Proving that the quotient map is a surjective group homomorphism

Given $$G$$ a group and $$K$$ a normal subgroup, the map $$f: G \to G/K$$ is a surjective group homomorphism with kernel $$K$$.

Here is my attempt.

Given $$aK \in G/K$$, $$f(a) = aK$$, so $$f$$ is surjective. Given $$a,b \in G$$, we have \begin{align*} f(ab) = (ab)K = aK bK = f(a) f(b), \end{align*} which is a well-defined multiplication of cosets since $$K$$ is normal. Finally, we have: \begin{align*} a \in \mathrm{ker}(f) \iff f(a) = eK = K \iff aK = K \iff a \in K. \end{align*}

How does this look? The only thing I am not completely sure about is the proof for the kernel. I'm not completely certain that every step I wrote down, specifically the last one, is reversible.

• It's perfect. However, if you are uncertain, just try to separately prove the two implications. Apr 12, 2021 at 17:37

Concerning your doubts: If $$a \in K$$, then $$ag \in K$$ for every $$g \in K$$ since $$K$$ is a subgroup. If $$aK = K$$ then $$a = a e \in K$$ since $$e \in K$$.