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Find an infinite group that has exactly two elements with order $4$?

Let $G$ be an infinite group for all $R_5$ (multiplication $\mod 5$) within an interval $[1,7)$. So $|2|=|3|=4$. Any other suggestions, please?

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    This is exercise 72 in chapter 4 of Gallian's Contemporary Abstract Algebra.2015-11-26

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The one we bump into most often is the multiplicative group of non-zero complex numbers. There are all sorts of minor variants of the idea, such as the complex numbers of norm $1$ under multiplication. One can disguise these groups as matrix groups, or geometric transformation groups.

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    Thank you very much for your reply. Actually, by "it" in "prove it" I meant that there are no other elements of order 4, I could prove that $i$ has order $4$. Sorry for confusion. In any case: Thank you very much for helping me, I understand it now. If I want to find elements of a given order in a group all I have to do is solve the equation $g^n = e$ where $n$ is the order I'm looking at.2015-12-01
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Pick $G$ an infinite group with no torsion (e.g. $\mathbb Z$ or $\mathbb Q$), then $\mathbb Z / 4 \mathbb Z \times G$ works.

As a non-abelian example, you can also take the free product $\mathbb Z / 4 \mathbb Z * G$.

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    A silly example is $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z} \times (3\ltimes 7)$. It would be nice if the group was sort of "generated" by the elements of order 4, but with only 2 that is impossible (they generate a cyclic group of order 4). So in some sense all examples are silly, but probably some are less silly than mine.2012-09-17
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Cyclic group generated by $1$ and $I =<1,i>.$