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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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    what about group of R/Z which is the isomorphic of U. does this group has only 2 elements with order 4?2012-09-16
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    Yes david, because $\mathbb R / \mathbb Z$ is in fact isomorphic to the multiplicative group of complex numbers of norm $1$.2012-09-16
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    Yes, that's isomorphic to the circle group I mentioned in the answer.2012-09-16
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    can you give a specific example of a cyclic group.2012-09-17
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    @david: The comment said **circle** group, which is a standard name for the group of complex numbers of norm $1$, And, as pointed out, it is isomorphic to your $\mathbb{R}/\mathbb{Z}$. There is no **infinite** cyclic group with the desired property. There are finite cyclic groups, like $\mathbb{Z}_4$, $\mathbb{Z}_8$, $\mathbb{Z}_{12}$, but you were asked for infinite.2012-09-17
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    Thank you this answer helped me a lot. There is only one thing I can't seem to do: while it's obvious to me that $i$ and $-i$ have order $4$ I don't see how to prove it. How can we be sure there are no other elements of order $4$ in $\mathbb C - \{0\}$?2015-11-26
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    @astudent: To show they have order $4$, all we need to do is to show that $i^4=1$, and $(-i)^4=1$, and that there is no $k$ with $1\le k\lt 4$ such that $i^k=1$ or $(-i)^k=1$. Just calculation. For showing there are no other elements of order $4$, consider the equation $z^4=1$, or equivalently $(z^2-1)(z^2+1)=0$. There are $4$ roots, $\pm 1$ and $\pm i$. But $1$ has order $1$ and $-1$ has order $2$, leaving only $\pm i$ with order $4$.2015-11-26
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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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    The free product has infinitely many elements of order 4. They form two conjugacy classes.2012-09-16
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    Right, I didn't think about conjugation. Could you give an example of a non-abelian group with exactly two elements of order 4?2012-09-16
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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 $I =<1,i>.$