If we have a short exacts sequence of abelian group with $A$ and $C$ finite, $0\to A\to B\to C$, is it true that $|B|=|A|\cdot |C|$ even if it does not split?
2026-04-21 16:01:54.1776787314
Order of middle group in short exact sequence
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Yes, it's simply Lagrange's theorem. By definition of short exact sequence, $C\cong B/\operatorname{im}(f)$, where $f\colon A\to B$ is the given map. Since $f$ is injective, $\lvert\operatorname{im}(f)\rvert=|A|$.