A standard infinite abelian group consist of finite abelian groups

Question

Let $C_1 = \{ 1\}$, $C_2 = \{ 1, -1 \}$, $C_3 = \{1, \omega, \omega^2\}$, …, $C_n = \{x: x^n=1\}$, i.e $x$ is $n$th root of unity for fixed $n\in\mathbb{N}$, the set of natural numbers, then clearly under multiplication of complex number as binary operation, each $C_i$ is an abelian group of order $i$.

In fact, it is cyclic also. But I am not getting what will happen if we take a set $G$ which is union of all these $C_i$s, i.e., $G = \bigcup_{i=1}^n C_i$, where $n\in\mathbb{N}$, under same binary operation is also a group? And if yes then it is of infinite order?

Also what is the difference between saying $H =\{ x: x \text{ is $n$th root of unity for fixed $n\in\mathbb{N}$}\}$ and $K =\{x: x\text{ is $n$th root of unity for some $n\in\mathbb{N}$}\}$?

Answer 1

Yes, $G$ constructed in this way is a group. More generally, any directed colimit (or even filtered colimit) of groups is computed in this way in terms of their underlying sets.

In fact, we can identify the group structure; there is an isomorphism

$$ \mathbb{Q}/\mathbb{Z} \to G : \frac{m}{n} \mapsto \exp\left( \frac{m}{n} 2 \pi i\right) $$

(check that this is well-defined!)

The inverse of this isomorphism can be expressed in terms of the complex argument function.

Answer 2

There is a big difference between $$ C_n=\{x\in\mathbb{C}:x^n=1\} $$ (for a fixed $n$) and $$ K=\{x\in\mathbb{C}:x^n=1\text{ for some $n\in\mathbb{N}$}\} $$ The set $K$ is precisely the union $$ K=\bigcup_{n\in\mathbb{N}}C_n $$ you want to investigate.

An analogy can be with $A_n=\{x\in\mathbb{N}:x\le n\}$ and $\mathbb{N}=\bigcup_{n\in\mathbb{N}}A_n$. Every set $A_n$ is finite, but their union is $\mathbb{N}$: $$ \mathbb{N}=\{x\in\mathbb{N}:x\le n\text{ for some $n\in\mathbb{N}$}\} $$

However this is just an analogy for explaining the difference.

Why is $K$ a group? Suppose $x,y\in K$. Then $x^m=1$ for some $m$ and $y^n=1$, for some $n$. In particular $$ (xy)^{mn}=x^{mn}y^{mn}=(x^m)^n(y^n)^m=1^n1^m=1 $$ so $xy\in K$. Also $1\in K$ and, if $x\in K$, also $x^{-1}\in K$ (why?). Since clearly $0\notin K$, we have that $K$ is a subgroup of the group $\mathbb{C}\setminus\{0\}$ (with respect to multiplication, of course).

^{Note: I assume throughout that $0\notin\mathbb{N}$, as results from your notation.}

This group is infinite. Indeed, if it were finite, say $K=\{x_1,\dots,x_r\}$, take for each $k$ a natural number $n_k$ such that $x_k^{n_k}=1$. Set $n=n_1n_2\dotsm n_k$: then $x_k^n=1$ for every $k$. Thus $K\subseteq C_n$, which is impossible, because $C_n$ does not contain all $2n$-th roots of unity.