Extension of a character across a cyclic quotient ($\tilde{\varphi }(h+ja)=\varphi (h){\lambda }^j,{\lambda }^k=\varphi (ka)$)

If $H$ is a subgroup of a finite abelian group, $a\notin H$, and $k$ is the least positive integer with $ka\in H$, then each character of a finite abelian group on $H$ has exactly $k$ extensions to $H+⟨a⟩$. The displayed root choice makes the extension well-defined. Starting with the trivial subgroup and adjoining generators proves that the number of characters equals the group order without using a decomposition into cyclic groups.