Question: Let $A$ be an abelian finite group and $B$ a subgroup of $A$. Let $\phi$ be an irreducible character of $B$. Show there is an irreducible character $\Phi$ of $A$ such that $\Phi|_B=\phi$. The number of possible $\Phi$ is $[A:B]$.
I was able to prove the existence of the irreducible character $\Phi$ but I don't know how to prove that the number of possible $\Phi$ is $[A:B]$. Any help on that part is appreciated.
Let me give you some hints and I let you fill out the details yourself. Look at the set $Irr(A/B)=\{\mu \in Irr(A): B \subseteq ker(\mu) \}$, the irreducible characters of $A/B$ basically. If $\phi \in Irr(B)$ and $\lambda$ is an extension, then for every $\mu \in Irr(A/B)$, $\lambda\mu$ is also an extension of $\phi$. Observe that $\#Irr(A/B)=|A:B|$. If $\mu$ runs over $Irr(A/B)$, then all the $\lambda\mu$ are different, since linear characters never take the value $0$.