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

Question

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 $ax=ay\Longrightarrow \mathit{l}_{a^{-1}}(ax)=\mathit{l}_{a^{-1}}(ay)\Longrightarrow x=y$

$\Box$

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Could someone verify if my solution is correct? And if not, give me some feedback :)? thank you

Answer 1

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

$$x=ex=(a^{-1}a)x=a^{-1}(ax)=a^{-1}(ay)=(a^{-1}a)y=ey=y\;.$$

Answer 2

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 $x=y \rightarrow f(x) = f(y)$.

Answer 3

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 $a^{-1}a=e$.