# infinite group with exactly $n$ elements of finite order.

Given $$n\in \mathbb{N}$$, I want to construct an infinite group with exactly $$n$$ elements of finite order.

Attempt:

I think we can take $$G=\mathbb{Z}_{n}\times \mathbb{Z}$$. $$(a,0)\in G$$ is of order equal to order of $$a$$.

Question 1: What are some other examples that I can construct?

Question 2: Are there some non Abelian groups having exactly $$n$$ elements of finite order? I guess I can take product of $$\mathbb{Z}_n$$ with some matrix ring.

For question 1, you can take any torsion-free (possibly nonabelian) group $$H$$ and set $$G$$ to be $$\mathbb{Z}_n\times H$$. For example, we can take $$H$$ to be the additive group of any division ring of characteristic $$0$$.

For question 2, you can take $$G=\mathbb{Z}_n\times F_k$$, where $$F_k$$ is a free group with $$k$$ generators (well, this answers question 1 too). If $$k=1$$, you get your example $$\mathbb{Z}_n\times\mathbb{Z}$$. If $$k>1$$, you get nonabelian examples. If $$k$$ is an uncountable cardinal, then you even get an example where $$G$$ is uncountable. However, it would be interesting if one can find an indecomposable group $$G$$ with this property.

Here is an example of infinite indecomposable group $$G$$ with exactly $$n$$ elements of finite order, where $$n>1$$ is an odd integer. Let $$n=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$$ be a prime factorization of $$n$$. Define $$G$$ to be the semidirect product $$\left(\prod\limits_{i=1}^r\mathbb{Z}_{p_i^{k_i}}\right)\rtimes\mathbb{Z}$$, whose underlying set of $$G$$ is simply the product of sets $$\left(\prod\limits_{i=1}^r\mathbb{Z}_{p_i^{k_i}}\right)\times \mathbb{Z}$$, and the multiplication in $$G$$ is given by $$(x,y)\cdot (z,w)=(x+(-1)^yz,y+w)$$ for all $$x,z\in\prod\limits_{i=1}^r\mathbb{Z}_{p_i^{k_i}}$$ and $$y,w\in\mathbb{Z}$$. In other words, we have $$(x,y)\cdot (z,w)=\Big(\left(x_1+(-1)^yz_1,x_2+(-1)^yz_2,\ldots,x_r+(-1)^yz_r\right),y+w\Big)\,,$$ where $$x=(x_1,x_2,\ldots,x_r)$$ and $$z=(z_1,z_2,\ldots,z_r)$$ with $$x_i,z_i\in\mathbb{Z}_{p_i^{k_i}}$$.

To show that $$G$$ is indecomposable, we prove by contradiction. Suppose otherwise that $$G$$ has a nontrivial (internal) direct product decomposition $$H\times K$$ for some subgroups $$H$$ and $$K$$. Write $$\pi:G\to\mathbb{Z}$$ for the canonical projection $$(x,y)\mapsto y$$, where $$x\in \prod\limits_{i=1}^r\mathbb{Z}_{p_i^{k_i}}$$ and $$y\in \mathbb{Z}$$. For simplicity, write $$N$$ for $$\prod\limits_{i=1}^r\mathbb{Z}_{p_i^{k_i}}$$. Then, $$\pi(H)$$ is a subgroup of $$\mathbb{Z}$$. Let $$\pi(H)$$ be generated by an integer $$q\geq 0$$.

If $$q$$ is odd, then it can be easily seen that $$H$$ contains the subgroup with the underlying set $$N\times q\mathbb{Z}$$. If there are any other elements of $$H$$, then we can show that $$\pi(H)$$ contains an element $$q'$$ such that $$0, which is a contradiction. Therefore, $$H$$ is precisely $$N\times q\mathbb{Z}$$ (and so $$q>1$$, otherwise $$H=G$$, but we assume that $$H$$ is a proper nontrivial subgroup of $$G$$). This also shows that $$K$$ is a subgroup of $$G$$ isomorphic to $$G/H\cong\mathbb{Z}_q$$, but this shows that $$\mathbb{Z}\cong q\mathbb{Z}\times \mathbb{Z}_q$$, which is absurd. Therefore, $$q$$ must be even, which we assume from now on.

Next, we look at $$K$$ instead. Since $$\pi(H)+\pi(K)=\mathbb{Z}$$, we conclude that $$\pi(K)=t\mathbb{Z}$$ for some integer $$t\geq 0$$ such that $$\gcd(t,q)=1$$. Because $$q$$ is even, $$t$$ must be odd. Therefore, the argument from the paragraph above works for $$K$$, if we replace $$H$$ by $$K$$, and we get a contradiction once again. So, the assumption that $$G$$ is decomposable is false.

There may be an example of an infinite indecomposable group $$G$$ with exactly $$n$$ element of finite order, where $$n$$ is an even positive integer. I do not know yet how to construct such a group $$G$$.

Edit. It turns out that the same construction seems to work for every even integer $$n>0$$ divisible by $$4$$, but I have not checked completely. This construction does not work for $$n\equiv 2\pmod{4}$$ simply because $$-1=1$$ in $$\mathbb{Z}_2$$ (and if you tried, you would get that $$\mathbb{Z}_2$$ is a direct product factor of $$G$$).

• What is an indecomposable group? Commented Oct 7, 2018 at 15:37
• A group $G$ is indecomposable if $G$ is not isomorphic to any group of the form $H\times K$, where $H$ and $K$ are nontrivial groups.
– user593746
Commented Oct 7, 2018 at 15:39