Cokernel of group homomorphism We know $H=\{ (1),(12) \}$ is subgroup of $S_3$. Consider inclusion '$\varphi: H \hookrightarrow S_3$', this is clearly group homomorphism. Prove that coker$\varphi$ is trivial. I can't understand how to apply universal property of cokernel to this homomorphism $\varphi$.
 A: The cokernel of $\varphi$ is, by definition, initial among all morphism $\psi: S_3\to G$ such that $\psi\circ \varphi$ is the trivial morphism. So you need to prove that if $\psi\circ \varphi$ is trivial then $\psi$ is trivial. Note that $\psi\circ \varphi$ is trivial if and only if $H\leq \ker(\psi)$, so it suffices to prove that any normal subgroup of $S_3$ containing $H$ must be equal $S_3$ itself.
A: Given a group homomorphism $f : A \to B$, it is easy to see that $\operatorname{coker}(f) = B / [f(A)]$, where $[ - ]$ denotes the normal closure of a subgroup. In the language of category theory, more formally, $\operatorname{coker}(f)$ is the qoutient homomorphism  $p : B \to B / [f(A)]$.
To see this, let $q : B \to C$ be any homomorphism such that $q \circ f = 0$. This means $f(A) \subset \ker(q)$, thus $[f(A)] \subset \ker(q)$ because kernels are always normal subgroups. Therefore $q$ can be written as $q = \bar q \circ p$ with a unique homomorphism $\bar q :  B / [f(A)] \to C$.
Now let us look at $[H]$ in $S_3$. The only non-trivial normal subgroup of $S_3$ is the alternating group $A_3$, but certainly $H \not\subset A_3$. Thus $[H] = S_3$ which proves your claim.
