Editing Lagrange's theorem (group theory) (section)

=== Counterexample of the converse of Lagrange's theorem ===
The converse of Lagrange's theorem states that if {{mvar|d}} is a [[divisor]] of the order of a group {{mvar|G}}, then there exists a subgroup {{mvar|H}} where {{math|1={{!}}''H''{{!}} = ''d''}}.

We will examine the [[alternating group]] {{math|''A''<sub>4</sub>}}, the set of even [[Permutation group|permutations]] as the subgroup of the [[Symmetric group]] {{math|''S''<sub>4</sub>}}.

:{{nowrap|{{math|1=''A''<sub>4</sub> = {{mset|''e'', (1 2)(3 4), (1 3)(2 4), (1 4)(2 3), (1 2 3), (1 3 2), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3)}}}}.}}

{{math|1={{!}}''A''<sub>4</sub>{{!}} = 12}} so the divisors are {{math|1, 2, 3, 4, 6, 12}}. Assume to the contrary that there exists a subgroup {{mvar|H}} in {{math|''A''<sub>4</sub>}} with {{math|1={{!}}''H''{{!}} = 6}}.

Let {{mvar|V}} be the [[Cyclic group|non-cyclic]] subgroup of {{math|''A''<sub>4</sub>}} called the [[Klein four-group]].

:{{math|1=''V'' = {{mset|''e'', (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}}}}.

Let {{math|1=''K'' = ''H'' ⋂ ''V''}}. Since both {{mvar|H}} and {{mvar|V}} are subgroups of {{math|''A''<sub>4</sub>}}, {{mvar|K}} is also a subgroup of {{math|''A''<sub>4</sub>}}.

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

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

Since any element of the form {{math|(''a b c'')}} squared is {{math|(''a c b'')}}, and {{math|1=(''a b c'')(''a c b'') = ''e''}}, any element of {{mvar|H}} in the form {{math|(''a b c'')}} must be paired with its inverse. Specifically, the remaining 5 elements of {{mvar|H}} must come from distinct pairs of elements in {{math|''A''<sub>4</sub>}} that are not in {{mvar|V}}. This is impossible since pairs of elements must be even and cannot total up to 5 elements. Thus, the assumptions that {{math|1={{!}}''K''{{!}} = 1}} is wrong, so {{math|1={{!}}''K''{{!}} = 2}}.

Then, {{math|1=''K'' = {{mset|''e'', ''v''}}}} where {{math|''v'' ∈ ''V''}}, {{mvar|v}} must be in the form {{math|(''a b'')(''c d'')}} where {{mvar|a, b, c, d}} are distinct elements of {{math|{{mset|1, 2, 3, 4}}}}. The other four elements in {{mvar|H}} are cycles of length 3.

Note that the cosets [[Generating set of a group|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 sets|disjoint]]. The index of a subgroup in a group {{math|1=[''A''<sub>4</sub> : ''H''] = {{!}}''A''<sub>4</sub>{{!}}/{{!}}''H''{{!}}}} is the number of cosets generated by that subgroup. Since {{math|1={{!}}''A''<sub>4</sub>{{!}} = 12}} and {{math|1={{!}}''H''{{!}} = 6}}, {{mvar|H}} will generate two left cosets, one that is equal to {{mvar|H}} and another, {{mvar|gH}}, that is of length 6 and includes all the elements in {{math|''A''<sub>4</sub>}} not in {{mvar|H}}.

Since there are only 2 distinct cosets generated by {{mvar|H}}, then {{mvar|H}} must be normal. Because of that, {{math|1=''H'' = ''gHg''<sup>−1</sup> (∀''g'' ∈ ''A''<sub>4</sub>)}}. In particular, this is true for {{math|1=''g'' = (''a b c'') ∈ ''A''<sub>4</sub>}}. Since {{math|1=''H'' = ''gHg''<sup>−1</sup>, ''gvg''<sup>−1</sup> ∈ ''H''}}.

Without loss of generality, assume that {{math|1=''a'' = 1}}, {{math|1=''b'' = 2}}, {{math|1=''c'' = 3}}, {{math|1=''d'' = 4}}. Then {{math|1=''g'' = (1 2 3)}}, {{math|1=''v'' = (1 2)(3 4)}}, {{math|1=''g''<sup>−1</sup> = (1 3 2)}}, {{math|1=''gv'' = (1 3 4)}}, {{math|1=''gvg''<sup>−1</sup> = (1 4)(2 3)}}. Transforming back, we get {{math|1=''gvg''<sup>−1</sup> = (''a'' ''d'')(''b'' ''c'')}}. Because {{mvar|V}} contains all disjoint transpositions in {{math|''A''<sub>4</sub>}}, {{math|''gvg''<sup>−1</sup> ∈ ''V''}}. Hence, {{math|1=''gvg''<sup>−1</sup> ∈ ''H'' ⋂ ''V'' = ''K''}}.

Since {{math|''gvg''<sup>−1</sup> ≠ ''v''}}, we have demonstrated that there is a third element in {{mvar|K}}. But earlier we assumed that {{math|1={{!}}''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 {{math|''A''<sub>4</sub>}} and the converse of Lagrange's theorem is not necessarily true.
[[Q.E.D.]]