Question

Let $\mathbf{G}$be an abelian group and $\mathbf{T}$the set of elements of finite order in $\mathbf{G}$. Prove that $\mathbf{T}$is a subgroup of order $\mathbf{G}$ (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 ${\left({\mathrm{a}}^{-1}\right)}^{\left|\mathrm{a}\right|}$ as:

$\begin{array}{rcl}{\left({\mathrm{a}}^{-1}\right)}^{\left|\mathrm{a}\right|}& =& {\left({\mathrm{a}}^{\left|\mathrm{a}\right|}\right)}^{-1}\\ & =& {\mathrm{e}}^{-1}\\ & =& \mathrm{e}\end{array}$

Thus, 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:

$\begin{array}{rcl}{\left(\mathrm{ab}\right)}^{\left|\mathrm{a}\right|\left|\mathrm{b}\right|}& =& \left({\mathrm{a}}^{\left|\mathrm{a}\right|\left|\mathrm{b}\right|}\right)\left({\mathrm{b}}^{\left|\mathrm{a}\right|\left|\mathrm{b}\right|}\right)\\ & =& {\left({\mathrm{a}}^{\left|\mathrm{a}\right|}\right)}^{\left|\mathrm{b}\right|}{\left({\mathrm{b}}^{\left|\mathrm{b}\right|}\right)}^{\left|\mathrm{a}\right|}\\ & =& {\mathrm{e}}^{\left|\mathrm{b}\right|}{\mathrm{e}}^{\left|\mathrm{a}\right|}\\ & =& \mathrm{e}\\ & & \end{array}$

Thus, T is closed under composition.

Therefore, T is closed under inverses and composition implies T is a subgroup of G.