I'm not looking for the solution, but a helpful first step.
So far, I realize that if , then ] is infinite. More generally, if , then is infinite.
Problem is, I don't see how to show anything more than
Thanks for your help.
I'm not looking for the solution, but a helpful first step.
So far, I realize that if , then ] is infinite. More generally, if , then is infinite.
Problem is, I don't see how to show anything more than
Thanks for your help.
Okay. I'm still not getting it. Maybe this example is too abstract.
Lets consider the integers mod 12. Let H = {0,3,6,9} and K = {0,4,8}. Thus, .
Thus would be defined as
0 = ({0,3,6,9},{0,4,8})
1 = ({1,4,7,10},{1,5,9})
2 = ({2,5,8,11},{2,6,10})
3 = ({0,3,6,9},{3,7,11})
4 = ({1,4,7,10},{0,4,8})
5 = ({2,5,8,11},{1,5,9})
6 = ({0,3,6,9},{2,6,10})
7 = ({1,4,7,10},{3,7,11})
8 = ({2,5,8,11},{0,4,8})
9 = ({0,3,6,9},{1,5,9})
10 = ({1,4,7,10},{2,6,10})
11 = ({2,5,8,11},{3,7,11})
Now, the claim is that for any , this function is one-to-one. I don't see how you could make this claim if you don't know anything about ?
I notice, though, that in the above example, returns the cosets . If this function is indeed one-to-one, there is one and only one with . Therefore, since there are finitely many such cosets, there are finitely many such g with .
I guess I understand where this is going, but I don't know how to prove the injectivity if you don't know .
After looking at this for a little while longer, I realized that maybe I'm overcomplicating this; maybe I see the logic about why you don't really need to know the intersection of H and K to prove anything.
Does this look about right?