Proving Cardinality of Cosets in Abstract Algebra

Question

*****************updated question**************

The question starts off like this:

Let $H$ be a subgroup of a finite group $G$. Prove that $|Ha| = |Hb|$ by proving that there exists a bijection $f : Ha \rightarrow Hb$. Find this bijection.

I know that I have to show injective and surjective but want to know if anyone thinks this is correct in this fashion

$$Let\ a\in G, \ y, x \in Ha,then \ x=ra, y=sa\ for\ some\ r,s \in H. $$ $$we\ define\ f:Ha \rightarrow Hb\ by \ f(x)=hb$$ $$suppose \ f(x)=f(y)$$ $$rb=sb\ for\ some\ r,s \in H$$ $$r=s$$ $$xa^{-1}=ya^{-1}\ (from\ x=ra\ and\ y=sa)$$ $$x=y$$ This shows injective. for the onto part this is really wanted to check $$let\ x\in Ha, y\in Hb\ and\ a,b \in G\ such\ that$$ $$x=sa,\ y=sb\ for \ some\ s \in H$$ $$then\ f(x)=sb$$ $$=yb^{-1}b\ (from \ y=sb)$$ $$=y$$ Thus, we have shown that $\forall y \in Hb, \exists\ x \in Ha \ such \ that f(x)=y $

here we have shown the function is also surjective and hence a bijection exist between $Ha$ and $Hb$ and therefore we know then $|Ha|=|Hb|$

Answer 1

Among the cosets there is $H=H1$. If, for every $a\in G$, you find a bijection $f_a\colon H\to Ha$, then the required bijection $Ha\to Hb$ can be obtained as the composition $f_b\circ f_a^{-1}$.

An obvious candidate for $f_a$ is the map $f_a\colon H\to Ha$, $f_a(h)=ha$; can you prove it's indeed bijective? If so, what's its inverse?

Answer 2

Let $x$ and $y$ be in $Ha$. Then $x=ra$ and $y=sa$ for some $r$ and $s$ in $H$. Then $f(x)=rb$ and $f(y)=sb$. If $f(x)=f(y)$, then $rb=sb$, so $r=s$, so $x=y$. Thus, $f$ is injective.