Let $A,B, C,A',B'$ be an abelian groups. Let $f:A\to B$ and $g:B\to C$ be group homomorphisms.

Question

Let $A,B, C,A',B'$ be an abelian groups. Let $f:A\to B$ and $g:B\to C$ be group homomorphisms.

Suppose $A'\cong A$, $B'\cong B$ and $C\cong C'$ as abelian groups and $imf\subset kerg$.

Let $f':A'\to B'$ and $g':B'\to C'$ be induced map. ($f'$ is a composition map $ A'\cong A\to B\cong B'$, and the same for $g'$ )

Then, How can we say there is an isomorphism between $kerf/imf \cong kerf'/img'$ ?

If both $kerf$ and $img$ is finite, it is clear $kerf/imf$ and $kerf'/img'$ has the same order, but I'm stuck with constructing isomorphism between them.

Answer 1

Preliminary fact.

Consider a commutative square of abelian groups as below, where the horizontal homomorphisms are isomorphisms: $\require{AMScd}$ $$\begin{CD} M@>\sim>>M'\\ @V\alpha VV @V\beta VV\\ N@>\sim>> N' \end{CD}$$

1. The unique homomorphism $\operatorname{ker}\alpha\to\operatorname{ker}\beta$ which makes the following square commutative is an isomorphism: $$\begin{CD} \operatorname{ker}\alpha@>>>\operatorname{ker}\beta\\ @VVV @VVV\\ M@>\sim>> M' \end{CD}$$
2. The unique homomorphism $N/\operatorname{im}\alpha\to N'/\operatorname{im}\beta$ which makes the following square commutative is an isomorphism: $$\begin{CD} N@>\sim>>N'\\ @VVV @VVV\\ N/\operatorname{im}\alpha@>>> N'/\operatorname{im}\beta \end{CD}$$

Answer.

In your problem, you obtain an isomorphism $h:\operatorname{ker}g\to\operatorname{ker}g'$, using point (1) on the following square: $$\begin{CD} B@>\sim>>B'\\ @Vg VV @Vg' VV\\ C@>\sim>> C' \end{CD}$$ Since $\operatorname{im}f\subset\operatorname{ker}g$ and $\operatorname{im}f'\subset\operatorname{ker}g'$, you get the following square:$$\begin{CD} A@>\sim>>A'\\ @Vf VV @Vf' VV\\ \operatorname{ker}g@>h>> \operatorname{ker}g' \end{CD}$$ Applying point (2) to the latter square gives the isomorphism you're looking for.