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

Solution 1:

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.