Q: If $H$ is a subgroup of $G$ ,why does $|Ha|=|H|=|aH|$ for all $a\in G$ Proof in the book notes:$Ha \rightarrow H$ is a bijection.
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I think there may exist $x,y \in H$ such that $xa=ya= \alpha \in H$,then $xa \rightarrow \alpha ,ya \rightarrow \alpha$
it become a surjection.
Where is the mistake?

A: I will elaborate on this bijection.  Indeed fix an element $a\in G$, and associate the map,
$$\varphi_a: H\longrightarrow aH $$ given by $$\varphi_a(h)=ah $$
I will show this is a bijection between $H$ and $aH$.  First, it is easy to see this map is surjective (onto).  Take $x\in aH$, then $x=ah$ for some $h\in H$ by definition. So we have $\varphi_a(h)=x$.  Thus our map is surjective.
Now I will show the map is injective (one to one).  Let $\varphi_a(h_1)=\varphi_a(h_2)$.  This implies that $ah_1=ah_2$.  But in $G$ the element $a$ has an inverse, so we see that,
$$a^{-1}(ah_1)=a^{-1}(ah_2)\Rightarrow h_1=h_2 $$
Therefore, $\varphi_a(h_1)=\varphi_a(h_2)\Rightarrow h_1=h_2$.  So our map is injective.  Therefore, since we have that $H$ and $aH$ are in one to one correspondence, we have $|H|=|aH|$.  $a$ was arbitrary so this argument holds for each $a$ in the group.  The same reasoning applies for multiplication by $a$ on the right.  I hope this elaboration was helpful!
