If for every homomorphism $\phi :A \to \mathbb{C}^{\times}$ we have $\phi(a) = 1$, show that $a = e$

Question

We are given a finite abelian group $A$ and $a \in A$ such that for every homomorphism $\phi :A \to \mathbb{C}^{\times}$ (where $\mathbb{C}^{\times}$ is the group of complex numbers $\mathbb{C} - \{0\}$ with multiplication) we have $\phi(a) = 1$. Show that $a = e$. Here is what I did and I would like to know if that is correct:
We know $A$ is isomorphic to $G=\mathbb{Z}_{m_1} \times ... \times \mathbb{Z}_{m_k}$ and therefore it is enough to prove it for $G$. If $g = (a_1,...,a_k) \in G$ satisfies the condition, define $f: G \to \mathbb{C}^{\times}$ in the following way:
$f(x_1,...,x_k) = e^{2\pi i(x_1/m_1+...+x_k/m_k)}$. It is easy to check that $f$ is indeed homomorphism. therefore, $1=f(g)=f(a_1,...,a_n)=e^{2\pi i(a_1/m_1+...+a_k/m_k)}$. Therefore, $x=a_1/m_1+...+a_k/m_k$ is an integer.

Now define $h:G \to \mathbb{C}^{\times}$ in the following way: $h(x_1,...,x_k) = e^{2\pi i(2x_1/m_1+x_2/m_2+...+x_k/m_k)}$ Once again $h(g)=1 $ and therefore $y = 2a_1/m_1+a_2/m_2+...+a_k/m_k $ is an integer. so $y-x = a_1/m_1$ is also an integer, and since $0 \leq a_1 \leq m_1-1$ we have $a_1=0$. Similarly $a_1=...=a_k=0$ and the problem is solved. Is the solution correct?