Lagrange's theorem raises the converse question as to whether every divisor of the order of a group is the order of some subgroup. This does not hold in general: given a finite group
G and a divisor
d of |
G|, there does not necessarily exist a subgroup of
G with order
d. The smallest example is
A4 (the
alternating group of degree 4), which has 12 elements but no subgroup of order 6. A "Converse of Lagrange's Theorem" (CLT) group is a finite group with the property that for every divisor of the order of the group, there is a subgroup of that order. It is known that a CLT group must be
solvable and that every
supersolvable group is a CLT group. However, there exist solvable groups that are not CLT (for example,
A4) and CLT groups that are not supersolvable (for example,
S4, the symmetric group of degree 4). There are partial converses to Lagrange's theorem. For general groups,
Cauchy's theorem guarantees the existence of an element, and hence of a cyclic subgroup, of order any prime dividing the group order.
Sylow's theorem extends this to the existence of a subgroup of order equal to the maximal power of any prime dividing the group order. For solvable groups,
Hall's theorems assert the existence of a subgroup of order equal to any
unitary divisor of the group order (that is, a divisor coprime to its cofactor).
Counterexample of the converse of Lagrange's theorem The converse of Lagrange's theorem states that if is a
divisor of the order of a group , then there exists a subgroup where . We will examine the
alternating group , the set of even
permutations as the subgroup of the
Symmetric group . : so the divisors are . Assume to the contrary that there exists a subgroup in with . Let be the
non-cyclic subgroup of called the
Klein four-group. :. Let . Since both and are subgroups of , is also a subgroup of . From Lagrange's theorem, the order of must divide both and , the orders of and respectively. The only two positive integers that divide both and are and . So or . Assume , then . If does not share any elements with , then the 5 elements in besides the
Identity element must be of the form where are distinct elements in . Since any element of the form squared is , and , any element of in the form must be paired with its inverse. Specifically, the remaining 5 elements of must come from distinct pairs of elements in that are not in . This is impossible since pairs of elements must be even and cannot total up to 5 elements. Thus, the assumptions that is wrong, so . Then, where , must be in the form where are distinct elements of . The other four elements in are cycles of length 3. Note that the cosets
generated by a subgroup of a group form a partition of the group. The cosets generated by a specific subgroup are either identical to each other or
disjoint. The index of a subgroup in a group is the number of cosets generated by that subgroup. Since and , will generate two left cosets, one that is equal to and another, , that is of length 6 and includes all the elements in not in . Since there are only 2 distinct cosets generated by , then must be normal. Because of that, . In particular, this is true for . Since . Without loss of generality, assume that , , , . Then , , , , . Transforming back, we get . Because contains all disjoint transpositions in , . Hence, . Since , we have demonstrated that there is a third element in . But earlier we assumed that , so we have a contradiction. Therefore, our original assumption that there is a subgroup of order 6 is not true and consequently there is no subgroup of order 6 in and the converse of Lagrange's theorem is not necessarily true.
Q.E.D. == History ==