Can we find a finite abelian group $(A,+)$ such that $A⊂ℝ$

Question

Let us consider the finite abelian group (ℤ/nℤ,+). My question is: Can we find a finite abelian group $(A,+)$ such that $A⊂ℝ$ and $(ℤ/nℤ,+)$ is isomorphic to $(A,+)$ . The same question for the groups of the form $ℤ/nℤ× ℤ/2ℤ$.

Answer 1

Apart from the trivial group, no.

If $x \in \mathbb{R}\neq 0$ then $x, x + x ,x + x + x ,\ldots$ is an infinite sequence of distinct elements of any group containing $x$. Hence any subgroup of $\mathbb{R}$ containing a non-zero element is infinite.

Answer 2

All groups here will be abelian, with additive notation.

Definition.

1. A group $G$ is said to be torsionfree if, for all $x\in G$, $x\ne0$ and $n>0$, we have $nx\ne0$.

2. A group $G$ is said to be torsion if, for all $x\in G$, there exists $n>0$ such that $nx=0$.

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A group can be neither torsion nor torsionfree. An example is $\mathbb{R}/\mathbb{Z}$, where $3(\frac{1}{3}+\mathbb{Z})=0+\mathbb{Z}$, but $n\sqrt{2}+\mathbb{Z}\ne0+\mathbb{Z}$ for all $n>0$. Prove that $\mathbb{R}$ (with respect to addition) is torsionfree. However, a finite group $G$ is always torsion, because $nx=0$ for all $x\in G$, where $n=|G|$.

Exercise: give an example of an infinite torsion group. (Hint: a subgroup of $\mathbb{R}/\mathbb{Z}$ will do.)

Exercise: a subgroup of a torsionfree group is torsionfree.

Exercise: the only group that is torsionfree and torsion is the trivial group $\{0\}$.

You should be able to conclude.