Homomorphisms from $S_3$ to $\mathbb{Z}/10\mathbb{Z}$

I want to check if my line of thought is correct.

We need to find all homomorphisms $$\phi: G=S_3\rightarrow H=\mathbb{Z}/10\mathbb{Z}$$. We already know that $$\phi(g)=\bar{0}$$ for all $$g\in G$$ is a possible homomorphism, so we can assume that $$\phi(g)=\bar{m}$$ for some $$0.

Now, if $$n$$ is the order of $$g$$, $$n=o(g)$$, we have that $$n\bar{m}=\bar{0}$$ so that $$o(\bar{m})\mid n$$ and, therefore, $$o(\bar{m})$$ is a common divisor of $$10$$ and $$6$$, hence either $$1$$ or $$2$$. Since we are excluding $$m=0$$, we get that $$m=5$$, that is, if $$\phi$$ is a non-trivial homomorphism from $$G$$ to $$H$$ we must carry some non-trivial element of $$G$$ to the element $$\bar{5}$$ in $$H$$.

Also, since $$o(\bar{m})\mid n$$ we must have that $$n$$ is an even number, and since $$n\mid 6$$ we can only have $$n=2$$. The only elements in $$S_3$$ with order $$2$$ are the tranpositions $$(12),(13),(23)$$. Hence, a non-trivial homomorphism must take some transposition to $$\bar{5}$$. Now, it is easy to check that the multiplication of two tranpositions, $$\tau_1\neq\tau_2$$ is a $$3\text{-cycle}$$. Hence:

$$\bar{0}=\phi(3\text{-cycle})=\phi(\tau_1\tau_2)=\phi(\tau_1)+\phi(\tau_2)$$

The only way to make this equation work, since $$\phi(\tau)=\bar{0},\bar{5}$$ is that both $$\phi(\tau_1)=\phi(\tau_2)=\bar{5}$$, hence if $$\phi$$ is a non-trivial homomorphism we must have that $$\phi(\tau)=\bar{5}$$ for all transpositions $$\tau\in S_3$$ and $$\phi(g)=\bar{0}$$ if $$g$$ is not a tranposition. It is easy to check that this is in fact a homomorphism, and because of the necessity of this conditions, we must have that $$\text{Hom}(G,H)=\{\text{ trivial },\phi\}$$.

Is there any flaw with my logic? Did I get all possible homomorphisms?

Thanks for any help

• Your logic seems flawless to me. Shortly: the nonidentity elements of $S_3$ have order either $2$ or $3$. A nontrivial homomorphism must send necessarily the former ones to $\bar 5$ (the only nonidentity element whose order divides $2$, i.e. is $2$) and latter ones to $\bar 0$ (as there are no nonidentity elements whose order divides $3$, i.e. is $3$). Very appropriate your note about checking this is indeed a homomorphism, as we are dealing with just a necessary condition.
– user1007416
Commented Apr 24, 2022 at 9:40

Your answer is certainly right. However, I would suggest that it's much easier to think about the normal subgroups of $$S_3$$. There are only three conjugacy classes, after all! We conclude that, besides the identity group and the whole of $$S_3$$, the only normal subgroup of $$S_3$$ is generated by three-cycles. This means that, besides the identity and the zero map, there is at most one homomorphism from $$S_3$$ into any other group $$H$$, up to automorphisms of $$H$$. $$\mathbb{Z}_{10}$$ has a unique element of degree 2, so in this case there is just one homomorphism.

• That's a great way to unravel de problem! Thank you! Just one question, when you mean "up to automorphisms" you mean that, in the case of $S_3$, there are "two" homomorphisms, besides ways of "shuffle" elements in $H$ in a way that preserves the group structure? Commented Apr 23, 2022 at 19:46
• The identity group can't be a valid homomorphism kernel, as $6\nmid 10$. Accordingly, the "identity" (map?) is not in the agenda.
– user1007416
Commented Apr 24, 2022 at 9:00