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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.

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    $\begingroup$ In any finite Abelian group, the converse will hold!! $\endgroup$ – Chinnapparaj R Feb 14 '17 at 14:31
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    $\begingroup$ @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. $\endgroup$ – user170039 Feb 14 '17 at 14:32
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    $\begingroup$ As a first step, the group will be solvable by Hall's theorem. $\endgroup$ – Tobias Kildetoft Feb 14 '17 at 14:34
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    $\begingroup$ @user170039 Every $p$-group by Sylow's thorem satisfies Lagrange's converse. I.e. it does not have to abelian. $\endgroup$ – freakish Feb 14 '17 at 14:35
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    $\begingroup$ @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. $\endgroup$ – Tobias Kildetoft Feb 14 '17 at 14:36
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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.

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

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