Lagrange's theorem (group theory)
Counterexample of the converse of Lagrange's theorem

The converse of Lagrange's theorem states that if d is a divisor of the order of a group G, then there exists a subgroup H where | H | = d.

We will examine the alternating group A, the set of even permutations as the subgroup of the Symmetric group S.

| A | = 12 so the divisors are 1, 2, 3, 4, 6, 12. Assume to the contrary that there exists a subgroup H in A with | H | = 6.

Let V be the non-cyclic subgroup of A called the Klein four-group.

Let K = H ⋂ V. Since both H and V are subgroups of A, K is also a subgroup of A.

From Lagrange's theorem, the order of K must divide both 6 and 4, the orders of H and V respectively. The only two positive integers that divide both 6 and 4 are 1 and 2. So | K | = 1 or 2.

Assume | K | = 1, then K = { e }. If H does not share any elements with V, then the 5 elements in H besides the Identity element e must be of the form (a b c) where a, b, c are distinct elements in {1, 2, 3, 4}.

Since any element of the form (a b c) squared is (a c b), and (a b c)(a c b) = e, any element of H in the form (a b c) must be paired with its inverse. Specifically, the remaining 5 elements of H must come from distinct pairs of elements in A that are not in V. This is impossible since pairs of elements must be even and cannot total up to 5 elements. Thus, the assumptions that | K | = 1 is wrong, so | K | = 2.

Then, K = { e, v } where v ∈ V, v must be in the form (a b)(c d) where a, b, c, d are distinct elements of {1, 2, 3, 4}. The other four elements in H 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 = | A |/| H | is the number of cosets generated by that subgroup. Since | A | = 12 and | H | = 6, H will generate two left cosets, one that is equal to H and another, gH, that is of length 6 and includes all the elements in A not in H.

Since there are only 2 distinct cosets generated by H, then H must be normal. Because of that, H = gHg (∀ g ∈ A). In particular, this is true for g = (a b c) ∈ A. Since H = gHg, gvg ∈ H.

Without loss of generality, assume that a = 1, b = 2, c = 3, d = 4. Then g = (1 2 3), v = (1 2)(3 4), g = (1 3 2), gv = (1 3 4), gvg = (1 4)(2 3). Transforming back, we get gvg = (a d)(b c). Because V contains all disjoint transpositions in A, gvg ∈ V. Hence, gvg ∈ H ⋂ V = K.

Since gvg ≠ v, we have demonstrated that there is a third element in K. But earlier we assumed that | K | = 2, 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 A and the converse of Lagrange's theorem is not necessarily true. Q.E.D.
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