divisible group is in general not a Q-Vector space

Let \(G\) be an n-divisible group for each \(n \in \mathbb{N}\).
Then \(G\) is in general not the additive group of a \(\mathbb{Q}\)-vector space.

Consider the quotient group \(\mathbb{Q}/\mathbb{Z}\).
Here \(\mathbb{Z}\) is a normal subgroup, since \(\mathbb{Q}\) is an abelian group.

It is n-divisible for each \(n \in \mathbb{N}\): for \([ \frac{a}{b}]\) we may consider \([\frac{1}{n} \cdot \frac{a}{b}]\) which gives \(\int_{i=1}^n [ \frac{1}{n} \frac{a}{b}] = [ \sum_{i=1}^n \frac{1}{n} \cdot \frac{a}{b}] = [ \frac{a}{b}]\).

Now we want to show that \(\mathbb{Q}/ \mathbb{Z}\) can't be equipped with \(\mathbb{Q}\)-vector space structure:
Suppose there exists one, then we get

\begin{align*} [0] =& [1] \\ =& [\frac{1}{2} + \frac{1}{2}] =& [ \frac{1}{2}] + [ \frac{1}{2} ] \\ =& 1 \cdot [ \frac{1}{2}] + 1 \cdot [\frac{1}{2}] \\ =& 2 \cdot [ \frac{1}{2} ] \\ \end{align*}

Now in particular \([0]\) is the neutral element of the addition, so in particular \(\lambda \cdot [0] = [0]\) for any \(\lambda \in \mathbb{Q}\).
So consider the calculation

\begin{align*} [0] =& 2 \cdot [\frac{1}{2}] && \vert \cdot \frac{1}{2} \\ [0] =& [\frac{1}{2}] \end{align*}

which is a contradiction, since \([0] \neq [\frac{1}{2}] \in \mathbb{Q}/\mathbb{Z}\)