Question

Let G be an abelian group and T its torsion subgroup (see Exercise 19 of section 7.3). Prove that $\mathit{G}\mathbf{/}\mathit{T}$has no nonidentity element of finite order.

Short answer

It is proved that$G/T$ has no nonidentity element of finite order.

Step-by-step solution

Definition of Torsion subgroup:

If T is a torsion subgroup of G, and$\mathit{g}\mathbf{\in }\mathit{G}$ then$\mathit{g}\mathbf{\in }\mathit{G}$ is torsion $\iff g^n:=e$ for all $n\in N$, where$\mathit{e}$ is the identity element of$\mathit{G}$.

Proving that G/T  has no nonidentity element of finite order

Suppose for any arbitrary $a\in T$, there is a nonidentity element $aT$in $G/T$.

Therefore, we can write

role="math" localid="1654509732957" $a^{i}T=T$, $i<\infty$for (since the order is given as finite)

Which implies,

$a^i\in T$

Since T is the torsion subgroup of G, from the definition of the torsion group, there exists an $n\in N$such that

$\begin{array}{c}{\left(a^i\right)}^n=e\\ a^{in}=e\end{array}$

Since e is the identity element of G, it implies aT is the identity element.

Therefore, our assumption is wrong, hence$G/T$ does not have any nonidentity element of finite order.