Inclusion induce an isomorphism

Question

I have a small question

If $E,F$ are two sub-modules of a module $G$,

how to prove that the inclusion $E \rightarrow E+F$ induces an isomorphism $(E/E \cap F) \simeq (E + F) / F$

Please

Thank you

Answer 1

The inclusion $E\to E+F$ induces a map $\phi: E\to (E+F)/F$ given by $x\mapsto x+F$. This defines a homomorphism with kernel $E\cap F$. That is, $$E/E\cap F \cong Im \ \phi.$$

To show $\phi$ is surjective, we let $x+y+F\in (E+F)/F$ where $x\in E$ and $y\in F$. Observe that $x+y+F=x+F$ in $(E+F)/F$ since $y\in F$ and notice that $$\phi(x)=x+F=x+y+F.$$ Thus, $\phi$ is surjective, which completes the proof.