Suppose $|G| = \infty$. Then there is an elment $g$ of $G$ with an $o(g) = \infty$

Question

Let $G$ a group. Suppose $|G| = \infty$. Then there is an elment $g$ of $G$ with an $o(g) = \infty$.

That's true? Intuitively it seems so, but I'm not sure.

Answer 1

In general this is not true.

For example, consider

1) $\Bbb{Q} / \Bbb{Z}$

2) $\{z \in \Bbb{C} : z^n=1\; \text{for some n $\in$ $\Bbb{N}$}\}$

Answer 2

It is false.

Consider the set $\Bbb G_n$ of sequences of elements from $\Bbb Z_n$, the finite cyclic group with $n$ elements where $2 \le n \in \Bbb N$. We may think of an element $g \in \Bbb G_n$ as an $\infty$-tuple

$g = (z_1, z_2, z_3, \ldots); \tag 1$

if

$h = (y_1, y_2, y_3, \ldots), \tag 2$

we may define an operation component-wise:

$gh = (z_1 + h_1, z_2 + h_2, z_3 + h_3, \ldots); \tag 3$

it is easy to see $\Bbb G_n$ is a group under this operation, and that for $g \in \Bbb G_n$,

$g^n = (nz_1, nz_2, nz_3, \ldots) = (0, 0, 0, \ldots), \tag 4$

where $(0, 0, 0, \ldots)$ is the identity element of $\Bbb G_n$. Thus every element of $\Bbb G_n$ is of finite order at most $n$, though clearly

$\vert \Bbb G_n \vert = \infty. \tag 5$