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 $$\varphi$$.

• One definition of the cokernel of a group homomorphism is the quotient of the codomain of the map by the normal closure of the image. The normal closure of the subgroup $H$ of $S_3$ is the whole of $S_3$, so the cokernel is trivial. Nov 12, 2020 at 13:18

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.

• I don't understand your question, can you clarify? Nov 12, 2020 at 16:54
• We want to prove that $\psi$ is trivial, so it suffices to prove that its kernel is equal to $S_3$. Nov 12, 2020 at 17:00
• We don't conclude that $im(\varphi)=S_3$. We want to prove that $\ker(\psi)=S_3$ if $H\leq \ker(\psi)$, but of course $H\neq \ker(\psi)$ since $H$ is not normal. Nov 12, 2020 at 17:30

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.

• @batuhan, Here normal closure of a subgroup $H$ means the smallest normal subgroup $N$ containing $H$.
– MAS
Nov 12, 2020 at 14:17