= Extension of a character across a cyclic quotient
{title2=$\widetilde\phi(h+ja)=\phi(h)\lambda^j,\quad\lambda^k=\phi(ka)$}
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 $H+\langle a\rangle$. 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.
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