Question

9.(b)Show that part (a) may be false if G is infinite. [Hint: Consider the group G(2) in Exercise 8.]

Short answer

If G is a finite abelian p-group then $\mathrm{pG}\ne \mathrm{G}$.

Step-by-step solution

Part (a) and group G(2)  

If G is a finite abelian p-group then $\mathrm{pG}\ne \mathrm{G}$ and the group $\mathrm{G}\left(2\right)$ is $\mathrm{G}\left(2\right)=\left\{\left[\frac{\mathrm{a}}{2^{\mathrm{n}}}\right]\in \raisebox{1ex}{$\xcancel{\mathrm{o}}$}\!\left/ \!\raisebox{-1ex}{$\cancel{\mathrm{c}}$}\right.|\mathrm{n}\in \mathrm{N},\cancel{2}\mathrm{a}\right\}$ .

Part (a) is false

Consider the group $\mathrm{G}\left(2\right)=\left\{\left[\frac{\mathrm{a}}{2^{\mathrm{n}}}\right]\in \xcancel{\mathrm{o}}/\cancel{\mathrm{c}}|\mathrm{n}\in \mathrm{N},2\cancel{|}\mathrm{a}\right\}$.

Since, all the elements is of the form $\left[\frac{\mathrm{a}}{2^{\mathrm{n}}}\right]$ with $2\cancel{|}\mathrm{a}$. Thus, $\left[\frac{\mathrm{a}}{2^{\mathrm{n}}}\right]$can be written as $2\left[\frac{\mathrm{a}}{2^{\mathrm{n}+1}}\right]$ implies $2\mathrm{G}\left(2\right)=\mathrm{G}\left(2\right)$.

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