Question

Let G be a nonempty set equipped with an associative operation such that, for all $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$, the equations$\mathit{a}\mathit{x}\mathbf{=}\mathit{b}$ and $\mathit{y}\mathit{a}\mathbf{=}\mathit{b}$ have solutions. Prove that G is a group.

Short answer

It is proved that, G is a group.

Step-by-step solution

Definition of a Group

A group is a non-empty set G equipped with a binary operation $\mathbf{*}$ that satisfies the following axioms:

1. Closure:if $a\in G$and $b\in G$, then$a\ast b\in G$.
2. Associativity: role="math" localid="1654275941829" $a\ast (b\ast c)=(a\ast b)\ast c$ for all $a,b,c\in G$.
3. There is an element $e\in G$(called theidentity element) such that $a\ast e=a=e\ast a$ for every $a\in G$.
4. For each $a\in G$, there is an element $d\in G$(called theinverse of a) such that $a\ast d=e$and $d\ast e=a$.

Proving that G is a group

Since it is given in the question that $ax=b$ and $ya=b$ have solutions, therefore G satisfies the properties of closure and associativity.

To prove that G is a group, it will be sufficient to prove that (i) G has an identity element, and (ii) inverse of its element exist.

Suppose $e\left(a\right)$is the solution of the equation $ax=b$.

As we know there exists any arbitrary b such that $ya=b$.

Since $e\left(a\right)$is the solution of an equation, it should satisfy this equation $yae\left(a\right)=be\left(a\right)$ implies $ya=b$.

So, we get $e\left(a\right)=e$.

Therefore, this proves that G has an identity element.

We also know that the equation $e\left(a\right)=e$will always have solution.

Therefore, G also has an inverse.

Since G satisfies all the axioms for a group, therefore it is a group.

Hence,G is a group.