Question

Let $\mathrm{G}$ be an abelian group and $\mathrm{T}$ the set of elements of finite order in $\mathrm{G}$. Prove that $\mathrm{T}$is a subgroup of order (called the torsion subgroup).

Short answer

It is proved that, $\mathrm{T}$ is a subgroup of order $\mathrm{G}$.

Step-by-step solution

T is closed under inverses

Let $\mathrm{a},\mathrm{b}\in \mathrm{T}$.

Then, find ${\left({\mathrm{\alpha }}^{-1}\right)}^{\left|\mathrm{a}\right|}$ as:

role="math" localid="1657364932590" $$\begin{gathered} {\left({\mathrm{a}}^{-1}\right)}^{\left|\mathrm{a}\right|}={\left({\mathrm{a}}^{\left|\mathrm{a}\right|}\right)}^{-1} \\ ={\mathrm{e}}^{-1} \\ =\mathrm{e} \end{gathered}$$

Thus, $\mathrm{T}$ is closed under inverses.

T is closed under composition

Let $\mathrm{a},\mathrm{b}\in \mathrm{T}$.

Then, find ${\left(\mathrm{ab}\right)}^{\left|\mathrm{a}\right|\left|\mathrm{b}\right|}$as:

role="math" localid="1657365670570" $$\begin{gathered} {\left(\mathrm{ab}\right)}^{\left|\mathrm{a}\right|\left|\mathrm{b}\right|}=\left({\mathrm{a}}^{\left|\mathrm{a}\right|\left|\mathrm{b}\right|}\right)\left({\mathrm{b}}^{\left|\mathrm{a}\right|\left|\mathrm{b}\right|}\right) \\ ={\left({\mathrm{a}}^{\left|\mathrm{a}\right|}\right)}^{\left|\mathrm{b}\right|}{\left({\mathrm{b}}^{\left|\mathrm{b}\right|}\right)}^{\left|\mathrm{a}\right|} \\ ={\mathrm{e}}^{\left|\mathrm{b}\right|}{\mathrm{e}}^{\left|\mathrm{a}\right|} \\ =\mathrm{e} \end{gathered}$$

Thus, $\mathrm{T}$ is closed under composition.

Therefore, $\mathrm{T}$ is closed under inverses and composition implies $\mathrm{T}$ is a subgroup of $\mathrm{G}$.