Subgroup of $\mathbb{R}$ that is a nontrivial direct sum of $n$ groups Given any integer $n\geq 2$, I need to find a subgroup of $\mathbb{R}$ which is a nontrivial direct sum of $n$ groups.
Initially, I was thinking of $\mathbb{Q}$, since it has the nice property of count ability, but them I remembered that every two nontrivial subgroups of $\mathbb{Q}$ has nontrivial intersection. 
So, I am really at a loss as to how to even start. What $n$ groups comprise the nontrivial direct sum of a subgroup of $\mathbb{R}$? And which subgroup of $\mathbb{R}$? 
I thank you ahead of time for any help you can give.
 A: Any finetely generated subgroup of $\mathbb{R}$ is torsion free, thus free. Thus, you are to find $n$ elements of $\mathbb{R}$ which are linearly independent over $\mathbb{Z}$. Hint: a bunch of irrationals will do the trick!
A: Consider a subgroup $\langle x_1, \ldots, x_t \rangle$ such that $x_1, \ldots, x_t \in \mathbb{R}$ are linearly independent over $\mathbb{Z}$. In this case $\langle x_1, \ldots, x_t \rangle \cong \langle x_1 \rangle \oplus \cdots \oplus \langle x_t \rangle$.
For example, the subgroup $H = \langle 1, \sqrt{2} \rangle$ works for $t = 2$. Here $H = \{a + b \sqrt{2} : a, b \in \mathbb{Z}\} \cong \mathbb{Z} \oplus \mathbb{Z}$.
Some families of $\mathbb{Z}$-independent real numbers are $\{ \log(p) : p \text{ prime} \}$ and $\{ \sqrt{p} : p \text{ prime} \}$.
A: There are two ways of solving this problem which came to my mind at first glance.
1) Probably a bit "cheating", as I assume you don't know the classification of finiteley generated abelian groups. However it says (in its simplest form) that a finitely generated abelian group $A$ is of the form $A \cong \mathbb{Z}^k \times T$, where $T$ is the subgroup of all torsion elements. As you are looking for a certain subgroup of $\mathbb{R}$ (I assume your operation is addition), it is nevessary to note that $(\mathbb{R}, +)$ does not contain any non-trivial elemtents of finite order. Indeed, can you think of a non-zero $x \in \mathbb{R}$, s.t. $x + \dots + x = 0$?
So if you take some elements $x_1, \dots, x_n \in \mathbb{R}$, you can be certain that they generate a subgroup isomorphic to $\mathbb{Z}^k = \mathbb{Z} \times \dots \times \mathbb{Z}$ for some $k$, i.e. $\langle x_1, \dots, x_n \rangle \cong \mathbb{Z}^k$. However you may not always be lucky and have $k < n$. And in your question you are asking for the "optimal" case where $k = n$. But this is exactly the case when your chosen $x_1, \dots, x_n$ happen to be linearly independant over $\mathbb{Z}$. Why? Suppose they are not. Then you can find integers $a_1, \dots, a_n \in \mathbb{Z}$ not all equal to zero, s.t. $a_1 x_1 + \dots + a_n x_n = 0$. Without loss let us assume that $a_n \neq 0$, then you can rearrange this to $x_n = -\frac{1}{a_n} (a_1 x_1 + \dots + a_{n-1} x_{n-1})$. In other words your $x_n$ can be expressed in terms of the other $x_1, \dots , x_{n-1}$, i.e. $x_n$ is an "unnecessary gnerator" for the group $\langle x_1, \dots, x_n \rangle$. Hence it would suffice to look at $\langle x_1, \dots, x_{n-1} \rangle$. I am sure you now get the point.
2) The more or less direct approach would be to look at one of the characterisations of direct products. I assume you know this, since this is probably something done in every first course in group theory (internal direct product). Namely, let $G$ be a group and $A,B \leq G$ be subgroups. Then $G \cong A \times B$ if and only if the following $3$ properties hold:
a) $A \cap B = \{ 0 \}$
b) Every element of $G$ can be expressed as a product of an element of $A$ and an element of $B$, i.e. $G = A + B := \{a+b | a \in A, b \in B\}$ (I took the symbol "+" since we are working with addition in this very example)
c) Both $A$ and $B$ are normal subgroups of $G$.
So take two non-trivial element $x,y \in \mathbb{R}$. Since they are non-torsion (see above paragraph) we have $A:=\langle y \rangle \cong \mathbb{Z}$ and $B:= \langle x \rangle \cong \mathbb{Z}$. As $\mathbb{R}$ is abelian all its subgroups are normal, in particular item c) holds. As we will form the group $A \cdot B = : G \subseteq \mathbb{R}$, we only have to handle item a). But can you think of two explicit elements $x$ and $y$ such that the intersection of the corresponding subgroups $A$ and $B$ is exactly the zero element? (I am most surely that you can). However if you have such two elements you have accomplished that $\mathbb{R} \supseteq G \cong A \times B \cong \mathbb{Z}^2$. For example take (the already mentioned) numbers of the form $\sqrt{j^2 + 1}$. If you take $x=\sqrt{1^2 + 1} = \sqrt{2}$ and $y =\sqrt{2^2 + 1} = \sqrt{5}$ you have an example. But now you can proceed inductively. Add another element, $\sqrt{10}$, and consider the subgroup generated by this element. You want it to have trivial intersection with your already formed group $G$. In any case this breaks down to the fact that these numbers are linearly independant over $\mathbb{Z}$, which you can easily show by simple algebra. Indeed suppose they have non-trivial intersection, i.e. there exist integers $a,b,c \in \mathbb{Z}$ (where $c \neq 0$), s.t. $a \sqrt{2} + b \sqrt{5} = c \sqrt{10}$ (you certainly see why this is "the same" as linear independancy of $\mathbb{Z}$). Square the whole thing and you will see the solution. You will get a contradiction and obtain that $a = b = c = 0$. I hope my answer helps you to understand why the other answers to this question were already enough help :)
A: If you want a specific example, consider for instance the subgroup generated by
the numbers 
$$
e^1, e^2, e^3,..., e^n
$$
In view of Lindemann–Weierstrass Theorem (Baker's Reformulation), these numbers are linearly independent over $\bar{{\mathbb Q}}$ and, hence, over ${\mathbb Q}$. Therefore, regarding ${\mathbb R}$ as a vector space over ${\mathbb Q}$, the subspace spanned by these exponentials is isomorphic to ${\mathbb Q}^n$. Hence, the subgroup they generate is isomorphic to ${\mathbb Z}^n$. The latter, I trust you know, is an $n$-fold direct product of nontrivial groups. 
As for your side question why ${\mathbb R}$ is torsion-free, consider the sum of strictly positive number $a_1+ ... + a_n$. Prove (by induction) that this sum is still strictly positive using Hilbert's axioms of real numbers. Now, specialize to the case $a_1=...=a_n=a>0$.  
