Subgroup of infinite group

Question

Are there any infinite order groups such that they have finite order non trivial subgroup? The simplest subgroup would be one where the element is its own inverse other than identity. Next case is if square of an element is its inverse. So are there any such infinite order groups? If such groups donot exist how to prove it?

Answer 1

The nonzero reals under multiplication have the subgroup $\{1,-1\}$.

Answer 2

The simplest example I can think of is the group of points on a grid which is of length 2 in one direction and an infinitely long strip in the orthogonal direction. Let's label such points $(x,y)$ with $x\in\Bbb{Z}, y\in \{0,1\}$.

Define the group operation "$+$" by $$ a+b = (a_x+b_x,a_y\oplus b_y) $$ with $\oplus$ being the exclusive or operation on $\{0,1\}$.

Now consider the nontrival subgroup consisting of $(0,0)$ and $(0,1)$.

Answer 3

The simplest example I can think of is $ \mathbf Q/ \mathbf Z $ (the additive group), which has cyclic subgroups of all orders.

This is in the same spirit as Hagen von Eitzen's example, since polar form gives a decomposition $ \mathbf C^{\times} \cong \mathbf R \oplus \mathbf R / \mathbf Z $, and the finite order subgroups come from the second summand.

Answer 4

Several good answers already. Here's another class of examples.

Let $S$ be an infinite sequence of nontrivial finite groups. Then the direct sum of the groups in $S$ is an infinite group in which every element has finite order, so generates a finite subgroup.

(The direct sum is the set of sequences for which the $i$th entry is a member of the $i$th group in $S$, such that all but a finite number of the entries is the identity of its group. This generalizes easily to sets $S$ of arbitrary cardinality.)