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If $G$ is an abelian finite group, how can I prove that there exist a chain of subgroups $$\{1\}\subseteq H_1 \subseteq H_2 \subseteq \ldots \subseteq H_n=G$$ So that $H_{i+1}/H_i$ is simple?

I don't even know how to start. The only thing that comes to my mind is that $H_i$ are normal groups (because $G$ is abelian). Also, I don't know how to use that $G$ is finite.

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Any finite group $G$ has a composition series. This can be shown by induction on $|G|=n$. When $n=1$, this is trivial. For the induction step, see here. The quotients are simple. So we do not need that $G$ is abelian.

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  • $\begingroup$ I have not seen the fourth isomorphism theorem and therefore I can't follow the proof. Isn't there an easier proof using that G is abelian? $\endgroup$ – John Keeper Apr 24 '18 at 2:02

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