Find a group G that has subgroups isomorphic to $\mathbb{Z}_{n}$ for all natural numbers n $\geq$ 2

Question

I was thinking of using the theorem: " If G is finite of order n, then $G \cong \mathbb{Z}_{n}$ " But cannot find an appropriate group.

Any help is appreciated, thanks.

EDIT: The correct version of the theorem is "If G is cyclic AND finite of order n, then $G \cong \mathbb{Z}_{n}$ "

Answer 1

An easy example is the multiplicative group of the complex numbers. For each $n$ it has exactly one cyclic subgroup of order $n$, consisting of the solutions of $z^n-1=0$.

Answer 2

It is not true that every finite group of order $n$ is isomorphic to $\Bbb Z_n$.

However, you can take $G = \displaystyle \prod_{n \ge 2} \Bbb Z_n$, where $\prod$ means direct product.

Answer 3

Let $\mathbb S = \mathbb R/1$, the set of all real numbers modulo $1$, i.e. the set of all sets of the form $[x] = \{x + n | n \in \mathbb Z\}$. Addition on $\mathbb S$ is defined by $[x]+[y] =[x+y]$. Then $\mathbb S$ is an abelian group with respect to addition. For any positive integer, $n$, the order of the group generated by $[\frac 1n]$ is $n$.

Answer 4

It's a bit silly, but you can use Cayley's theorem that states that every finite group is embedded in some symmetric group $S_n$, and then $$ G = \prod_{n \in \mathbb N} S_n $$ will be a group that contains any finite group as a subgroup. This also is true for the symmetric group of the natural numbers $S(\mathbb N)$, since it contains all $S_n$ as subgroups.