# Explanation why an abelian tower admits a cyclic refinement

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. 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.

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.
• 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? – yunone May 17 '11 at 13:54
• @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. – Alex B. May 17 '11 at 16:08