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.

  • 2
    $\begingroup$ It's perfect. However, if you are uncertain, just try to separately prove the two implications. $\endgroup$
    – Berci
    Apr 12, 2021 at 17:37

1 Answer 1


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 $e \in K$.


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