Proof of a group (in the context of finite groups)

Question

my task is the following:

Let $G$ be a finite set with an inner connection $\circ: G \times G \rightarrow G$, which is associative and for which a neutral element exists in $G$. In addition, for all $a,b,c \in G$, it applies that from $a \circ b = a \circ c$ also $b = c$ follows. Show that $ G $ is a group.

My idea was:

The following axioms must apply to a group:

1. Closure
2. Associativity
3. Neutral element
4. Existence inverse

The first 3 axioms are already given according to the task definition (please correct me if I am wrong). So I only have to show the inverse.

Here's what I'm thinking: Be $a \in G$ arbitrary. We want to show that there is a $a′\in G$ with $a \circ a′$ being the neutral element of $G$. Consider the left multiplication with a, i.e. the map $l_a : G \to G$ defined by $ x \rightarrow a \circ x$.

Which property of $l_a$ can I deduce from the precondition $a \circ b = a \circ c \implies b = c$ and how can I continue?

Could someone help me please. Thanks in advance.

Answer 1

You have to use that $l_a(x)=a \circ x$ is bijective since $G$ is finite, you know that it is injective since $a \circ x=a\circ y$ implies $x=y$. There exists $a^{-1}$ such that $a \circ a^{-1}=e$, use the associativity to show that $a^{-1}$ is a left inverse of $a$:

$$\begin{align} (a^{-1} \circ a) \circ a^{-1}&=a^{-1} \circ(a \circ a^{-1})\\ &=a^{-1} \circ e\\ &=a^{-1}. \end{align}$$

Since $e \circ a^{-1}=(a^{-1} \circ a) \circ a^{-1}$, we deduce that $a^{-1}\circ a =e$.

Answer 2

Fix a neutral element $e$.

Fix $a \in G$. The function $x \mapsto a \circ x$ is an injection (given). Because $G$ is finite, every injection is a bijection. Therefore, there exists some $x$ such that $a \circ x = e$. Thus, every element has a right inverse. This is sufficient for every element to have an inverse.

Then $G$ is a group.