(Pontrjagin Duality) If $G$ is a (discrete) abelian group, its Pontrjagin dual is the group
$$
G^{*}=\operatorname{Hom}_{\mathrm{Z}}(G, \mathbb{R} / \mathbb{Z}) .
$$
(More generally, the Pontrjagin dual of a locally compact abelian topological group $G$ consists of all the continuous homomorphisms from $G$ into the circle group $S^{1} \cong \mathbb{R} / \mathbb{Z}$.)
(i) If $G$ is an abelian group and $a \in G$ is nonzero, prove that there is a homomorphism $f: G \rightarrow \mathbb{R} / \mathbb{Z}$ with $f(a) \neq 0$.
(ii) Prove that $\mathbb{R} / \mathbb{Z}$ is an injective abelian group.
(iii) Prove that if $0 \rightarrow A \rightarrow G \rightarrow B \rightarrow 0$ is an exact sequence of abelian groups, then so is $0 \rightarrow B^{*} \rightarrow G^{*} \rightarrow A^{*} \rightarrow 0$.
(iv) If $G$ is a finite abelian group, then $G^{*} \cong \operatorname{Hom}_{\mathrm{Z}}(G, Q / \mathbb{Z})$.
(v) If $G$ is a finite abelian group, prove that $G^{*} \cong G$.
(vi) Prove that every quotient group $G / H$ of a finite abelian group $G$ is isomorphic to a subgroup of $G$.
Remark. The analogous statement for nonabelian groups is false: if $Q$ is the group of quaternions, then $Q / Z(Q) \cong$ $\mathbf{V}$, where $Z(\mathbf{Q})$ is the center of $\mathbf{Q}$ and $\mathbf{V}$ is the four-group. But $\mathbf{Q}$ has only one element of order 2 while $\mathbf{V}$ has three elements of order 2, so that $\mathbf{V}$ is not isomorphic to a subgroup of $\mathbf{Q}$. Part (vi) is also false for infinite abelian groups: since $\mathbb{Z}$ has no element of order 2 , it has no subgroup isomorphic to $\mathbb{Z} / 2 \mathbb{Z}=\mathrm{I}_{2}$.