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Let $G / H = M_0 \supseteq ... \supseteq M_n$ be a normal tower of $G / H$ with $M_{i} / M_{i + 1}$ being cyclic. Then, define the cannonical homomorphism: $$\Phi: G \rightarrow G / H;\quad \Phi: g \rightarrow gH$$ Then, how to show that tower $$G = G_0 = \Phi^{-1}(M_0) \supset ... \supseteq G_n = \Phi^{-1}(M_n)$$ of $G$ is in fact normal and cyclic. The normal part is trivial since we are considering homomorphism. Reducing to $$\Phi: G_{n} \rightarrow M_{n}$$ proves $$M_{n + 1} \text{ normal in } M_{n} \Rightarrow G_{n + 1} = \Phi^{-1}(M_{n + 1}) \text{ normal in } G_{n}$$ But why $G_{n} / G_{n + 1}$ cyclic? The author state it like it's trivial. I'm starting with graduate algebra so please provide an answer in detail.

I have worked similar problems in: Solvability of $G$ w.r.t $H$ and $G / H$. But getting no answers. By the way, this is from Theorem 6.6 Serge Lang's Algebra.

Arturo Magidin
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Duck Gia
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    Stop reading Lang. It is not a good book to learn from, and it is clear that you are simply not at a level that would let you bulldoze through the difficulties. Get a better texbook to learn first, then you can try Lang. – Arturo Magidin Dec 04 '24 at 16:01
  • @ArturoMagidin Sorry but I might not :D – Duck Gia Dec 05 '24 at 04:27

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If $M\unlhd G/H$, denote by $\bar{M}=\Phi^{-1}(M)$. As you said, $\bar{M}\unlhd G$. Moreover, $H\leq \bar{M}$ and $\bar{M}/H\simeq M$ (see for example the answer of this: The correspondence theorem for groups). Therefore, $$\frac{\bar{M}_{i}}{\bar{M}_{i+1}}\simeq\frac{(\bar{M}_{i}/H)}{(\bar{M}_{i+1}/H)}\simeq\frac{M_i}{M_{i+1}},$$ that is cyclic. The first isomorphism above is the "third isomorphism theorem" for groups (see this Understanding the three isomorphism theorems).

Tod
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