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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 $0<m\leq 9$.

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

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    $\begingroup$ 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. $\endgroup$
    – user1007416
    Apr 24, 2022 at 9:40

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Fernando Nazario
    Apr 23, 2022 at 19:46
  • $\begingroup$ The identity group can't be a valid homomorphism kernel, as $6\nmid 10$. Accordingly, the "identity" (map?) is not in the agenda. $\endgroup$
    – user1007416
    Apr 24, 2022 at 9:00

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