- #1
Bashyboy
- 1,421
- 5
Here is a link to a proof which I am trying to understand.
http://groupprops.subwiki.org/wiki/Left_cosets_partition_a_group
The claim I am referring to is number 4, which is
Any two left cosets of a subgroup either do not intersect, or are equal.
Assuming that I am skeptical, then for all I know there are three cases: (1) ##aH \cap bH \ne \emptyset## but ##aH \ne bH##; (2) ##aH = bH##; or (3) ##aH \cap bH = \emptyset##.
The truthfulness of (2) and (3) is made reasonable by simple examples and calculations involving certain groups, such as ##D_4##. However, working with these examples, it is not clear that (1) is true or false; consequently, it remains as a possibility.
In the proof given in the link, they start out by supposing that ##aH## and ##bH## are not disjoint, that they could have some elements in common. Continuing on in the proof, we see that by supposing this is true, this undoubtedly leads to the cosets being equal; in doing this, they also show that it is not possible for ##aH \cap bH \ne \emptyset## but ##aH \ne bH##.
But they don't treat whether it is possible for ##aH \cap bH = \emptyset## to be true. Why is that?
http://groupprops.subwiki.org/wiki/Left_cosets_partition_a_group
The claim I am referring to is number 4, which is
Any two left cosets of a subgroup either do not intersect, or are equal.
Assuming that I am skeptical, then for all I know there are three cases: (1) ##aH \cap bH \ne \emptyset## but ##aH \ne bH##; (2) ##aH = bH##; or (3) ##aH \cap bH = \emptyset##.
The truthfulness of (2) and (3) is made reasonable by simple examples and calculations involving certain groups, such as ##D_4##. However, working with these examples, it is not clear that (1) is true or false; consequently, it remains as a possibility.
In the proof given in the link, they start out by supposing that ##aH## and ##bH## are not disjoint, that they could have some elements in common. Continuing on in the proof, we see that by supposing this is true, this undoubtedly leads to the cosets being equal; in doing this, they also show that it is not possible for ##aH \cap bH \ne \emptyset## but ##aH \ne bH##.
But they don't treat whether it is possible for ##aH \cap bH = \emptyset## to be true. Why is that?