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

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.

• If $g_{1}, g_{2} \in G^{\prime}$, both have finite order. What can you then say about $o(g_{1}g_{2})$? It's the $lcm \{ g_{1}, g_{2} \}$. Isn't that finite? Commented Mar 17, 2014 at 23:48
• What if $g_1$ and $g_2$ are inverse? You can only say that the order of the product divides the LCM of the orders.
– Seth
Commented Mar 18, 2014 at 3:11
• Isn't that enough to claim that they are finite? Commented Mar 18, 2014 at 3:20
• Yes. (space filler so that this is long enough to be a comment)
– Seth
Commented Mar 18, 2014 at 20:41

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.

• Thanks for the hint. I got it now. Commented Mar 18, 2014 at 2:37

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