# The Group of Rational Numbers is Not Finitely Generated ## Problem 461

(a) Prove that the additive group $\Q=(\Q, +)$ of rational numbers is not finitely generated.

(b) Prove that the multiplicative group $\Q^*=(\Q\setminus\{0\}, \times)$ of nonzero rational numbers is not finitely generated. Add to solve later

## Proof.

### (a) Prove that the additive group $\Q=(\Q, +)$ is not finitely generated.

Seeking a contradiction assume that the group $\Q=(\Q, +)$ is finitely generated and let $r_1, \dots, r_n$ be nonzero generators of $\Q$.
Express the generators as fractions
$r_i=\frac{a_i}{b_i},$ where $a_i, b_i$ are integers.

Then every rational number $r$ can be written as the sum
$r=c_1r_1+\cdots+c_k r_n$ for some integers $c_1, \dots, c_n$.

Then we have
$r=\frac{m}{b_1\cdots b_n},$ where $m$ is an integer (which you can write down explicitly using $a_i, c_i$).

Let $p$ be a prime number that does not divide $b_1\cdots b_n$, and choose $r=1/p$.

Then we must have
$\frac{1}{p}=\frac{m}{b_1\cdots b_n}$ for some integer $m$.
Then we have
$pm=b_1\cdots b_n$ and this implies $p$ divides $b_1\cdots b_n$, which contradicts our choice of the prime number $p$.

Thus, the group $\Q$ cannot be finitely generated.

### (b) Prove that the multiplicative group $\Q^*=(\Q\setminus\{0\}, \times)$ of nonzero rational numbers is not finitely generated.

Suppose on the contrary that the group $\Q^*=(\Q\setminus\{0\}, \times)$ is finitely generated and let
$r_i=\frac{a_i}{b_i}$ be generators for $i=1, \dots, n$, where $a_i, b_i$ are integers.

Then every nonzero rational number $r$ can be written as
$r=r_1^{c_1}\cdots r_n^{c_n}=\frac{a_1^{c_1}\cdots a_n^{c_n}}{b_1^{c_1}\cdots b_n^{c_n}}$ for some integers $c_n$.

Let $p$ be a prime number that does not divide $b_1\cdots b_n$, and consider $r=1/p$.
Hence $\Q^*$ is not finitely generated. Add to solve later

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