example of an infinite group

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 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
@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. – André Nicolas Sep 17 '12 at 17:06

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