Homomorphisms from finite abelian group into $\mathbb{C}^*$ I am trying to solve a concrete problem in finite abelian groups. I can see its quick solution by using fundamental theorem of finite abelian groups, wich I learned around $15$ years ago, without raising questions to myself - and preferring to "read, understand step-by-step proof" given in classroom notes or book.
But, now, I am trying to look by asking elementary questions, in some concepts, and looking whether we can give simple argument.
My concrete question is this one: can we prove the following statement without using classification of finite abelian groups?

Let $G$ be any finite abelian group of order $16$ and of exponent $4$. Let $x\in G$ be an element of order $4$. Then ehere exists a homomorphism $f:G\rightarrow \mathbb{C}^*$ such that $f(x)=i$.

Why this question: It is mentioned in almost all books about finite abelian $p$-groups, that if $x$ is an element of maximum order in it, then $\langle x\rangle$ has a complement in the group (if group is non-cyclic). The proof of this result proceeds by element-wise computations. And, I am looking same problem in terms of mappings/homomorphisms, which is nowadays heavily used approach in algebra. If we get a homomorphism $f$ mentioned in the problem, then $\ker f$ is a complement.
 A: For finite groups, in the spirit of of Ravi's comments, you can prove the following result in an elementary way (without using the theory of injective modules or the structure theorem for a finite abelian groups):
Thm. Let $G$ be a finite abelian group, and let $H$ be a subgroup. Then any morphism $H\to\mathbb{C}^*$ extends to a morphism $G\to\mathbb{C}^*$.
The proof is by induction on $m=[G:H]$. If $m=1$, it's OK. If $>1$, pick $g\in G\setminus H$, and let $n$ the order of $gH$ in $G/H$. In other words, $n$ is the smallest positive integer such that $g^n\in H$. Let $\chi:H\to\mathbb{C}^*$ a morphism, and set $t=\chi(g^n)$. We can write $t=w^n$ for some $w\in\mathbb{C}^*$. Let $H'=\langle H,g\rangle.$  Now every element $h'$ of $H'$ can be written $h'=hg^k,h\in H,k\in\mathbb{Z}$. Set $\chi'(h')=\chi(h) w^k$. One can show that this does not depend on the decomposition of $h'$ and that $\chi'$ is a morphism. Now $g\in H'\setminus H$, so $[G:H']<[G:H]$ and we may apply induction.
Now to solve your question, since $x$ has order $4$, we have a morphism $H=\langle x\rangle \to\mathbb{C}^*$ sending $x$ to $i$, and we can apply the theorem.
