Recently I have been reading about groups from Michael Artin's Algebra. There I read a statement that there always is a bijective map from a subgroup $H$ to a coset $aH$. As to why there exists such a bijective map, no proof was given. Injectivity is quite obvious.
Here I have tried to come up with a proof for surjectivity and wanted to verify it.
Let $H$ be a subgroup of $G$ where $H=\{h_1,h_2,\ldots,h_n\}$
Let $aH$ be a left coset of $H$ for $a\in G$ and let $h_x$ be an arbitary element of $H$
Since $H$ is a subgroup of $G$, $$h_x\in G$$ $$\Rightarrow h^{-1}_x\in G$$ $$a\in G$$ $$\Rightarrow ah^{-1}_x\in G$$ Now let $ah^{-1}_x=b$ $$ah^{-1}_xh_x=bh_x$$ $$a=bh_x$$ $\therefore b\in aH$
Now as $h_x$ was an arbitary element of $H$, for every $h_x\in H$, there exist a $b\in G$ such that $a=bh_x$
$\therefore aH$ contains as many elements as in $H$. Now as there is an injective mapping from $H$ to $aH$, the same mapping must be surjective.
Please verify this proof and offer suggestions or alternate methods for this proof.
Your proof is okay, but I don't see the need for the subscript $x$ for your element of $H$. It's also too long.
Here's the nub of the argument . . .
The bijection is $$\begin{align} \lambda_a: H&\to aH,\\ h&\mapsto ah \end{align}$$
with inverse
$$\begin{align} \lambda^{-1}_a: aH&\to H,\\ k&\mapsto a^{-1}k. \end{align}$$