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Exercise 7: Show that the elements of finite order in an abelian group $G$ form a subgroup of $G$

I just solved this exercise but I can't find example of a group which is abelian and has finite (except the $e$) and infinite order elements. Are there any well known such groups?

For not abelian group with other properties group of matrixes I know.

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What about the group $\mathbb{Z}_2\times\mathbb{Z}$ ? $(1,0)$ has finite order and $(1,1)$ has infinite order.

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The multiplicative group of units $U$ from $(\mathbb{C},+,\cdot)$ is an abelian group. Of the real numbers in $U$, $-1$ has order $2$, and besides $e=1$, all other real numbers will have infinite order. Also, $i \in U$ and $i$ has order 4.

$-1$ and $i$ are both examples of $2^{nd}$ and $4^{th}$ roots of unity, respectively. That is, they are solutions to $z^2=1$ and $z^4=1$, respectively.

In general, an $n^{th}$ root of unity in $\mathbb{C}$ is a complex number $z$ satisfying $z^n=1$. For every $n \in \mathbb{N}$ there are $n$ distinct complex numbers which satisfy $z^n=1$, $z=1$ being a solution for every $n \in \mathbb{N}$.

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Any infinite finitely generated abelian group that is not a free group will satisfy this property. This follows from the classification theorem, as any such group will have the form

$\mathbb{Z}_{m_{1}}\times\cdots\mathbb{Z}_{m_{r}}\times\mathbb{Z}^{n}$

for some $m_{1},\ldots,m_{r}\geq 2$ and some $n\geq 1$, and $m_{1}\vert m_{2}\vert\cdots\vert m_{r}$.

The example $\mathbb{Z}_{2}\times \mathbb{Z}$ given by pritam above is the the easiest such example and is a great one to show why the given property is satisfied.

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