Question

Let G be a nonempty set equipped with an associative operation with these properties:

$\mathbf{\left(}\mathbf{i}\mathbf{\right)}$There is an element$\mathit{e}\mathbf{\in }\mathit{G}$ such that$\mathit{e}\mathit{a}\mathbf{=}\mathit{a}$ for every $\mathit{a}\mathbf{\in }\mathbf{\text{\hspace{0.17em}}}\mathit{G}$.

$\left(ii\right)$For each $\mathit{a}\mathbf{\in }\mathbf{\text{\hspace{0.17em}}}\mathit{G}$, there exists$\mathit{d}\mathbf{\in }\mathit{G}$ such that$\mathit{d}\mathit{a}\mathbf{=}\mathit{e}$ .

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 that satisfies the following axioms:

1. Closure:if$a\in G$ and $b\in G$, then $a\ast b\in G$.
2. Associativity: $a\ast (b\ast c)=(a\ast b)\ast c$ 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 $a\ast d=e$androle="math" localid="1654275155764" $d\ast e=a$ .

Proving that G is a group

According to the properties of G given in the question, we can conclude that G already satisfies the axioms of Closure, Associativity, and Inverse.

Therefore, if we prove that it also has an identity element then G will be a group.

As given in the question $ea=a$.

Taking$da=e$.

Find role="math" localid="1654275267804" $ad$as:

$\begin{array}{l}ad=a\left(ed\right)\\ ad=a\left(dad\right)\\ ad=\text{\hspace{0.17em}\hspace{0.17em}}adad\\ ad={\left(ad\right)}^2\mathrm{...........}\left(1\right)\end{array}$

Now considering $e=ad$as:

role="math" localid="1654275406028" $\begin{array}{c}e=ad\\ e=aed\\ e=adad\\ e={\left(ad\right)}^2\mathrm{.........}\left(2\right)\end{array}$

From equation (1) and (2), $e=ad$

Since $da=e$,$da=e=ad$.

This implies that G has an identity element.

Since G satisfies all the properties of a group, G is a group.

Hence, it is proved that G is a group.