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So I've got to find all epimorphisms $C_3\times S_3 \rightarrow C_6$.

One idea I have is to find all normal subgroups of $C_3\times S_3$. Each such group can be the kernel of my epimorphism. Since $|C_3\times S_3| = 18$ and $|C_6| = 6$ and $C_3\times S_3/\ker\phi \cong C_6$, I'm looking for normal subgoups of order $3$. One such normal subgroup is $N_1 = \langle(1, id)\rangle$, and the other is $N_2 = {0}\times A_3$. But then $C_3\times S_3 / N_1$ is not cyclic, so we discard it.

Is this the right approach? I'm wondering if these are all the normal subgroups and if this indeed is the right way to proceed here.

jrg
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  • Yours feels like an ok way. In general it may not be easy to list all the normal subgroups. An alternative way to proceed might begin with the observation that $C_6$ is abelian. This implies that every commutator in $C_3\times S_3$ must be in the kernel. Are you familiar with commutators, the derived subgroups and such? – Jyrki Lahtonen Dec 01 '24 at 21:07
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    "Epimorphism" means "right cancellable morphism". Although in the category of all groups (resp. finite groups), being surjective is equivalent to being an epimorphism this is far from obvious, and the two terms should not be carelessly used interchangeably. – Arturo Magidin Dec 01 '24 at 21:23
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    It should be noted that some authors use this language irrespective of their category theory meaning. I learned these terms in my introductory course that was very basic. I would not get onto OP for this when it may not be their doing. – Cameron L. Williams Dec 02 '24 at 06:00
  • I wasn't aware of the deeper meaning of 'epimorphism', the definition in my algebra course was 'surjective homomorphism', thanks for pointing that out! – jrg Dec 02 '24 at 09:48
  • @jrg there's no deeper meaning – suckling pig Dec 06 '24 at 19:43
  • We call epimorphisms epimorphisms because they're epic. Someone got some type of a flunk in mathematics, and thus has his own ridiculous version. – suckling pig Jan 06 '25 at 20:12

3 Answers3

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Any morphism from $C_3\times S_3$ to $C_6$ restricts to a morphism from $\{e\}\times S_3$ to $C_6$. Because $S_3$ is not abelian, but $C_6$ is, this morphism is not injective (if you know about abelianizations, you can use them here to simplify). The only normal subgroups of $S_3$ are $\{e\}$, $A_3$, and $S_3$.

If the kernel of the morphism i to $C_6$ were to contain $S_3$, then the image would have cardinality at most $3$, which means it is not surjective. So any surjective map has kernel containing $\{e\}\times A_3$. This has to be exactly the kernel, because of order considerations.

That means a surjective morphism is completely determined by what happens to the generator of $C_3$, which must be an element of order $3$ in $C_6$ (there are two choices), and what happens to the elements of order $2$ in $S_3$: they all must map to the same element of order $2$ (and there is only one choice).

So there are at most two possibilities. Now check whether they give morphisms or not (if the image contains elements of order $2$ and of $3$, then it is surjective).

Arturo Magidin
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That approach looks good to me for this particular instance, but in general it can be difficult to list normal subgroups. There's another technique that we can apply here that takes advantage of the fact that the target of the homomorphism is abelian. (I see that Jyrki Lahtonen pointed out the same in the comments.)

It follows immediately from the definitions for any abelian group $A$ and group homomorphism $\varphi: G \to A$, we have $$\varphi(ghg^{-1}h^{-1}) = 1_A$$ for all $g, h \in G$, i.e., the commutator subgroup $[G, G]$ of $G$ satisfies $[G, G] \subseteq \ker \varphi$.

Now, what is the commutator subgroup of $C_3 \times S_3$?

The commutator subgroup of $C_3 \times S_3$ is $\{0\} \times A_3$, which already has $3$ elements, hence $\ker \varphi = \{0\} \times A_3$.

Remark In fact, the above argument implies that any homomorphism $\phi: G \to A$ factors through a homomorphism $\tilde \varphi : G / [G, G] \to A$, i.e., has the form $\varphi = \tilde\varphi \circ \pi$, where $\pi$ is the canonical quotient map $G \to G / [G, G]$. (By construction, $G / [G, G]$ is abelian, and we call it the abelianization of $G$.) Thus, we can classify all homomorphisms $G \to A$ by classifying the homomorphisms $G / [G, G] \to A$ of abelian groups and precomposing them with $\pi$.

Travis Willse
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The only way a homomorphism $h$ can be surjective is if the kernel has order $3.$

The commutator is $A_3.$ Therefore it's the kernel, since the image is abelian.

So there's only the canonical projection (aka abelianization). And, since there's a non-trivial automorphism of $C_6,$ one more.

suckling pig
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