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.
Best Answer
Your proof is correct.
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 $e \in K$.