Question

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 $pG\ne G$

Step-by-step solution

Part (a) and groupG(2)

If G is a finite abelian p-group then $pG\ne G$ and the group $G\left(2\right)$ is$G\left(2\right)=\left\{\left[\frac{a}{2^n}\right]\in \square /\square |n\in N,2⟩a\right\}.$

Part (a) is false

Consider the group $G\left(2\right)=\left\{\left[\frac{a}{2^n}\right]\in \square /\square |n\in N,2⟩a\right\}$.

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

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