Quotients of $(\mathbb{Q}/\mathbb{Z})^n$ Is every quotient of $(\mathbb{Q}/\mathbb{Z})^n$ by a finite subgroup $T$ isomorphic to $(\mathbb{Q}/\mathbb{Z})^n$?. It seems true for $n = 1$ by inducting on the number of generators of $T$. So if $n > 1$, it should still be true if $T$ is a direct sum $\bigoplus_{i=1}^n T_i$, where the $i$-th component $T_i$ is a subgroup of the $i$-th component of $(\mathbb{Q}/\mathbb{Z})^n$. Any subgroup $T$ of $(\mathbb{Q}/\mathbb{Z})^n$ is contained in a subgroup $T'$ which is such a direct sum, so there is a surjection
$$(\mathbb{Q}/\mathbb{Z})^n/T \twoheadrightarrow (\mathbb{Q}/\mathbb{Z})^n/T' \cong (\mathbb{Q}/\mathbb{Z})^n.$$
So $(\mathbb{Q}/\mathbb{Z})^n$ is a quotient of its own quotient $(\mathbb{Q}/\mathbb{Z})^n/T$. Can we find an injective map $(\mathbb{Q}/\mathbb{Z})^n/T \to (\mathbb{Q}/\mathbb{Z})^n$ whose image is of finite index, and therefore all of $(\mathbb{Q}/\mathbb{Z})^n/T$? Or this statement not even true?
 A: If $G$ is an abelian group such that $G/\langle g \rangle \cong G$ for all $g \in G$, then $G/T \cong G$ for every finitely generated subgroup $T \subseteq G$. This is a simple induction on the size of a generating set of $T$.
Let $g \in (\mathbb{Q}/\mathbb{Z})^n$. Choose a preimage $g' \in \mathbb{Q}^n$. Then $(\mathbb{Q}/\mathbb{Z})^n / \langle g \rangle \cong \mathbb{Q}^n / (\mathbb{Z}^n + \langle g' \rangle)$. The group $\mathbb{Z}^n + \langle g' \rangle$ is finitely generated and torsionfree, hence free. The rank is clearly $\geq n$, but also $\leq n$ as the group embeds into $\mathbb{Q}^n$. So the group is isomorphic to $\mathbb{Z}^n$. If $(v_1,\dotsc,v_n)$ is a $\mathbb{Z}$-basis of it, then this is a $\mathbb{Q}$-basis of $\mathbb{Q}^n$ (since it is linearly independent and has the correct number of elements). This implies $\mathbb{Q}^n / (\mathbb{Z}^n + \langle g' \rangle) = (\bigoplus_{i=1}^{n} \mathbb{Q} \cdot v_i) / (\bigoplus_{i=1}^{n} \mathbb{Z} \cdot v_i) \cong (\mathbb{Q}/\mathbb{Z})^n$.
