Question

Let G be a nonempty finite set equipped with an associative operation such that for all$\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathit{c}\mathbf{,}\mathit{d}\mathbf{\in }\mathit{G}$:

ifrole="math" localid="1654270958980" $\mathit{a}\mathit{b}\mathbf{=}\mathit{a}\mathit{c}\mathbf{,}$ then$\mathit{b}\mathbf{=}\mathit{c}$ and if $\mathit{b}\mathit{d}\mathbf{=}\mathit{c}\mathit{d}$, then $\mathit{b}\mathbf{=}\mathit{c}$.

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$a\ast d=e$ and $d\ast e=a$.

Proving that G is a group

As given in the question $a,b,c,d\in G$, if $ab=ac,$then $b=c$and if $bd=cd$then $b=c$.

According to the question, G have the property of closer and associativity.

So, we need to prove that G satisfies other axioms.

Consider a finite set $\{a,a^2\mathrm{....}{a}^x\mathrm{...}{a}^y\}\subseteq G$

Since G is finite, for some $a^x=a^y,y>x\ge 0$

We take $e=a^{y-x}$, now we need to show that this is an identity element.

Suppose

$c=eb$

Multiplying with role="math" localid="1654271863251" $a^x$as:

$\begin{array}{c}{a}^{x}c=a^{x}eb\\ a^{x}c=a^x{a}^{y-x}b\\ a^{x}c=a^{x}b\\ c=b\end{array}$

Which is true, hence it is proved that $e=a^{y-x}$ is identity element in G.

This implies, G has identity element.

Again, for some $a\in G,$$a^x=a^y,y>x$, let us consider $e=a^{y-x}$.

Since $y>x$ and G is finite, $a^1=a^{y-x+1}$.

Here,$y-x+1>1$.

Now, we find inverse as:

$$\begin{gathered} ea=a \\ e{a}^{y-x+1}=a \\ e=a^{x-y-1}a \\ e=a^{-1}a \end{gathered}$$

Therefore, g has inverse$a^{x-y-1}$ for element a.

Since G satisfies both axioms, hence it is a group.

Thus, G is a group.