Details on the explanation of why there is no functor $\mathsf{Group}\to\mathsf{Ab}$ sending groups to their centers?

Question

The hint to the problem is to consider homomorphism $S_2 \rightarrow S_3 \rightarrow S_2$

I found this answer by Arturo Magidin here Why is there no functor $\mathsf{Group}\to\mathsf{AbGroup}$ sending groups to their centers?

Consider for example $G=C_2$, $H=S_3$, $K=C_2$, and the maps $f\colon G\to H$ sending the nontrivial element of $G$ to $(1,2)$, and $g\colon H\to K$ by viewing $S_3/A_3$ as the cyclic group of order $2$.

Since $Z(G) = Z(K) = C_2$, and $Z(H) = \{1\}$, such a putative functor $\mathcal{F}$ would give that $\mathcal{F}(f)\colon C_2\to\{1\}$ is the zero map $\mathbf{z}$, and $\mathcal{F}(g)\colon \{1\}\to C_2$ is the inclusion of the trivial group into $C_2$. But $g\circ f=\mathrm{id}_{C_2}$, so $$\mathrm{id}_{C_2} = \mathcal{F}(\mathrm{id}_{C_2}) = \mathcal{F}(gf) = \mathcal{F}(g)\mathcal{F}(f) = \mathbf{z}$$ where $\mathbf{z}\colon C_2\to C_2$ is the zero map.

Thus, no such functor $\mathcal{F}$ can exist.

but there are several things I don't understand. I guess $S_2=C_2$?

1. What does the author mean with "viewing $S_3/A_3$ as the cyclic group of order $2$" ? Can you explain what the element of that quotient are and why and how is that related to the cyclic group of order 2? and why can you use it instead of S_3 as the hint suggests?

2. Is $C_2$ the same as $S_2$?

3. why is the map $C_2 \to \{1\} $ the zero map ? The image is not 0 as in https://mathworld.wolfram.com/ZeroMap.html

4. why is $g\circ f: C_2 \rightarrow C_2 = id_{C_2}$? Just because a map has the same domain and codomain it does not mean it is the identity.

5. why is $Z(H)=\{1\}$, i.e why the other elements of H don't conmute?

Hope someone can clear up these points. Many thanks

Edit:

$ g\circ f:S_2 \rightarrow S_3 \rightarrow S_2$

$(1,2) \mapsto (1,2) \mapsto (1,2)$

$ e \mapsto e \mapsto ?$

$? \mapsto (1,3) \mapsto ? $

$ ? \mapsto (2,3)\mapsto ?$

$ ? \mapsto (1 2 3)\mapsto ?$

$ ? \mapsto (1 3 2)\mapsto ?$

Answer 1

1. Elements of a quotient $G/H$ can be viewed as cosets $gH$. In particular for our case $S_3/A_3$ has 2 elements and so has to be the unique (cyclic) group of order 2.

2. Yes, there is only 1 group of order 2 and $|S_2|=2=|C_2|$

3. The zero map in this context just refers to a map that sends everything to the indentity.

4. You are correct that the domain and codomain being equal does not imply the map is the identity. The screenshot you posted has explicit definitions for the maps $g$ and $f$. I encourage you to write down where each element of $C_2$ gets sent the compositum of both maps to check that $g\circ f$ is indeed the identity.

5. You can check this directly for every element of $H=S_3$..

   For example: $(12)(123)=(23)\neq (13)=(123)(12)$

   You write down similar equations by permuting the indices to show that no non-identity element of $S_3$ is in the center.