Show that $\forall a,x,y \in G:ax=ay\Longrightarrow x=y$ with $(G,\cdot)$ being a group 
Let $(G,\cdot)$ be a group, with $G$ being a finite set.
Show that $\forall a,x,y \in G:ax=ay\Longrightarrow x=y$

Since $(G,\cdot)$ is a group $a \in G \Longrightarrow a^{-1} \in G$ with $a^{-1}a=e$
So we now define the automorphism (bijection!):
$\mathit{l}_{a^{-1}}:G\longrightarrow G:g\mapsto a^{-1}g$
(that the map is bijective was proven in the chapter before)
Now:
$$\mathit{l}_{a^{-1}}(ax)=a^{-1}ax=x$$
$$\mathit{l}_{a^{-1}}(ay)=a^{-1}ay=y$$
Since the map is injective $\forall a,b \in G:a=b \Longrightarrow\mathit{l}_{a^{-1}}(a)=\mathit{l}_{a^{-1}}(b)$
So $ax=ay\Longrightarrow \mathit{l}_{a^{-1}}(ax)=\mathit{l}_{a^{-1}}(ay)\Longrightarrow x=y$
$\Box$

Could someone verify if my solution is correct? And if not, give me some feedback :)? thank you
 A: It’s basically correct apart from some terminology (for which see FiMePr’s answer), but you’re working much harder than necessary: if $ax=ay$, then
$$x=ex=(a^{-1}a)x=a^{-1}(ax)=a^{-1}(ay)=(a^{-1}a)y=ey=y\;.$$
A: The general idea is correct, but some statements are false :
First, the map $g \mapsto a^{-1} g$ is not an automorphism of the group $G$. It is only a bijection.
Then, your notations are a bit perilous : you write $\forall a,b \in G : a= b \rightarrow I_{a^{-1}}(a) = I_{a^{-1}}(b) $ The symbol $a$ appears twice here. This is not a good idea in general. To solve this, you can simply write : $\forall x,y \in G : x= y \rightarrow I_{a^{-1}}(x) = I_{a^{-1}}(y) $
Finally, you do not need injectivity of a map $f$ to prove the implication $x=y \rightarrow f(x) = f(y)$.
A: Suppose $a,x,y$ are arbitrary in a group $G$ such that
$$ax=ay.\tag{1}$$
Multiply on the left of $(1)$ by $a^{-1}\in G$ like so:
$$a^{-1}(ax)=a^{-1}(ay).\tag{2}$$
Apply associativity to $(2)$ to get
$$(a^{-1}a)x=(a^{-1}a)y.\tag{3}$$
Now $(3)$ gives $x=y$, since $a^{-1}a=e$.
