On p. 18 of the revised third edition of Serge Lang’s Algebra, he says that the preimage of an Abelian/Cyclic tower is an Abelian/Cyclic tower. I want to know if my proof is correct and why there is a detail that Lang mentions which I am not using.


Let $f:G\longrightarrow G'$ be a group homomorphism. Let $$G'=G'_0\trianglerighteq\dotsc\trianglerighteq G_n'$$ be a normal tower in $G'$ and let $G_i=f^{-1}(G_i')$. Then clearly the $G_i$ form a normal tower. If the $G_i'$ form an abelian tower (resp. cyclic tower), the $G_i$ form an abelian tower (resp. cyclic tower), because we have an injective homomorphism $$\bar{f}:G_i/G_{i+1}\longrightarrow G_i'/G_{i+1}'$$ for each $i$, and because a subgroup of an abelian group (resp. cyclic group) is abelian (resp. cyclic).

I'm trying to understand why he mentions that the subgroup of an abelian/cyclic group is abelian/cyclic, since in my proof I didn't occupy this at all (unless I didn't notice). My proof is the following.

If $$G'=G'_0\trianglerighteq\dotsc\trianglerighteq G_n'$$ is an abelian tower, then each $G_i'/G_{i+1}'$ is abelian and therefore \begin{align*} \bar{f}(x_i G_{i+1} y_i G_{i+1})&=\bar{f}(x_i G_{i+1})\bar{f}(y_i G_{i+1})\\ &=\bar{f}(y_i G_{i+1})\bar{f}(x_i G_{i+1})\\ &=\bar{f}(x_i G_{i+1} y_i G_{i+1}), \end{align*} which by injectivity implies that $x_i G_{i+1} y_i G_{i+1}=x_i G_{i+1} y_i G_{i+1}$ and hence each $G_i/G_{i+1}$ is abelian. This proves that the preimage of an abelian tower is an abelian tower.

Now suppose $$G'=G'_0\trianglerighteq\dotsc\trianglerighteq G_n'$$ is a cyclic tower. Then each $G_i'/G_{i+1}'=\langle a_i' G_{i+1}' \rangle$ and therefore \begin{align*} \bar{f}(x_i G_{i+1})&=f(x_i)G_{i+1}'\\ &=(a_i')^{k_{x_i'}} G_{i+1}'. \end{align*} By definition, every element of $G_i$ is the preimage of some element in $G_i'$, so $a_i'=f(a_i)$ for some $a_i$ and then $$f(x_i G_{i+1})=f\big(a_i^{k_{x_i'}} G_{i+1}\big),$$ which by injectivity implies that $x_i G_{i+1}=a_i^{k_{x_i'}} G_{i+1}$ and hence each $G_i/G_{i+1}$ is cyclic. This proves that the preimage of a cyclic tower is a cyclic tower.

Is this proof correct? Why does Lang mention that the subgroup of an abelian/cyclic group is abelian/cyclic if it is not used in the proof? I want to know what I'm ignoring.

  • 1
    $\begingroup$ Nothing mysterious, you are just implicitely proving in your proof that subgroups of abelian groups are abelian and same for cyclic groups... $\endgroup$
    – reuns
    Feb 8 at 0:55

1 Answer 1


Your proof for the cyclic case is not quite right.

But first I will try to guess what Lang meant. Here is a very simple lemma.

Lemma. Let $A$, $B$ be two groups and $g: A\to B$ be an injective homomorphism. If $B$ is an abelian/cyclic group, then $A$ is also an abelian/cyclic group.

Proof. Since $A$ is isomorphic to $g(A)$ and $g(A)$ is a subgroup of $B$, then $g(A)$ and $A$ are abelian/cyclic groups.

Since $\overline{f}: G_{i}/G_{i+1}\to G'_i/G'_{i+1}$ is an injective homomorphism everything follows from the lemma.

I dare to hope that this is what Lang meant.

Now where you have made an inaccuracy. If $a'\in G'_i$, then it does not follow at all that there exists an element $a\in G$ that $f(a)=a'$. The homomorphism $f$ does not have to be surjective.

  • $\begingroup$ simple and clean, thank you. $\endgroup$
    – Nerhú
    Feb 8 at 18:19

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