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Now that school is wrapping up, I'm trying to crack down and get better at algebra. This proposition from Lang's Algebra loses me at the end.

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Here is my understanding so far: (Please excuse me if a lot of the things I say are very obvious/wrong, I'm trying to be detailed for my own understanding.) If $|G|=1$, then $\{e\}$ is the desired cyclic tower. So suppose the result holds for $|G|\leq n-1$. Suppose $|G|=n$. Letting $G'$ be as above, $|G'|=|G|/|X|\lt|G|$, so by the induction hypothesis, there exists a cyclic tower in $G'$, say $$ G'=G/X\supset G_1'\supset G_2'\supset\cdots\supset G_m'. $$ I'm not quite sure what Lang means by "its inverse image is a cyclic tower in $G$ whose last element is $X$." Is there some assumed homomorphism $f\colon G\to G'$, and then the inverse image of the tower would be $$ f^{-1}(G')\supset f^{-1}(G_1')\supset\cdots\supset f^{-1}(G_m')? $$ Why is the last element of the tower $X$, as Lang claims? Also, Lang says a normal tower is cyclic if each factor group $G_i/G_{i+1}$ in the tower is cyclic. Does this mean all the $G_i$ themselves are cyclic, or is it possible for the factor group to be cyclic, but the normal subgroup being modded out is not? Thanks for any explanation.

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The assumed homomorphism $G\rightarrow G'=G/X$ is the quotient map: $g\mapsto gX$. One of the isomorphism theorems says that this map establishes a bijection between subgroups of $G/X$ and subgroups of $G$ that contain $X$. Moreover, this bijection preserves inclusions, normality, and quotients. In other words if $\bar{U}$ is a subgroup of $G'$ with subgroup $\bar{U}'$, and if the corresponding subgroups of $G$ are $U$ and $U'$, then $U'\leq U$, it is normal in $U$ if and only if $\bar{U}'$ is normal in $\bar{U}$, and if they are, then $\bar{U}/\bar{U}'\cong U/U'$. All this is easy to prove, and Lang is using all of this in his proof.

The subgroup of $G$ corresponding to the trivial subgroup of $G'$ is indeed $X$ - the smallest subgroup of $G$ containing $X$.

As for the last question, it means exactly what it says: all factor groups are cyclic. It would be pointless to introduce this notion if it was simply a reformulation of "$G$ is cyclic". E.g. $1\leq \mathbb{Z}/2\mathbb{Z} \leq \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ is a cyclic tower where the top term is not cyclic.

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  • $\begingroup$ Much appreciated, thank you Alex. I think I understand now, so the last element of the tower for $G'$ is $\{X\}$. If $f$ is the quotient map, then $f^{-1}(\{X\})=X$ so this gives some cyclic tower $G\supset G_1\supset\cdots\supset X$, and inserting $\{e\}$ to the end, gives $G\supset G_1\supset\cdots\supset X\supset\{e\}$, which is still cyclic since $X/\{e\}=X$. Is this the right understanding? $\endgroup$ – yunone May 17 '11 at 13:54
  • $\begingroup$ @yunone: Yes, that's right. A minor nitpick: the last element in the tower for $G'$ is the trivial group (such a chain of subgroups of $G'$ exists by inductive hypothesis). The pre-image of the trivial group under $f$ is $X$. Once you get the cyclic chain $G\supset G_1\supset\ldots\supset X$, you complete it as you say, by inserting $\{e\}$ at the end, and this works because $X$ itself is cyclic. $\endgroup$ – Alex B. May 17 '11 at 16:08

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