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I'm trying to prove the converse of Lagrange's theorem holds for when $G$ is an abelian group such that $|G|=n$.

What I have is if $k > 1$ and $k|n$, there is a subgroup of prime order $p$ ($p$ is such that $p|k$) of $G$, say $N$, which is normal in $G$.

Consider the canonical epimorphism $\phi: G \to G/N$. Then $G/N$ is abelian group of order $n/p <n$, and I have by induction hypothesis that the converse holds for every abelian group of order less than $n$ so there is a subgroup of order $k/p$ of $G/N$, say $H$. How do I prove that $|\phi^{-1}(H)|=k$?

Mark
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  • @DietrichBurde I'm not asking for the full proof here, only asking a question related to my own, different proof. – Mark Dec 31 '15 at 13:35
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    I see. Probably every proof will use the ideas of the many duplicates, though (see also here). See also here, this might be more related to your approach. – Dietrich Burde Dec 31 '15 at 13:37

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If $\overline{g} \in G/N$ is a coset with representative $g$, then $\phi^{-1}(\overline{g}) = g+N$ as sets. Hence the preimage of any coset has size $p$. Preimages of different cosets are disjoint, so the preimage of a set of size $k/p$ has size $k$.

Alex G.
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