In the revised third edition of Serge Lang’s Algebra, I am studying Abelian towers of groups. It is shown that the inverse image of an Abelian/cyclic tower is Abelian/cyclic. I tried to formulate a proof that the image of an Abelian/cyclic tower is Abelian/cyclic under surjective group homomorphisms and I’d like to know if there are any errors.
Proof:
Let $G$ be a group with an Abelian tower $G=G_0 \supset G_1 \supset \cdots \supset G_k$. Let $f:G\to \overline{G}$ be a surjective group homomorphism. Of course, $f(G_{i+1})<f(G_i)$ because $G_{i+1}\subset G_i$ and images of subgroups via group homomorphism are subgroups. Since for all suitable $i$ we have $G_{i+1} \lhd G_i$, it follows from the normality test that $f(G_{i+1}) \lhd f(G_i)$. This is because for all $g_{i}’\in f(G_i)$, letting $x\in g_{i}’f(G_{i+1})g_{i}’^{-1}$ implies for some $g_i\in G_i$, $x=f(g_i)f(g_{i+1})f(g_i)=f(g_ig_{i+1}g_i^{-1})$ and by normality, $x=f(s)$ for some $s \in G_{i+1}$. Thus, $g_i’f(G_{i+1})g_i’^{-1}\subset f(G_{i+1})$.
For each suitable $i$, consider the composite homomorphism from $G_i\to f(G_i) \to f(G_i)/f(G_{i+1})$ given by applying $f$ and then the mapping that takes $g_i’$ to the coset $g_i’f(G_{i+1})$. The composite homomorphism is surjective. Also, the kernel of this composite homomorphism is precisely $G_{i+1}$, so by the first isomorphism theorem, $G_i/G_{i+1} \cong f(G_i)/f(G_{i+1})$. By assumption, the LHS of this isomorphism is Abelian/cyclic, so the RHS is Abelian/cyclic as well.
Therefore, this shows that the tower in $\overline{G}$ given by $\overline{G}=f(G) \supset f(G_1) \supset \cdots \supset f(G_k)$ satisfies both the normality condition for Abelian/cyclic towers and the requirement that for all suitable $i$ successive factor groups are Abelian/cyclic.
Of course, any help would be appreciated.
Your proof is almost correct, except that there is no reason for the kernel of $G_i \twoheadrightarrow f(G_i) \twoheadrightarrow f(G_i)/f(G_{i+1})$ to be equal to $G_{i+1}$ (think about the case when your tower is of length $1$). You only know that is must contain $G_{i+1}$. But this is enough to conclude: you then have an induced surjection (which does not have to be an isomorphism) $G_i/G_{i+1} \twoheadrightarrow f(G_i)/f(G_{i+1})$, and a quotient of a cyclic (resp. abelian) group is again cyclic (resp. abelian).