at 14:34
• @user170039 Every $p$-group by Sylow's thorem satisfies Lagrange's converse. I.e. it does not have to abelian. – freakish Feb 14 '17 at 14:35
• @freakish And more generally, all nilpotent groups satisfy this, so the group is somewhere between nilpotent and solvable. There is a book with a name along those lines, so this might be covered in that, though I don't think I ever managed to track down a copy myself. – Tobias Kildetoft Feb 14 '17 at 14:36

Such groups are called Lagrangian, or CLT-groups. They have been studied often in the literature. There is no complete classification, but many interesting criteria. Two (out of many) references are the following:

• H. G. Bray: A note on CLT groups, Pacific Journal of Mathematics 27 (1968), no. 2., 229-231.

• F. Barry, D. MacHale, A. N. She: Some Supersolvability conditions for finite groups., Math. Proceedings of the Royal Irish Academy 167 (1996), 163--177.

Definition: A finite group $G$ is called Lagrangian if and only if for each positive divisor $d$ of $|G|$ there exists at least one subgroup $H\le G$ with $|H|=d$.

It is easy to see that every Lagrangian group is solvable, and conversely every supersolvable group is Lagrangian. The inclusions are strict. In fact, every group $G=A_4\times H$ with a group $H$ of odd order is solvable, but not Lagrangian; and for any Lagrangian group $G$, the group $(A_4\times C_2)\times G$ is Lagrangian, but not supersolvable. The classical counterexample to Lagrange's Theorem is $A_4$.

For example, no group $S_n$ or $A_n$ with $n\ge 5$ is Lagrangian. This follows from the fact that $A_n$ and $S_n$ are not solvable for $n\ge 5$. There are some more interesting facts, which can be easily found in the literature. For example, we have:

Proposition: If $(G:Z(G))<12$ for the index, then $G$ is supersolvable, hence Lagrangian.

The group $A_4$ shows that the above result is best possible. We have $(A_4:Z(A_4))=12$.

In the paper of Barry et al. the following result is shown:

Proposition: If $|[G,G]|<4$, then $G$ is supersolvable, hence Lagrangian.

Again $A_4$ shows that this result is best possible.

Proposition: If $|G|$ is odd and $|[G,G]|<25$, then $G$ is supersolvable, hence Lagrangian.

In fact, $[G_{75},G_{75}]\simeq C_5\times C_5$ has order $25$, so that this result is best possible. Here $G_{75}$ denotes the unique non-abelian group of order $75$.

Denote the number of different conjugacy classes of $G$ by $k(G)$.

Proposition: If $\frac{k(G)}{|G|}>\frac{1}{3}$, then $G$ is supersolvable, hence Lagrangian.

Because of $\frac{k(A_4)}{|A_4|}=\frac{1}{3}$ the result is best possible. It means that if the average size of a conjugacy class of $G$ is less than $3$, then $G$ is Lagrangian.

Proposition: If $|G|$ is odd and $\frac{k(G)}{|G|}>\frac{11}{75}$, then $G$ is supersolvable, hence Lagrangian.

In fact, $\frac{k(G_{75})}{|G_{75}|}=\frac{11}{75}$, so that the result is best possible.

Finally, let us mention a result of Pinnock ($1998$), which is related to Burnside's $p^aq^b$-theorem on the solvability of groups of such order.

Proposition: Let $G$ be a group of order $pq^b$ with primes $p,q$ satisfying $q\equiv 1 \bmod p$. Then $G$ is supersolvable, hence Lagrangian.

• I have changed the formatting a bit (as you can see). But feel free to rollback it if you don't want this change. – user170039 Feb 15 '17 at 3:32
• On the contrary, it looks much better. Thank you! – Dietrich Burde Feb 15 '17 at 9:08
• Sir what this notation $|[G,G]|$ means – Akash Patalwanshi Feb 19 '20 at 13:23
• @AkashPatalwanshi $|H|$ denotes the order of $H$. Then take $H=[G,G]$, the commutator subgroup. – Dietrich Burde Feb 19 '20 at 13:51