Cokernel within the Category of Groups

Question

This question is in regards to Aluffi's Algebra: Chapter $0$, II.$8.22$

$\textbf{8.22: }$Let $\varphi: G \rightarrow G'$ be a group homomorphism, and let $N$ be the smallest normal subgroup containing im $\varphi$. Prove that $G'/N$ satisfies the universal property of ${\rm coker}~\varphi$ in $\textbf{Grp}$.

My scratch work/thinking:

For $G'/N$ to satisfy the universal property of ${\rm coker}~\varphi$ there must be a group homomorphism \begin{equation*} \pi: G' \rightarrow G'/N \end{equation*} which is initial with respect to all morphisms $\alpha: G' \rightarrow L$ such that $\alpha \circ \varphi = 0$ . Thus, it follows that im $\varphi \subseteq \ker \alpha$. Now, this is where I get lost.

I think I understand that due to the universal property of quotients with respect to group homomorphisms results in the unique induced group homomorphism $\widetilde{\alpha}: G'/N \rightarrow L$. Does the 'initial' role come from $N$ being the smallest normal subgroup which contains ${\rm im}~\varphi$?

Side question: Precisely what does it mean for every homomrphism $\alpha: G' \rightarrow L$, such that $\alpha \circ \varphi$ is the trivial map, must factor through $G'/N$?

Answer 1

So, first off, $\pi$ is going to be the obvious projection $g \mapsto gN$.

Now, we just need to check initiality. Let $\alpha: G' \to L$ be such that $\alpha \circ \varphi = 0$. Then we need a map $\beta: G'/N \to L$ such that $\alpha = \beta \circ \pi$. There's an obvious choice for this: define $\beta(gN) = \alpha(g)$. We need to check that it's well-defined: if $gN = hN$, then $gh^{-1}\in N$.

Suppose that $\alpha(g)\neq \alpha(h)$. then $gh^{-1} \not\in \ker(\alpha)$. But $\ker\alpha$ is a normal subgroup of $G'$, and since $\alpha\circ\varphi = 0$, $\ker\alpha\supseteq\mathrm{im}\varphi$, so $\ker\alpha\supseteq N\ni gh^{-1}$, a contradiction. Thus, $\alpha(g) = \alpha(h)$, and $\beta$ is well-defined.

Finally, uniqueness of $\beta$ is clear, so we're done.

Essentially, the answer to your question is "yes".

For your side-question, it means that for all such $\alpha$, there are maps $\delta: G' \to G'/N$ and $\varepsilon: G'/N \to L$ such that $\alpha = \varepsilon\delta$.

Answer 2

The answer is straightforward if you are familiar with Theorem 7.12 in Aluffi, which is stated as:

Let $H$ be a normal subgroup of group G, and $\pi: G \to G/H$ be defined as $\pi(g) = gH$. Then for every group homomorphism $\varphi : G \to G'$ such that $H \subseteq \ker\varphi$ there exists a unique group homomorphism $\tilde{\varphi}:G/H \to G'$ so that $\tilde{\varphi} \circ \pi = \varphi$.

The requirement that $H \subseteq \ker\varphi$ makes sure $\varphi$ sends related elements in G to the same element in $G'$, then universal property of quotient is applied to prove the theorem. For this problem, let $\alpha:G' \to L$ be any homomorphism from $G'$ to arbitrary group $L$ such that $\alpha \circ \varphi = 0$. Then $\ker\alpha \supseteq \mathrm{im}\varphi$.

By Theorem 7.12, to show there exists a unique homomorphism $\bar{\alpha}:G'/N \to L $ such that $\bar{\alpha} \circ \pi = \alpha$, we only need to show $N \subseteq \ker\alpha$ . But $\ker\alpha$ is a normal subgroup of $G'$ cantaining $\mathrm{im}\varphi$, and $N$, by definition, is the smallest subgroup of $G'$ containing $\mathrm{im}\varphi$. Thus $N \subseteq \ker\alpha$, and we're done.