Question

Question: If $\mathit{G}$is an infinite abelian group, do the elements of infinite order in G (together with 0) form a subgroup?[Hint: Consider $\mathbf{\mathbb{Z}}\mathbf{\oplus }{\mathbf{\mathbb{Z}}}_{\mathbf{3}}$.]

Short answer

Answer

No, the elements of infinite order in $G$does not form a subgroup

Step-by-step solution

Given that

Given that G is an infinite abelian group.

Consider the infinite abelian group as,$\mathrm{\mathbb{Z}}\oplus {\mathrm{\mathbb{Z}}}_3$ .

Prove infinite order does not form a subgroup

Consider an element ${\left(1,\left[1\right]\right)}^n,{\left(-1,\left[1\right]\right)}^n\in \mathrm{\mathbb{Z}}\oplus {\mathrm{\mathbb{Z}}}_3$.

For all positive integer n, ${\left(1,\left[1\right]\right)}^n=\left(n,\left[n\right]\right)\ne \left(0,\left[0\right]\right)$and${\left(-1,\left[1\right]\right)}^n=\left(-n,\left[n\right]\right)\ne \left(0,\left[0\right]\right)$.

So, they indeed have infinite order, but $\left(1,\left[1\right]\right)=\left(-1,\left[1\right]\right)=\left(0,\left[2\right]\right)$has order 3.

Therefore, it does not form a subgroup.

Hence, the elements of infinite order in$G$does not form a subgroup.