Abstract Algebra. Let $\mathit{G} $ be an abelian group. Show that the elements of finite order in $\mathit{G}$ form a subgroup of $\mathit{G}$.

Question

Let $\mathit{G} $ be an abelian group. Show that the elements of finite order in $\mathit{G}$ form a subgroup of $\mathit{G}$, called the torsion subgroup of $\mathit{G}$.

let $g \in G$

I know that the order of the element is the smallest integer $n$ such that $g^{n}=e$, the identity element of the group.

My approach is that let $G'=\left \{ g_{1}, g_{2}, ..., g_{n} \right \} $, $g_{x} \in G, x\in \mathbb {N}$ be an arbitrary set of elements of finite order.

Then, show that the set $G'$ is closed under the binary operation, has an unique identity, has inverse for each element, and is associative.

However, I do realize that $G'$ does not have to be finite; moreover, I don't have any clue how to show closure with the information given by the problem (I don't know where to start).

Could you please point me in the right direction?

Thanks.

Answer 1

Not every set of elements of finite order is a subgroup. What is true though is that the product and inverse of elements with finite order also has finite order, so the collection of ALL elements of finite order is a subgroup.

For product of elements of finite order take let $g$ and $h$ be of order $n$ and $m$ respectively. Try to find some large number $k$ such that $(gh)^k=e$.

Hint: in an abelian group exponents distribute over multiplication.

I'll leave proving that inverses of elements of finite order have finite order to you.

Answer 2

Let $U$ be the set of all elements of finite order.

• The neutral element $1$ has order 1. So $1\in U$.
• Let $g,h\in U$. Then there are positive integers $n,m \geq 1$ with $g^n = 1$ and $h^m = 1$. So $$(gh)^{nm} \overset{gh = hg}{=} g^{nm} h^{nm} = (g^n)^m (h^m)^n = 1^m 1^n = 1.$$ Hence $\operatorname{ord}(gh) \le nm$ and therefore $gh \in U$.
• Let $g\in U$. Then there is a positive integer $n$ with $g^n = 1$. Multiplication with $(g^{-1})^n$ yields $$\underbrace{(g^{-1})^n g^n}_{=1} = (g^{-1})^n.$$ So $\operatorname{ord}(g^{-1})\le n$ and $g^{-1}\in U$.

Therefore, $U$ is a subgroup