Today I'm obsessed with universal properties and I'm trying to prove that the universal property of quotient groups implies that the subgroup is killed by the cannonical epimorphism. I hope this feeling will pass soon but I don't see it going before resolving this question, haha.
Let $G$ be a group with a normal subgroup $N\leq G$. We call a quotient such a map $\pi:G\to G/N$ (to some group denoted $G/N$) that for every map $\varphi:G\to C$ blind to the subgroup $N$ ($\ker \varphi\supseteq N$), there exists a unique $\tilde\varphi:G/N\to C$ such that $\tilde\varphi\circ\pi=\varphi$: $$\forall\varphi\in\hom(G,C):N\subseteq\ker\varphi\to\exists!\tilde{\varphi}\in\hom(G/N,C):\tilde{\varphi}\circ\pi=\varphi.$$
The previous sentece in a commutative diagram:
I want to prove that the map $N\to G/N$ here (which I denoted by "!0") is the zero map, i.e. that $\forall n\in N:\pi(n)=e_{G/N}$.
(Yes, here $G/N$ is not the classical quotient, but any group satisfying the universal property)
The natural approach seems to take $C$ to be the classical model of the quotient, i.e. the set $C:=\{[g]:g\in G\}$ of the cosets $[g]=\{gn:n\in N\}$ with $\varphi(g)=[g]$.
The problem is that now, for $n\in N$, I can only know something about $\tilde\varphi(\pi(n))$ - that it vanishes $\tilde\varphi(\pi(n))=\varphi(n)=[n]=e_{C}$, but nothing about the argument inside the $\tilde\varphi$ - $\pi(n)$, which I want to be $e_{G/N}$.
I'd know it vanishes too only if $\tilde\varphi$ were an injection (which I know it must be, because we took one "best" quotient possible, i.e. $\tilde\varphi$ must be an isomorphism), but I don't know how to prove that, kind of for the same reason I struggle with the above - once I have a $g\in G$ or $c\in C$, I can only go the whole way forward or the whole way backward, always skipping $G/N$.
For example, if I try to prove injectivity of $\tilde\varphi$, I take $a,b\in G/N$ with $\tilde\varphi(a)=\tilde\varphi(b)$, then by surjectivity of $\varphi$ I know $\tilde\varphi(a)=\tilde\varphi(b)=\varphi(g)=[g]$ for some $g\in G$. At this point I expect that $g$ is a common preimage of my $a$ and $b$, i.e. $\pi(g)=a=b$, but I see no way to make that connection, because the only thing I know here is that after applying $\tilde\varphi$ I'd get the same element, $\tilde\varphi(\pi(g))=\varphi(g)=\tilde\varphi(a)=\tilde\varphi(a)$, where I'd need the hypothesis for injectivity of $\tilde\varphi$.
I would love a hint out of here. Not a full solution, though I expect it to be so simple as to be impossible to break down. I even asked ChatGPT, which confidently used the goal in its proof lol.
I even started wondering whether my claim could be wrong, in which case I'd love a counterexample (a sample $G/N$ and $\pi:G\to G/N$). In that case, though, wouldn't it be inappropriate to call it a universal property of the quotient group?
This is not the correct definition of the quotient group. You missed the condition $N \subseteq \ker(\pi)$. You ask if this condition is automatic, but of course it's not. Consider $\pi := \mathrm{id} : G \to G$ otherwise. Any homomorphism $ G \to C$ factors uniquely through $\pi$ here. Also notice that with your incorrect definition quotient groups would not be unique.
One of the many correct definitions of the quotient group is the following: it is an initial object in the category $\mathscr{C}$ of homomorphisms $\varphi : G \to C$ with $N \subseteq \ker(\varphi)$. Keep in mind that an initial object of $\mathscr{C}$ is actually supposed to be an object in $\mathscr{C}$.
What is more interesting (and the answer is not trivial at all):
I leave this as an exercise ;)