Proving $S_3$ is not a coproduct of cyclic groups

Question

Aluffi II.6.11 suggests proving the above.

Here's the sketch of my proof by contradiction.

Assume that $S_3$ is a coproduct of a family $\mathcal{C}$ of cyclic groups $C^i$. Using the universal property for coproducts (and considering morphisms $\sigma_i : S_3 \rightarrow C^i$ such that $\sigma_i \iota_i = 1$, where $\iota_i$ is the injection function) shows that each $C^i$ cannot be larger (as a set) than $S_3$.

Next, since $S_3$ has three elements of order 2 and two elements of order 3, it can be shown that any cyclic group that maps its generator onto one of those elements shall have order 2 or 3 respectively.

So we're down to a corpoduct of a certain number of $C_2$ and $C_3$.

Now, by considering both elements of order 3 in $S_3$, it can be shown that having two $C_3 \in \mathcal{C}$ with different injection functions leads to contradiction, so we have at most one $C_3$. We also have to have at least one $C_2$ mapping onto some element of order 2, otherwise there's a certain freedom in defining the behavior of some morphisms from $S_3$. Similarly, we also have to have at least one $C_3$.

Now, considering the group $C_2 \times C_3$ along with a pair of morphisms $\varphi_2 : C_2 \rightarrow C_2 \times C_3, \varphi_2([n]_2) = ([n]_2, [0]_3)$ and $\varphi_3 : C_3 \rightarrow C_2 \times C_3, \varphi_3([n]_3) = ([0]_2, [n]_3)$ it can be shown that there is no valid homomorphism $\sigma : S_3 \rightarrow C_2 \times C_3$ satisfying the corresponding universal property for coproducts, hence the contradiction with the original assumption.

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Overall, this looks quite clunky. Does it look reasonable though? Is there a better way to prove the claim (perhaps limited to the little amount of algebra and category theory that might be expected by this point)?

Answer 1

The problem states

“Since direct sums are coproducts in Ab, the classification theorem for abelian groups mentioned in the text says that every finitely generated 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 this is not the case, by showing that $S_3$ is not a coproduct of cyclic groups.”

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I think that your approach is a reasonable way of going about it given what information you have available. Certainly if $C_n$ is a factor of the coproduct then $n\in\{2,3\}$ as you mentioned, since these are the only nontrivial orders of elements in $S_3$.

I think the simplest way to move forward from here is to note that there should be a map $f:S_3\to C_3$ such that

$$C_3 \xrightarrow{i} S_3 \xrightarrow{f} C_3$$

is the identity map, if we assume that $S_3$ is a coproduct with a $i:C_3\to S_3$ factor. You can then reach a contradiction by showing that no such $f$ exists.

Answer 2

I have a much simpler way of contemplating this, but as a beginner, it might be flawed, so please let me know if it's not reasonable:

By the universal property of coproduct, there is a unique homomorphism from the direct sum of cyclic groups to S3. Now if we prove that the homomorphism has a inverse (i.e. that the homomorphism is actually a isomorphism), we are done.

To establish isomorphism, we need same number of elements (6), which constrains that C2*C3 the only possibility, because 2*3=6. Then, comparing the order of the 6 elements in both groups, we found out they are not isomorphic to each other.

This seems too simple a solution, so I doubt my thoughts here. Much appreciated for any comments.

Edit: This not the correct way to think, thanks to the comment.

Answer 3

Here is a maybe simpler proof, but which involves more concepts that may be unknown.

$\newcommand{\Z}{\mathbb{Z}}$Suppose that a group $G$ is the coproduct of cyclic groups: $$G = \Z/n_1\Z * \dots * \Z/n_k\Z.$$ Then the abelianization of $G$, i.e. its quotient by its commutator subgroup, is isomorphic to the direct sum $G_{\mathrm{ab}} \cong \Z/n_1\Z \oplus \dots \oplus \Z/n_k\Z$. So if we know the abelianization, we know what cyclic groups must be involved.

Now it's well-known that the commutator subgroup of $S_3$ is the alternating group $A_3$, and the quotient $(S_3)_{\mathrm{ab}}$ is the cyclic group $\Z/2\Z$. Therefore, if $S_3$ were a coproduct of cyclic groups, then it would be isomorphic to $\Z/2\Z$. This is obviously false because $S_3$ has $6$ elements whereas $\Z/2\Z$ only has two. Therefore $S_3$ cannot be isomorphic to a coproduct of cyclic groups.