Preimage of an Abelian/Cyclic tower is an Abelian/Cyclic tower

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.

Excerpt:

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.

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

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.

• simple and clean, thank you. Feb 8 at 18:19