An Integer Linear Combinaton of Rationals is a Cyclic Group

Question

Let $r_i$ be rational. Prove $G=(\sum_{i=1}^nn_ir_i|n_i\in\mathbb{Z})$ is a cyclic group under addition.

This is related to the question: Prove that group $G=\{n_1r_1+n_2r_2 | n_1,n_2\in\mathbb{Z}\}$ is cyclic

If I represent $r_i=\frac{p_i}{q_i}$ and expand everything out I get a big expression, the numerator of which is a big linear combination.

1. I think I get a term in the numerator that can be interpreted as a GCD, but I'm having trouble establishing that an arbitrary linear combination is a GCD.

2. Even if I assume this, I'm having some trouble precisely stating how the induction would work. Partly because of the mess of terms in the numerator.

Any thoughts?

Answer 1

First, some obvervations: For two subgroups $A, B$ of an abelian group, the set $A+B:= \{a+b \mid a \in A, b \in B\}$ is again a subgroup

We will prove by induction on $n$, that for any $n$ rational numbers $r_1, \dots, r_n$, the set $\{\sum_{i=1}^nn_ir_i|n_i\in\mathbb{Z}\}$ is a cyclic group.

The case $n=1$ is trivial and $n=2$ is treated in the linked question, so I won't repeat the argument here.
In the induction step, we have $$\{\sum_{i=1}^{n+1}n_ir_i|n_i\in\mathbb{Z}\}= \{n_{n+1}r_{n+1} \mid n_{n+1} \in \Bbb Z\}+\{\sum_{i=1}^nn_ir_i|n_i\in\mathbb{Z}\}$$

By the induction hypothesis, $\{\sum_{i=1}^nn_ir_i|n_i\in\mathbb{Z}\}$ is cyclic, so there exists a $s \in \Bbb Q$ such that $\{\sum_{i=1}^nn_ir_i|n_i\in\mathbb{Z}\} = \{ks \mid k \in \Bbb Z\}$, thus

$$\{n_{n+1}r_{n+1} \mid n_{n+1} \in \Bbb Z\}+\{\sum_{i=1}^nn_ir_i|n_i\in\mathbb{Z}\}=\{n_{n+1}r_{n+1} \mid n_{n+1} \in \Bbb Z\}+\{ks \mid k \in \Bbb Z\}= \{n_{n+1}r_{n+1}+ks \mid n_{n+1}, k \in \Bbb Z\}$$

This group is cyclic by the case $n=2$.

Answer 2

Let $r_i = \frac{p_i}{q_i}$ and $q = \prod_{i=1}^n q_i$.

The group isomorphism $G \to G'$ given by $x \mapsto qx$ sends $G$ to a subgroup of $\mathbb Z$. Then use the fact that every subgroup of $\mathbb Z$ is cyclic.