In the following post: $G$ a group, $G=N \times K$, $M$ normal subgroup of $N \Rightarrow$ $M$ normal subgroup of $G$
Dan Shved states in his response
We can build a map $\varphi: G \to (N/M) \times K$ defined like this: $$ \varphi(nk) = (nM, k) \qquad \textrm{for all }\ n \in N,\ k \in K $$ It is easy to see that $\varphi$ is well defined and is a homomorphism, and $M$ is its kernel.
The kernel for $\varphi$ would be $$\varphi(nk) = (nM, k)=(M,e)\in M\times \{e\}\cong M.$$ But in set builder notation, do I write
$$\text{Ker }\varphi=\{nk\in G: \varphi(nk)=(nM,k)=(M,e), n\in N, k\in K\}?$$
Thank you in advance
The function, for any $k\in G$, given by
$$\begin{align} \rho_k: G&\to G,\\ n&\mapsto nk \end{align}$$
is a bijection; therefore, nothing is lost in writing
$$\ker (\varphi)=\{g\in G\mid \varphi(g)=(M, e)\}.$$