I wanted to try and find a bijection between them but I wasn't sure quite how to do it.Suppose that G is a group, H is a subgroup of G, and . Show that .
(Do not assume that G is finite).
Alternatively, is there a way to prove it without using bijections?
Not really... You have to show that the map is both one-one and onto.
This is equivalent to doing the following steps:
One-One: Show that
Onto: Show that
This is wrong, be careful.
The question is asking you to prove they have the same number of elements , its not asking you to prove the sets are equal.
The argument you are using is claiming the sets and are the same. But that's not necessarily true here.
Show
Injective:
Suppose . (Need To Show: ).
Since using the property of groups that they have inverses means:
Therefore the function is injective.
Surjective:
I'm not really sure how to show it is surjective.
I Need To Show: holds for all elements in .
Can I define to be where ? Since all the 's are different this would be a surjection (i'm wondering whether this is a reasonable assumption).
Combining this result with the proof that it is injective would prove is bijective.
You are right
Surjectivity is not an assumption.
For the sake of clarity, let me explain the details:
Every element of is of the form for some . Now maps to this . Thus every element in the range is mapped onto by some element in the domain and hence the map is onto.