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If $G$ is a finite abelian group and $H$ is a subgroup. Show $\exists K\le G$ such that $G/K\cong H$.

I observe that it is enough to find $K\le G$ such that $H\cap K=\{e\}$ and $G=HK$. As in that case $G=H\oplus K$.

But how to find such $K$, can anyone give any idea to do this? Thanks for help in advance.

DeltaEpsilon
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    Fundamental Theorem of Finitely Generated Abelian Groups? – ndhanson3 Jul 28 '21 at 07:48
  • Such a $K$ as you hypothesize doesn't necessarily exist. Let $G= C_4, H= C_2$. – Robert Shore Jul 28 '21 at 07:51
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    https://math.stackexchange.com/questions/198435/splitting-exact-sequences-of-finite-abelian-groups – Tsemo Aristide Jul 28 '21 at 07:55
  • @ndhanson Yes, my intuition also says the same. But how to argue using Fundamental theorem of finitely generated abelian group? I'm not able to show it rigoursly – DeltaEpsilon Jul 28 '21 at 08:13
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    The standard way of doing this is to use dual groups (sometimes called character groups). Show that every finite abelian group is isomorphic to its dual and then note that subgroups of the group correspond to quotients of the dual and vice versa. – Derek Holt Jul 28 '21 at 10:58

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