Abstract Algebra/Group Theory/Subgroup/Coset/a Group is Partitioned by Cosets of Its Subgroup

=Theorem= Let G be a Group. Let H be a Subgroup of G.


 * Then, Cosets of Subgroup H partition Group G.

=Proof=

Overview: G is partition by the cosets if


 * 1) The cosets are subsets of G
 * 2) Each element of G is in one of the cosets.
 * 3) The cosets are disjoint

 Cosets of H are Subsets of G

 * 0. Choose $$ g \in G $$


 * 1. Choose $$ k \in gH $$

By definition of gH
 * 2. $$ \exists \; h \in H: k = g \ast h $$

As Subgroup H is Subset of G
 * 3. $$ h \in G $$

By 2., and Closure on G justified by 0. and 3.,
 * 4. $$ k = g \ast h \in G$$

 Each Element of G is in a Coset of H

 * {| Style ="width:60%"

|-    |-     |-
 * 1. $$e_{G} \in H$$
 * subgroup inherits identity (usage 2)
 * 2. Choose $$ g \in G$$
 * 3 $$g \ast e_{G} \in gH$$
 * 3 $$g \ast e_{G} \in gH$$

|-
 * definition of gH
 * 4. $$ g = g \ast e_{G} \in gH $$
 * 4. $$ g = g \ast e_{G} \in gH $$


 * eG is identity of G (usage 3)
 * }

The Cosets of H are Disjoint

 * 0. Suppose 2 different cosets of H are not disjoint


 * 1. Let the 2 cosets be g1 H and g2 H where $$ {\color{Blue}g_1}, {\color{OliveGreen}g_2} \in G $$

Since they are not disjoint
 * 2. $$ \exists u \in G: u \in {\color{Blue}g_1}H \text{ and } u \in {\color{OliveGreen}g_2}H $$

By Definition of the Cosets,
 * 3. $$ \exists {\color{Blue}h_1}, {\color{OliveGreen}h_2} \in H: {\color{Blue}g_1} \ast {\color{Blue}h_1} = u = {\color{OliveGreen}g_2} \ast {\color{OliveGreen}h_2} $$


 * Let $$ z = {\color{OliveGreen}g^{-1}_2} \ast {\color{Blue}g_1} = {\color{OliveGreen}h_2} \ast {\color{Blue}h^{-1}_1} \in H $$


 * 4. Choose $$ k \in g_1 H $$

By Definition of g1H
 * 5. $$ \exists \; h_{k} \in H: k = g_1 \ast h_{k} $$


 * 6. $$ z \ast h_{k} \in H $$


 * 7. $$ g_2 \ast (z \ast h_{k}) \in g_2 H $$


 * 8. $$ g_2 \ast g^{-1}_2 \ast g_1 \ast h_{k} \in g_2 H $$


 * 9. $$ g_1 \ast h_{k} \in g_2 H $$


 * 10. $$ k \in g_2 H $$


 * 11. $$ g_1 H \subseteq g_2 H $$

As we can exchange g_1 and g_2 and apply the same procedure
 * 12. $$ g_2 H \subseteq g_1 H $$


 * 13. $$ g_1 H = g_2 H $$ contradicting that the two coset are different (0.)

Thus, two Cosets of H are either identical or are disjoint. Hence, the Cosets of H are disjoint.