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Let $G$ be a group of infinite order . Does there exist an element $x$ belonging to $G$ such that $x$ is not equal to $e$ and the order of $x$ is finite?

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  • $\begingroup$ Not necessarily, but in some examples, yes. $\endgroup$ – Tobias Kildetoft Jan 13 '14 at 9:52
  • $\begingroup$ If yes! Give me an example! $\endgroup$ – Adimathematica Jan 13 '14 at 9:56
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I can do better: consider the following set under the operation of multiplication: $$\lbrace x=e^{i\pi t},t \in \mathbb{Q} \rbrace$$ The set is infinite, but every element has finite order (namely, if $t=a/b$ in lowest terms, the order of $x$ is $2b$).

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    $\begingroup$ Or just $\mathbb Q/\mathbb Z$ $\endgroup$ – Hagen von Eitzen Jan 13 '14 at 10:05
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Easy example: Take the cross product of $\mathbb{Z}$ with your favourite non-trivial finite group $H$, $G=\mathbb{Z}\times H$. It in infinite as it contains $\mathbb{Z}$ as a subgroup, but it contains elements of finite order as it contains your favourite non-trivial finite group $H$ as a subgroup.

For example, if $H=C_2$ is cyclic or order two then $G=\mathbb{Z}\times C_2$ contains an element of order two. Using $C_k$, the cyclic group of order $k$, gives an infinite group with an element of order $k$.

Interestingly, the other two examples which use $\mathbb{R}$ and $\mathbb{Q}$ are not "finitely generated". A group $G$ is finitely generated if there exists a finite subset $S$ of $G$ such that every element of $G$ is a product of elements from $S$. So in the examples of $\mathbb{R}$ under multiplication, and of rotations of the circle, there exists no such set. In my above example, the set $S$ could consist of the copy of the finite subgroup $H$ and the element $1\in\mathbb{Z}$ (more formally, $S=\{(1, e)\}\cup\{(0, h): h\in H\}$). Note that there are lots of choices for the set $S$: the above is just a single example.

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No. The only element of $\mathbb{Z}$ that has finite order is $0$

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    $\begingroup$ Why must $G$ be $\mathbb{Z}$? $\endgroup$ – preferred_anon Jan 13 '14 at 9:54
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    $\begingroup$ @DanielLittlewood I thought the OP was making the conjectrue that : Let $G$ be an infinite group .... $\endgroup$ – Amr Jan 13 '14 at 9:55
  • $\begingroup$ Huh? ${}{}{}{}$ $\endgroup$ – anon Jan 13 '14 at 10:32
  • $\begingroup$ @anon Isn't the OP asking for a proof or a counterexample to the statement " for every infinite group $G$, there exists a non-identity element $x$ with finite order: ? I answered this question negatively by giving a counterexample. $\endgroup$ – Amr Jan 13 '14 at 11:03
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    $\begingroup$ It was unclear from the original question what the OP was asking, so it is not surprising that there have been different interpretations! But the comment by the OP made it clear that they really wanted an example rather than a counterexample :-) $\endgroup$ – Derek Holt Jan 13 '14 at 11:11
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Sometimes no, for example in $(\mathbf{Z},+)$. Sometimes yes, for example in $(\mathbf{R}^*,\times)$.

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  • $\begingroup$ What is the element(not equal to e) which has finite order? $\endgroup$ – Adimathematica Jan 13 '14 at 9:59
  • $\begingroup$ @Adimathematica $-1\in\Bbb R^\times$ has order two. $\endgroup$ – anon Jan 13 '14 at 10:31
  • $\begingroup$ I should have noticed it . Anyways thanks! $\endgroup$ – Adimathematica Jan 13 '14 at 10:34

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