Question

Show that Part (a) will be false if G is infinite.

Short answer

It is shown that, part (a) is false if G is infinite.

Step-by-step solution

Referring to assumption made in Part (a)

We had considered a finite set $\mathbf{\{}\mathbf{a}\mathbf{,}{\mathbf{a}}^{\mathbf{2}}\mathbf{.}\mathbf{...}{\mathbf{a}}^{\mathbf{x}}\mathbf{.}\mathbf{..}{\mathbf{a}}^{\mathbf{y}}\mathbf{\}}\mathbf{\subseteq }\mathit{G}$ . Since G is finite therefore for some role="math" localid="1654274320188" $a^x=a^y,y>x\ge 0$, We take $e=a^{y-x}$, and $y-x+1\ge 0$.

Given that G is infinite.

Proving that part (a) Is false if G is infinite

Since throughout our assumption we considered that G is finite.

However, if G is infinite then, a few assumptions of part(a) are not possible.

For example, consider a non-negative integer N equipped with addition.

Since our group is infinite, this is an infinite set with an associative operation.

So, the only identity element it can have is zero, but it cannot have an inverse.

Hence, from the above example, we can say that if group G is infinite part (a) is false.

Thus,part (a) is false if G is infinite.