# Number of the group homomorphisms $G\to S_3$, where $|G|=8$

I've thought of the following solution to this assignment: find the number of the group homomorphisms $$G\to S_3$$, where $$|G|=8$$.

Let's call $$\mathbb{h}$$ the sought number. Any such a homomorphism is equivalent to an action of $$G$$ on $$X:=\{1,2,3\}$$. The only allowed orbit equations are $$3=1+1+1$$ and $$3=1+2$$, because $$3=3$$ is ruled out by the orbit-stabilizer theorem and the fact that $$3\nmid 8$$. The former corresponds to the trivial homomorphism, so it counts one. The latter corresponds to two stabilizers of order $$4$$, hence normal (index $$2$$), hence equal (as they are conjugate); therefore, each subgroup of order $$4$$ of $$G$$ gives rise to one such an action, or rather to three, considered that the singleton's role ("$$3=1+\dots$$") can be played by any of the three elements of $$X$$. So, said $$\mathbb{n}_4$$ the number of subgroups of order $$4$$ of $$G$$, we'd have: $$\mathbb{h}=3\mathbb{n}_4+1 \tag 1$$ Unless it is a coincidence, $$(1)$$ would account for the well-known four homomorphisms $$C_8\to S_3$$ (where $$\mathbb{n}_4=1$$) and the ten $$D_4\to S_3$$ (where $$\mathbb{n}_4=3$$).

Is my argument correct and then $$(1)$$ valid for every $$G$$ of order $$8$$?

Edit. I think that the argument and $$(1)$$ work. In fact, the subgroups of order $$4$$ of $$G$$ are the only proper, normal (index $$2$$) subgroups suitable as kernels of nontrivial homomorphisms $$\phi\colon G\to S_3$$ (kernels of order $$2$$ would lead to $$|\phi(G)|=4(\nmid 6)$$ and the trivial kernel to $$|\phi(G)|=8(> 6)$$). For each kernel of order $$4$$, say $$H_4$$, we have that the elements of $$G\setminus H_4$$ have order either $$2$$ or $$4$$ or $$8$$; therefore, they must be mapped to one same $$2$$-cycle of $$S_3$$; in fact, they cannot be mapped to any $$3$$-cycle (as $$3\nmid 2,4,8$$), nor to different $$2$$-cycles only (as $$\phi(G)$$ wouldn't be a subgroup of $$S_3$$). So, for each $$H_4$$, there are three homomorphisms, each mapping all the elements of $$G\setminus H_4$$ to a given $$(ij)$$.

• A question should be written in such a way that it can be understood even by someone who did not read the title. Commented Sep 27, 2021 at 7:30
• I haven't checked it, but in principle, there's only five groups of order 8 to check... Commented Sep 27, 2021 at 7:54
• And most of them have been already checked at this site, e.g. here for $(1)$. Commented Sep 27, 2021 at 8:59
• You have the quaternion group, dihedral group, cyclic group $C_8$, $C_4 \times C_2$, and $C_2 \times C_2 \times C_2$. Which ones do you still need to check? Commented Sep 27, 2021 at 13:47
• @GeoffreyTrang, $Q_8$, $C_4\times C_2$ and $C_2\times C_2\times C_2$. And in case they all would fit with $(1)$ (as $C_8$ and $D_4$ do), I'd like to know whether this happens because $(1)$ is correctly gotten, or it's just a lucky strike as my argument is incorrect.
– user943729
Commented Sep 27, 2021 at 14:01

## 1 Answer

I think your argument is basically correct.

More generally: Let $$G$$ and $$H$$ be groups.

The number of homomorphisms $$\psi: G \rightarrow H$$ with $$\operatorname{Im} \psi \cong C_2$$ is equal to $$tk$$, where $$t$$ is the number of elements of order $$2$$ in $$H$$, and $$k$$ is the number of non-trivial homomorphisms $$G \rightarrow C_2$$.

(In your case $$t = 3$$.)

For non-trivial homomorphisms $$\psi, \psi': G \rightarrow C_2$$ you have $$\psi = \psi'$$ if and only if $$\operatorname{Ker} \psi = \operatorname{Ker} \psi'$$. So $$k$$ is equal to the number of subgroups of index $$2$$ in $$G$$. (Using the notation in your case, $$k = n_4$$.)

What remains is for you to calculate the number of subgroups of index $$2$$ for the groups of order $$8$$ (of which there are five). This you will need to do case-by-case.