TITLE: Generating the middle group in short exact sequences, not split
QUESTION [1 upvotes]: Assume that we have a short exact sequence:
$$1 \rightarrow A \xrightarrow{f} B \xrightarrow{g} C \rightarrow 1$$
Assume the generating sets $\{a_i\}$ of $A$ and $\{c_j\}$ of C are given. I want to construct a generating set for $B$.
If the sequece is split, this is really easy since in this case $B \cong A \oplus C$ and therefore we can just map the set of generators of $A$ and $C$ to generators of $B$ by this isomorphism. I have a strong suspicion that a similar technique works for non-split sequences, although I can't recall such a statement.
I know that by exactness, we have that $ C \cong B /{\rm im}(f)$, but i have my doubts if my technique works to find a generating set of $B$. My attempt is the following:
We map the generators of $A$ via $f$ to $B$, and denote those by $b_i$. Then for each generator $c_j$ of $C$, we choose an element of $B$ in the preimage $g^{-1}(c_j) \notin{\rm im}(f)$, calling this  $\tilde{b_j}$. Now we take the union $\{b_i\} \cup \{\tilde{b_j}\}$ and claim that this generates $B$.
I guess that this works, but I'm unable to prove it.
So my question is:
Does this work, and how would one prove that this is in fact a generating set of $B$?

REPLY [1 votes]: You should consider the subgroup generated by your elements. It contains the generators of $f(A)$ and therefore contains $f(A)$. Its image in $C$ under $g$ contains the generators of $C$ and therefore is equal to $C$. By standard isomorphism theorems (or checking yourself) this means that your subgroup is the whole $B$.
{"set_name": "stack_exchange", "score": 1, "question_id": 4169218}