So far I have $$ A = \ker ( \mathbb{Q}/\mathbb{Z} \xrightarrow{\cdot 3} \mathbb{Q}/\mathbb{Z} ) = \{ [\frac{a}{b}] \in \mathbb{Q}/\mathbb{Z} \mid [\frac{3a}{b}]=0 \}. $$ Since we in other words need that $3a/b$ is an integer, the subgroup of $\mathbb{Q}$ that would satisfy this is $\frac{1}{3} \mathbb{Z}$, so I have come to the conclusion that $$ A =\{ [0], [\frac{1}{3}], [\frac{2}{3}] \} \cong \mathbb{Z}/3$$ since the map is $\mathbb{Q}/\mathbb{Z} \xrightarrow{\cdot 3} \mathbb{Q}/\mathbb{Z}$ as opposed to $\mathbb{Q} \xrightarrow{\cdot 3} \mathbb{Q}$. Is this correct? I was also tempted to write something like $\frac{1}{3} \mathbb{Z}/\mathbb{Z}$, but this looks strange. Is it the case that $\frac{1}{3} \mathbb{Z}/\mathbb{Z} \cong \mathbb{Z}/3$? I guess it is since there are only three numbers of the form $\cdots/3$ mod $\mathbb{Z}$.
Thank you.
Correct computation, and the notation $$ \frac{1}{3}\mathbb{Z} / \mathbb{Z} $$ is excellent.
Yes, this group is isomorphic to $\mathbb{Z}/3\mathbb{Z}$. To show this formally, you can take the map $$ \frac{1}{3} \mathbb{Z} \to \mathbb{Z}/3\mathbb{Z} $$ given by $x \mapsto 3x$ and check that its kernel is the subgroup $\mathbb{Z}$ of $\frac{1}{3}\mathbb{Z}$, then appeal to the first isomorphism theorem.