# Finite Order Elements of Infinite Abelian Group form Normal Subgroup

## Theorem

Let $G$ be an infinite abelian group.

Let $H \subseteq G$ be the subset of $G$ defined as:

- $H := \set {x \in G: x \text { is of finite order in } G}$

Then $H$ forms a normal subgroup of $G$.

### Corollary

All the elements of the quotient group $G / H$ are of infinite order except the identity.

## Proof

Let $e$ be the identity element of $G$.

From Identity is Only Group Element of Order 1, $\order e = 1$ and so $H \ne \O$.

Let $a \in H$.

Then by Order of Group Element equals Order of Inverse:

- $\order a = \order {a^{-1} }$

and so $a \in H$.

Let $a, b \in H$.

From Order of Product of Abelian Group Elements Divides LCM of Orders of Elements:

- $\order {a b} \divides \lcm \set {\order a, \order b}$

where:

- $\order a$ denotes the order of $a$
- $\divides$ denotes divisibility
- $\lcm$ denotes the lowest common multiple.

Thus $a b$ is also of finite order.

Thus by definition:

- $a b \in H$

By the Two-Step Subgroup Test it follows that $H$ is a subgroup of $G$.

By Subgroup of Abelian Group is Normal, $H$ is normal in $G$.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $14$