For G a group and H a subgroup of G, I'm required to show that $|aH| = |bH|$ for all $a,b\epsilon G$
Proof: Define a function $f:aH\to bH$ such that $f(ah) = bh$ for some $h\epsilon H$ Then we show, f is 1-1 and onto, i.e. f is a bijection 1) f is 1-1: let $f(ah)=f(h')$ then $bh = bh'$ multiplying by $b^{-1}$ to the left we get $ h= h'$ Hence, f is injective
2) I have to show that f is onto but I don't know how. Can anyone let me know as well as verify the aforementioned proof for injectivity of f?
One-one: If $f(ah)=f(ah')$, then $bh=bh'$ and the cancellation property gives $h=h'$.
Onto: For all $bh\in bH$, $f(ah)=bh$. This covers every element of $bH$.