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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. – a student Nov 26 '15 at 5:25

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? – david Sep 16 '12 at 5:41
    
Yes david, because $\mathbb R / \mathbb Z$ is in fact isomorphic to the multiplicative group of complex numbers of norm $1$. – Niccolò Sep 16 '12 at 5:43
    
Yes, that's isomorphic to the circle group I mentioned in the answer. – André Nicolas Sep 16 '12 at 5:44
    
can you give a specific example of a cyclic group. – david Sep 17 '12 at 16:57
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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 $4$. – André Nicolas Nov 26 '15 at 5:37

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. – Jack Schmidt Sep 16 '12 at 6:33
    
Right, I didn't think about conjugation. Could you give an example of a non-abelian group with exactly two elements of order 4? – Niccolò Sep 16 '12 at 21:13
    
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. – Jack Schmidt Sep 17 '12 at 20:41

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