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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 ?$

some_math_guy
  • 3,230
  • Clearly $|S_3/A_3|=2$. There is only one group of order $2$, so $S_3/A_3\cong C_2$. 2. Yes, again, $|S_2|=2$. There is only one group of order $2$. And so on.
    • – Dietrich Burde Sep 19 '23 at 13:51
  • @DietrichBurde 1) The hint said to consider $S_3$, not $S_3/A_3$, so is this an alternative solution or why is it legal to do so? 2) for higher n, $S_n$ and $C_n$ are different things right? – some_math_guy Sep 19 '23 at 13:57
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    You know that $|S_n|=n!$ and $|C_n|=n$. This cannot be equal for $n\ge 3$. Of course $Z(S_3)=1$, which has been shown here already - see for example this duplicate. – Dietrich Burde Sep 19 '23 at 13:59
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  • $A_n$ is a subgroup of index $2$ in $S_n,$ so $S_n/A_n\cong C_2.$ And $g:H=S_3\to S_3/A_3\cong K$ is the canonical projection. 2. $S_2\cong C_2$ by definition of $S_n.$ 3. Any morphism to the trivial group (the group with a single element, the identity element, which is often denoted by $e,1,0,\dots$) is "the zero map". 4. Look at how $f,g$ were defined. 5. Work a little.
    • – Anne Bauval Sep 19 '23 at 14:03
  • @AnneBauval The definition of g is $g: S_3\rightarrow C_2$ by viewing " $S_3/A_3$ as the cyclic group" I don't understand why is the autor writes $S_3$ as domain but then in an indirect manner implies that he wants $S_3/A_3$ as domain. From the words it looks to me that they are not defining $g:S_3 \rightarrow C_2$ but $g:S_3/A_3 \rightarrow C_2$ or better $g:C_2 \rightarrow C_2$. Otherwise I cannot compose, if I don't have but 2 elements in the set in the center, while $S_3$ has originally 6 – some_math_guy Sep 19 '23 at 14:49
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    Don't use images to quote material from this site! You shouldn't use images to convey mathematical information anyway. Using it to quote from this site is downright lazy. You are quoting from my answer here. Quoting without attribution. And see here about using images. – Arturo Magidin Sep 19 '23 at 14:51
  • As told in my previous comment: $g:S_3\to S_3/A_3\cong C_2.$ – Anne Bauval Sep 19 '23 at 14:53
  • @ArturoMagidin Sorry I forgot to put the link of the question,and I did say I found this answer, I don't see what is the problem – some_math_guy Sep 19 '23 at 14:53
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    The problems are: (i) you are using images, when you shouldn't. (ii) You are quoting an answer without attribution, when you shouldn't. (iii) You are referencing material in this site without linking, when you shouldn't. Pretty much everything is the problem here. – Arturo Magidin Sep 19 '23 at 14:54
  • Regarding whether $C_2=S_2$, since you are using the category-theory tag, you are probably familiar with the fact that the most important equivalence relation in any category is isomorphism. For this problem, it does not matter whether $C_2=S_2$ or whether $C_2$ and $S_2$ are the *same* (a 4-letter word which should be banned in any mathematical discussion). You should not get hung up on that issue. What matters is that $C_2$ and $S_2$ are isomorphic. – Lee Mosher Sep 19 '23 at 15:09
  • @LeeMosher Could you please explain the following: The definition of g is $g: S_3\rightarrow C_2$ by viewing " $S_3/A_3$ as the cyclic group" I don't understand why is the autor writes $S_3$ as domain but then in an indirect manner implies that he wants $S_3/A_3$ as domain. From the words it looks to me that they are not defining $g:S_3 \rightarrow C_2$ but $g:S_3/A_3 \rightarrow C_2$ or better $g:C_2 \rightarrow C_2$. I cannot compose if the domain is $g\circ f$ , if I don't have but 2 elements in the set in the center which is ok when I use $S_3/A_3$ or $C_2$, while $S_3$ has originally 6 – some_math_guy Sep 19 '23 at 15:12
  • it looks like a contradiction to say both things thag g is define with domain g, and then that we defining viewing it as the cyclic group $C_2$ – some_math_guy Sep 19 '23 at 15:14
