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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.

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    Do you need one subgroup $G\subseteq\Bbb R$ such that $G\cong \bigoplus_{i = 1}^n H_{i,n}$ (for some collection of groups $\{H_{i,n}\mid (i,n)\in\{1,\dots,n\}\times\Bbb N_{\geq 2}\}$) for any $n$, or do you need for each $n$ a $G_n\subseteq\Bbb R$ such that $G_n\cong\bigoplus_{i = 1}^n H_i$ (for some groups $H_i$, $i = 1,\dots, n$)? The question itself appears to be asking for the latter, but your attempt suggests you might be looking for the former. – 2017-01-16
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    @Stahl go by what the question is asking. I just have no idea what I'm doing. – 2017-01-16
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    Do you know that $\mathbb R$ is a vector space over $\mathbb Q$? Any subspace is in particular a subgroup. If a subspace is a direct sum of $\mathbb Q$-subspaces it is also a direct sum of these as subgroups. To be concrete you could look at $\mathbb Q$ and $\sqrt{2}\cdot\mathbb Q$ or going one little (unnecessary) step further at $\mathbb Z$ and $\sqrt{2}\cdot\mathbb Z$ (to have subgroups isomorphic to $\mathbb Z$). That takes care of $n=2$. To generalize these concrete examples to $n>2$ you can use additionally square roots of other primes or take the (real) $n$-th root of $2$ instead. – 2017-01-27

4 Answers 4

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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 $a_1=...=a_n=a>0$.

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    Maybe for the purposes of this question it is easier to just use the fact that $e$ is transcendental. If $\sum n_k e^k = 0$ for some $n_k \in \mathbb{Z}$, then $n_k = 0$ for all $k$ since $e$ is transcendental. Then this example generalizes by replacing $e$ with any transcendental number. – 2017-01-25
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    @spin: True.... – 2017-01-25
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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!

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    like $j+\sqrt{2}$ from $j=1$ to $n$? – 2017-01-16
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    @JessyunBourne No, this choice won't work: if $1+\sqrt 2$ and $2+\sqrt 2$ are in the subgroup, then their difference is $1$ and also belongs to the subgroup, and $\sqrt 2$ itself also. Thus, you'll get the subgroup generated by $1$ and $\sqrt 2$, which is free of rank 2. Rather, I was thinking about something like $\sqrt{j^2+1}$. – 2017-01-16
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    I'll give it a go and let you know if I have any difficulties. – 2017-01-16
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    I've been working on this on and off for the past 9 days, but to no avail. I don't understand 1) why finding $n$ elements of $\mathbb{R}$ which are linearly independent over $\mathbb{Z}$ does the job, and 2) How to explicitly show that it does. I put out a 50 point bounty on this question, so if you'd like to edit your answer to include more in the way of details/explanations for how/why things work, I'll award it to you and accept your answer. But, as it is, it has not been particularly helpful. – 2017-01-25
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    I also cannot find any reference to or proof of your claim that any finitely generated subgroup of $\mathbb{R}$ is torsion free. – 2017-01-25
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    @Jessy All subgroups of the reals are torsion free. – 2017-01-25
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    @Jessy The reals themselves are torsion free. This should be well known. – 2017-01-25
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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 $\{ \sqrt{p} : p \text{ prime} \}$.

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    I'm not interested in $t=2$. I'm interested in general $t$. Also, I'm looking for an answer that explicitly shows that the individual summands are normal subgroups with trivial intersection, etc. – 2017-01-25
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    For general $t$, I gave families you can use. For seeing the direct sum, you can use the usual condition: if $H_1, \ldots, H_t$ are normal subgroups with $H_i \cap \sum_{j \neq i} H_j = 0$ for all $i$, we have $\langle H_1, \ldots, H_t \rangle \cong H_1 \oplus \ldots \oplus H_t$. – 2017-01-25
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    can I use $\sqrt{j^{2}+1}$ for those? for $j=1$ to $n$? – 2017-01-25
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    @JessyunBourne: You should show that they are independent over $\mathbb{Z}$. I think it's easier for the families I suggested. – 2017-01-25
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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 :)

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    I wanted to tell you that this answer helped me TREMENDOUSLY! You were right that the other answers were already enough to help, but I had been conceiving of what this problem was actually asking all wrong. I thank you so much for clarifying! – 2017-01-29
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    one thing though - when you square both sides of $a\sqrt{2} + b\sqrt{5} = c\sqrt{10}$, you get that $2a^{2} + 2\sqrt{10}ab + 5b^{2} = 10c^{2}$. Is the contradiction supposed to be that since $a$ and $b$ need to be integers, and $\sqrt{10}$ is not an integer, $a$ and $b$ must both equal $0$? Which then forces $c$ to equal $0$? – 2017-01-29
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    More or less. You can, e.g. work through all possible cases ($a=0$ and $b\neq 0$ and so on ...) and you will get a contradiction in a second by the fact that all the squareroots are irrational. However, in my opinion the "nicest" numbers to choose for this problem are the ones sugggested by @Moishe Cohen, namely $e^k$, as transcendence of $e$ is a nice fact to bring in at this point (and is exactly $\mathbb{Z}$-linear independance). Indeed any transdental number (an its $n$-powers) would work by exactly this reason. – 2017-01-29