Group/Normal subgroup/Residue class group/Fact/Proof
Proof
Since the canonical projection shall be a group homomorphism, the operation must fulfill
![{\displaystyle {}[x][y]=[xy]\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1ab87a0ba8284f428d7efbef3b2148f40d40bef)
We have to show that this rule gives a well-defined operation on , that is, is independent of the choice of representatives. Hence, we have to show for and that holds. Due to the condition, we can write and with . Therefore,
![{\displaystyle {}[x]=[x']}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e5406f8a38ea7d2bf4f2e1660bc4f25a9278bc6)
![{\displaystyle {}[y]=[y']}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e9087fc64dc5231f3b03084437b09fbc46c602d)
![{\displaystyle {}[xy]=[x'y']}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4abdb835196fa2640a5dc3ca0542374c36afd05)
This means . From this, the group property, the homomorphism property of the projection and the uniqueness follows.