How to show $\mathbf{Q} $ is not free

Question

We know that torsion-free plus finitely generated $\rightarrow$ free and that $\mathbf{Q}$ is torsion-free is easy.

But how to show $\mathbf{Q}$ is not finitely generated and not free?

Answer 1

For the non-free part:

Take any two nonzero elements $x, y ∈ ℚ$ and show they satisfy $λx + μy = 0$ for some nonzero $λ, μ ∈ ℤ$, hence any two elements are linearly dependent. Thus, since $ℚ$ is not cyclic, it cannot have a basis.

For the non-finitely-generated part:

If $ℚ$ was finitely generated then without loss of generality (by finding the common denominator), there were a denominator $s ∈ ℤ$ and numerators $r_1, …, r_n ∈ ℤ$ such that $$ℚ = ℤ\frac{r_1}{s} + \cdots + ℤ\frac{r_n}{s} = (ℤr_1 + \cdots+ ℤr_n)\frac{1}{s} = ℤ\frac{r}{s}$$ for $r = \gcd (r_1, …, r_n)$. Thus, since $ℚ$ isn’t cyclic, it isn’t finitely generated.

Answer 2

Show any finitely generated subgroup of $\mathbf Q$ is cyclic. Since $\mathbf Q$ is not cyclic, it cannot be finitely generated.

$(1)$ It cannot be free: it is not cyclic, so any putative basis has at least two elements. But if $x,y$ are elements of the basis, we know $\mathbf Z^2\simeq \langle x,y\rangle=\langle x'\rangle\simeq\bf Z$ which is impossible. So no basis exists.

$(2)$ It cannot be free: if it were, say $\mathbf Q\simeq\mathbf Z^{(I)}$, then $\mathbf Q\simeq\mathbf Q\otimes_{\mathbf Z}\mathbf Q\simeq \mathbf Q\otimes_{\mathbf Z}\mathbf Z^{(I)}\simeq \mathbf Q^{(I)}$. This forces $|I|=1$, which is absurd.

Answer 3

If an abelian group $G\ne\{0\}$ is free, then it is isomorphic to $\mathbb{Z}^{(X)}$ (direct sum of copies of $\mathbb{Z}$), for some set non empty set $X$.

Then $\mathbb{Z}$ is an epimorphic image of $G$. Since $\mathbb{Z}$ is not divisible, $G$ can't be divisible. But $\mathbb{Q}$ is divisible.

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The group $\mathbb{Q}$ is not finitely generated: if $a_1/b_1, a_2/b_2,\dots,a_n/b_n$ are elements in $\mathbb{Q}$, then we can assume $b_1=b_2=\dots=b_n=b$ (the $a_i$ and $b_i$ are integers, of course); then $$ \frac{1}{2b}=\sum_{i=1}^nc_i\frac{a_i}{b} \qquad(c_1,\dots,c_n\in\mathbb{Z}) $$ implies $\frac{1}{2}\in\mathbb{Z}$, which is false; so the set $\{a_1/b,\dots,a_n/b\}$ doesn't generate $\mathbb{Q}$.

Answer 4

Here is a more fundamental approach to showing that $\mathbb{Q}$ is not finitely generated, based on a general method.

Here's the general method. In a group $G$ with a finite generating set $\{g_1,…,g_K\}$, for any increasing sequence of subgroups $H_1 \subset H_2 \subset \cdots$ whose union $\cup_i H_i$ equals $G$, for sufficiently large $i$ the subgroup $H_i$ equals $G$, because each $g_k$ is contained in some $H_{i_k}$ and so the whole generating set $\{g_1,…,g_K\}$ is contained in $H_I$ where $I = \text{max}\{i_k\}$.

Here's the application to $\mathbb{Q}$. Take $H_i$ to be the cyclic subgroup of $\mathbb{Q}$ generated by $\frac{1}{n_i}$ where $n_i$ is the product of the $i^{\text{th}}$ powers of the first $i$ prime numbers. Since $n_i$ divides $n_{i+1}$, $H_i$ is a subgroup of $H_{i+1}$, so the sequence is increasing $H_1 \subset H_2 \subset \cdots$. Since each integer divides some $n_i$, the union $\cup_i H_i$ equals $\mathbb{Q}$.