Let for .Let H be a subgroup of G and suppose that the number n of right cosets of H is G is finite. Prove that there are exactly n left cosets of H in G.
(Hint: Consider the set of inverses of the elements in a right coset).
Let for some .
Then for .
This is equal to ________________(*)
Since inverses are unique, and is a group (closure axiom) we have thatb [tex] (h^k)^{-1}=h^i[/tex] for for some .
Hence we have
Since , we have exactly 's to multiply by.
Thus, we have left cosets of in .
I think this is right (can someone check =S), but step (*) applies only to multiplicative groups. Is it a fact that addition is always commutative?
I got caught up on the question before that was about cyclic groups!
As for the function being a bijection:
We have lots of . We also have lots of (proof as follows):
Since we have lots of , we have at most lots of . We will only have less if, for some , (since inverses are unique).
Therefore we have lots of .
Also, since inverses are unique, is unique to a . Therefore the function is injective.
We have a mapping from a set of elements to a set of elements which is injective. By the pigeonhole principle, the function is a bijection.
Thanks again Tonio. I have to admit that I have never seen that before. =S
However, doesn't that create a contradiction?
Suppose we have with .
Since is a group and is closed, .
This implies that .
This is also so the function isn't injective.
What's wrong with this argument?
(Also, sorry i'm being the bee in your bonnet! This question is really difficult for me).
I haven't been getting anywhere with this function idea. I can't seem to prove anything with it!
I stumbled across this:
Suppose we have right cosets ( , is a group and ).
Cosets are either equal or disjoint so they partition into a finite number of cosets (since is finite).
By Lagrange's theorem, ("order of H divides order of G", it looks confusing how i've written it!) so .
This proves the theorem (apparently).
I understand what it's done, but I fail to see how this actually works! Is it because we have at the end and not ?
We did it today! I decided to try using it for this question (since i've pretty much had it with that function!).
Was what i've written correct? I'm still trying to understand it properly. =S
Let be the number of right cosets of a group . Then, since is a subgroup, or since we know that .
be the number of left cosets of the same group. Then .
Then so so we have the same number of left and right cosets!!
Please say that's right!!