  • Again, do not get hung up on whether one can "view $S_3/A_3$ as the cyclic group" $C_2$. Instead, use category theoretic language to translate that stement into a rigorously true fact: the group $S_3/A_3$ is isomorphic to the group $C_2$. Even more generally, it is a fact that any two groups of order $2$ are isomorphic to each other. If that fact is not familiar to you, then that fact seems to be what your question is really about. Can you prove that fact? – Lee Mosher Sep 19 '23 at 15:15
  • The author (me) writes it that way because people who are familiar with basic group theory will understand. There is a canonical map from $S_3$ to $S_3/A_3$; and this quotient is isomorphic to $C_2$ via a unique isomorphism. Composing the canonical map with the unique isomorphism gives you a map from $S_3$ to $C_2$ by "vieweing $S_3/A_3$ as the cyclic group of order $2$." – Arturo Magidin Sep 19 '23 at 15:16
  • And, by the way, your question 3 is another example not to get hung up on: any two groups of order 1 are isomorphic to each other; for example, the additive group ${0}$ and the multiplicative group ${1}$. – Lee Mosher Sep 19 '23 at 15:24
  • I think I already got it. The problem started considering : $g \circ f$ as a composition of 2 functions $f:C_2 \rightarrow S_3$ and $S_3 \rightarrow C_2$, so: $g \circ f:C_2 \rightarrow S_3 \rightarrow C_2 $ but then a third arrow was added, or better $g$ was written as$ g = i \circ \pi$, with $i$ the isomorphism and $\pi$ the projection to the quotient so we actually had $g\circ f = i \circ \pi \circ f:C_2 \rightarrow S_3 \rightarrow S_3/A_3 \rightarrow C_2 $ so my problem was that I didn't see it like that, but just as as someone had erased $S_3$ and put instead $S_3/A_3$, ... – some_math_guy Sep 19 '23 at 15:34
  • so it didn't fit in the flow of the solution. It might be clear for experience people, but these details are important for beginners – some_math_guy Sep 19 '23 at 15:35
  • You are talking about functors, so you are talking about categories. The trivial group is a zero object in both $\mathsf{Grp}$ and $\mathsf{Ab}$. Morphisms to, from, and through the zero object are called "zero morphisms" or "zero maps. Hence 3. – Arturo Magidin Sep 19 '23 at 17:27
  • @ArturoMagidin I see, thank you Sorry, I am still thinking on 1), I am trying to make sense of the composition $g\circ f$ being well- defined where $f: S_2 \rightarrow S_3 $ and $g: S_3 \rightarrow S_2 $, by seeing which elements map to which elements (see Edit at the end of the post). Since $S_3$ has 6 elements (before the quotienting), it's not possible to compose them, unless the domain of $g$ gets restricted to the image of f, but then g doesn't really have codomain the whole group $S_3$ or the first function is multivalued, then it's not a function. – some_math_guy Sep 19 '23 at 17:57
  • of course there will be no problem if the maps where $f: S_2 \rightarrow S_3/A_3 $ and $g: S_3/A_3 \rightarrow S_2 $ from the beginning, but doesn't work to get the centers we want, because then the center of $S_3/A_3$ would homomorphic to $S_2$ instead of the center of $S_3$ being {1} as we want – some_math_guy Sep 19 '23 at 18:02
  • "Since $S_3$ has 6 elements [...] it's not possible to compose them, unless the domain of $g$ gets restricted to the image of $f$". That is absolute, complete, and utter nonsense. You can compose any two group morphisms whenever the codomain of the first equals the domain of the second (or, for that matter, a subgroup of the domain of the second). The real problem is that you area apparently completely ignorant about what $g$ actually is. What you wrote at the end of the post demonstrates this. Even though I spelled out that second map already. – Arturo Magidin Sep 19 '23 at 18:08
  • @MarianoSuárez-Álvarez It's been years since group theory I forgot a lot and not used it at all, plus it was a very reduced syllabus actually, never saw the center for instance or the genrator of that quotient – some_math_guy Sep 19 '23 at 20:05