Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
This is exercise 72 in chapter 4 of Gallian's Contemporary Abstract Algebra. – a student Nov 26 at 5:25

2 Answers 2

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.

share|cite|improve this answer
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 $\mathbb Z / 4 \mathbb Z * G$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.