Show that, even if G is not finite, the set of left cosets of H in G is in 1-1 correspondence with the set of?

Question

Show that, even if G is not finite, the set of left cosets of H in G is in 1-1 correspondence with the set of right cosets of H in G. (Hint: correspond the inverse of Ha to aH)

Answer 1

As suggested, let's attempt to define a function from the left cosets of H to the right cosets of H by
f: aH -> H(a^-1)
First we have to prove this is well defined; if aH = bH, then we need to show that H(a^-1) = H(b^-1). (I'm assuming you already know that the left (or right) cosets of H form a partition of G.) Suppose that aH = bH, then for some h ∈ H we know b = ah (this is from the same proof that shows the cosets of H partition G; briefly, if ah_1 = bh_2 then b = a(h_1 h_2^-1) and h_1 h_2^-1 is in H). Then b^-1 = h^-1 a^-1. So Hb^-1 = H(h^-1 a^-1) = Ha^-1 since h^-1 is in H. So f is well defined.

Now we need to show that f is a bijection. Clearly f is onto since for any right coset Ha of H, Ha = f(a^-1 H). So we must show that f is 1-1, i.e. that if Ha^-1 = Hb^-1 then aH = bH. But this is just the same as before - if Ha^-1 = Hb^-1 then there is an h ∈ H such that b^-1 = ha^-1, and so b = a h^-1 and bH = (a h^-1) H = aH since h^-1 ∈ H.

So there exists a bijection between the left and right cosets of H in G.

Answer 2

archives via ability of induction on the order 2n. assume your certainty is definitely for all abelian communities of order 2m with m unusual and m