Finite Order Elements of Infinite Abelian Group form Normal Subgroup/Corollary

Corollary to Finite Order Elements of Infinite Abelian Group form Normal Subgroup

Let $G$ be an infinite abelian group.

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

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

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 Finite Order Elements of Infinite Abelian Group form Normal Subgroup, $H$ forms a normal subgroup of $G$.

Hence $G / H$ is defined as the quotient group.

From Quotient Group is Group, $\order {e H} = H$ is the identity element of $G / H$.

From Identity is Only Group Element of Order 1, $\order {e H} = 1$.

Let $x H$ be of finite order in $G / H$, where $x \in G$.

Let $\order {x H} = m$.

Then $\paren {x H}^m = H$ by definition of finite order element.

Thus:

$\exists n \in \Z_{>0}: \paren {x^m}^n = e$

and so $x \in H$.

That is:

$x H = H$

and so $x H$ is the identity element of $G / H$.

Thus if $x H$ is of finite order in $G / H$, it has to be the identity element.

$\blacksquare$

Sources

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