I am doing Aluffi's problem II.6.11 which states that:
Since direct sums are coproducts in Ab, the classifications theorem for abelian groups mentioned in the text says that every finitely generated \emph{abelian group} is a coproduct of cyclic groups in Ab. The reader may be tempted to conjecture that every finitely generated group is a coproduct in Grp. Show that this is not the case by proving that $S_3$ is not a coproduct of cyclic groups.
(Just like Proving $S_3$ is not a coproduct of cyclic groups.)
I got to do a great part of the problem: I reduced it to proving that $C_2$ can't be a factor in such coproduct.
If someone wants to see, I did it in the following way:
Suppose that $S_3$ satisfies the universal property of a coproduct of cyclic groups. That is, there exist morphisms $i_j$ from some cyclic group (lets denote it by $C^j$) to $S_3$. By the universal property, it exists a unique morphism $\sigma:S_3\to C^j$ such that $\text{id}_{C^j}=\sigma i_j$. The fact that $\text{id}_{C^j}$ is surjective then implies that so is $\sigma$. That is, $|C^j|\leq 6$ for all $j$.
By Proposition 4.1, for all $g\in S_3$, the order of $\sigma(g)$ divides $|g|$. Since $S_3$ has elements of order $1$, $2$, and $3$, the order of $\sigma(g)$ can only be one of those $3$ numbers. As $\sigma$ is surjective, it follows that $C^j$ is either equal to $C_2$ or $C_3$.
Now, $S_3$ being a coproduct of some number of $C_2$'s and $C_3$'s implies that there is a injective morphism $i$ and surjective morphism $\varphi$ such that
$$C_3\overset{i}{\to}S_3\overset{\varphi}{\to}C_3$$
is equal to the identity. But there is no surjective morphism $S_3\to C_3$. In fact, recall that $S_3=\{e,x,y,xy,y^2,xy^2\}$ is the group generated by $x,y$ such that $x^2=e$, $y^3=e$ and $yx=xy^2$. By Proposition 4.1, the order of $\varphi(x)$ divides $2$. Since $C_3$ has no element of order $2$, $\varphi(x)=0$. Similarly, $\varphi(xy)=\varphi(xy^2)=0$. It follows that $\varphi(y)=\varphi(x\cdot xy)=0$ and thus $\varphi$ is not surjective.
Now, the same method that I utilised to prove that $C_3$ can't be a factor doesn't work for $C_2$ since it does exist a injective morphism $i$ and a surjective morphism $\varphi$ such that
$$C_2\overset{i}{\to}S_3\overset{\varphi}{\to}C_2$$
is the identity.
How should I finish the proof, knowing that at this point the reader knows very little about category theory and nothing about free products (which make the problem trivial anyway)?
Obs: The linked post surely does not answer my question.
Best Answer
A handy trick for proofs like this with finite structures is to count homomorphisms. For instance, notice that there are $4$ homomorphisms $C_2\to S_3$ (and one homomorphism $C_1\to S_3$), so if $S_3$ were a coproduct of cyclic groups of order $2$ (and trivial groups) then the number of homomorphisms $S_3\to S_3$ would be a power of $4$. But there are $10$ homomorphisms $S_3\to S_3$ (the trivial homomorphism, the quotient $S_3\to C_2$ followed by any of the $3$ injections $C_2\to S_3$, and the $6$ inner automorphisms).