Question

Let $\mathbf{G}$ be an abelian group and $\mathbf{T}$ the set of elements of finite order in $\mathbf{G}$. Prove that

Every nonzero element of the quotient group $\mathbf{G}\mathbf{/}\mathbf{T}$ has infinite order.

Short answer

It is proved that, every non-zero element of the quotient group $\mathrm{G}/\mathrm{T}$ has infinite order.

Step-by-step solution

Step 1:Group G and T

Given that G is an abelian group and T is the set of elements of finite order in G.

G/T has infinite order

Let $\left[\mathrm{a}\right]\in \mathrm{G}/\mathrm{T}$ is a non-zero element.

Then, $\mathrm{a}\notin \mathrm{T}$ and $\left[\mathrm{a}\right]$ has finite order m.

The element ${\left[\mathrm{a}\right]}^{\mathrm{m}}$ can be written as:

$\begin{array}{rcl}{\left[\mathrm{a}\right]}^{\mathrm{m}}& =& \left[{\mathrm{a}}^{\mathrm{m}}\right]\\ & =& \left[{\mathrm{a}}^{\left|\begin{array}{c}\mathrm{a}\end{array}\right|}\right]\\ & =& \mathrm{e}\end{array}$

Thus, ${\mathrm{a}}^{\mathrm{m}}=\mathrm{e}$ implies ${\mathrm{a}}^{\mathrm{m}}\in \mathrm{T}$ has finite order n that is, ${\mathrm{a}}^{\mathrm{mn}}=\mathrm{e}$.

Hence, $\mathrm{a}\in \mathrm{T}$ is a contradiction to $\mathrm{a}\notin \mathrm{T}$.

Therefore, every non-zero element of the quotient group $\mathrm{G}/\mathrm{T}$ has infinite order.