Question

Let G be an abelian group and let T be the set of elements of G with finite order. Prove that T is a subgroup of G ;it is called the torsion subgroup. (This result may not hold if G is nonabelian; see Exercise 20 of Section 7.2.)

Short answer

Answer

It is proved that T is a subgroup of G.

Step-by-step solution

Step by Step Solution Step 1: Show that  T  is closed under group operation

We have thedefinition of torsion subgroup:

Let G be an abelian group and T be the set of elements of G with finite order. Then, T is called the torsion subgroup of an abelian group G .

The elements of T are called torsion elements.

Let $e\in TthisimpliesthatT\ne \varphi$

We will show that T is closed under group operations and inverse.

$Letx,y\in Tand\exists m,n\in \mathbb{N}suchthat:$

X^m= e

And,

Yⁿ = e

Thus,

$\begin{array}{l}{\left(xy\right)}^{mn}=x^{mn}{y}^{mn}\dots \left(\because Gisabelian\right)\\ ={\left(x^m\right)}^n{\left(y^n\right)}^m\\ =e^n{e}^m\\ =e\end{array}$

Hence, the element also has a finite order.

Here, $xy\in T$

Hence, T is closed under group operations.

Show that  T is closed under inverse

Now, for x^-1:

$\begin{array}{c}{\left(x^{-1}\right)}^m=x^{-m}\\ ={\left(x^m\right)}^{-1}\\ =x^{-1}\\ =e\end{array}$

Therefore,x^-1 has finite order.

Thus, $x^{-1}\in T$.

Here, T is closed under inverse, thus T is closed under group operations and inverse.

Hence, T is a subgroup of G .