# Find the number of group homomorphisms from $\mathbb{Z}_8$ to $S_3$.

Find the number of group homomorphisms from $$\mathbb{Z}_8$$ to $$S_3$$.

My try:

I know $$\mathbb{Z}_8$$ is a cyclic group of order $$8$$, and that the number of group homomorphisms from a cyclic group to any other group is determined where the generators of the cyclic group get mapped.

It is easy to notice that the generators of $$\mathbb{Z}_8$$ are

$$\langle 1\rangle,\langle 3\rangle,\langle 5\rangle$$ and $$\langle 7\rangle$$.

I also know that the order of the image divides the order of $$S_3$$ (as it forms a subgroup of $$S_3$$). Also if $$f$$ be the homomorphism, then $$f(0)= (1)(2)(3)$$. So till now I got $$f(0)=(1)(2)(3)$$, $$|f(2)| \mid 6$$, $$|f(3)| \mid 6$$, $$|f(4)| \mid 6$$, $$|f(5)| \mid 6$$, $$|f(6)| \mid 6$$, $$|f(7)| \mid 6$$, $$|f(8)| \mid 6$$. So the orders of $$|f(i)|$$ can be $$1$$ or $$2$$ or $$3$$ or $$6$$.

But how can I proceed from here?

• Do you know Lagrange's theorem? Commented Sep 20, 2021 at 14:51

Let $$\varphi\colon\Bbb Z_8\longrightarrow S_3$$ be a homomorphism. Since $$1$$ has order $$8$$, the order of $$\varphi(1)$$ must divide $$8$$. That is, it must be $$1$$, $$2$$, $$4$$ or $$8$$. But $$S_3$$ has no element whose order is $$4$$ or $$8$$, and the only $$\varphi$$ such that $$\varphi(1)=\operatorname{Id}$$ is the trivial homomorphism. Finally, $$S_3$$ has exactly three elements whose order is $$2$$ (the three transpositions), and if $$\tau$$ is any of them you can define $$\varphi(n)=\tau^n$$; it is a group homomorphism and it maps $$1$$ into $$\tau$$.
Therefore, there are $$4$$ such homomorphisms.
Suppose that $$\phi:\mathbb{Z}_8\to S_3$$ is a homomorphism then $$\phi(\mathbb{Z}_8)\leq S_3$$ using properties of homomorphism $$\phi(\mathbb{Z}_8)$$ is cyclic group $$\phi(Z_8)\approx\mathbb{Z}_3$$ or $$\mathbb{Z}_2$$ Since $$|\phi(\mathbb{Z_8})| \;divides \;|\mathbb{Z}_8|=8$$ $$\Rightarrow \phi(\mathbb{Z}_8)\approx \mathbb{Z}_2$$ there are 3 element of order 2 in $$S_3$$ so we have 3 homomorphism and one trivial homomorphism=4 homomorphism are there
The subgroups of $$\Bbb Z_8=\{0,1,2,3,4,5,6,7\}$$ are: \begin{alignat}{1} &H_1:=\{0\} \\ &H_2:=\{0,4\} \\ &H_3:=\{0,2,4,6\} \\ &H_4:=\Bbb Z_8 \\ \end{alignat} and they are all normal, as $$\Bbb Z_8$$ is cyclic and hence Abelian. The subgroups $$H_1$$ and $$H_2$$ can't be kernels of homomorphisms $$\Bbb Z_8\to S_3$$, because the respective images would have order $$8(>6)$$ and $$4(\nmid 6)$$. The subgroup $$H_3$$ is the kernel of three distinct homomorphisms: they map the elements of $$H_3$$ to $$()$$ (by definition of kernel), and the other elements (all of order $$8$$) to one same $$2$$-cycle: in fact, they can't be mapped to $$3$$-cycles (because $$3\nmid 8$$), nor to different $$2$$-cycles (because the image set wouldn't be a subgroup of $$S_3$$). So, for $$H_3$$ as kernel we have the three homomorphisms: $$0\mapsto (), 1\mapsto(ij), 2\mapsto (), 3\mapsto(ij), 4\mapsto (), 5\mapsto(ij), 6\mapsto (), 7\mapsto(ij)$$ The case of kernel equal to $$H_4$$, i.e. to the whole $$\Bbb Z_8$$, corresponds to the trivial homomorphism, where all the elements are mapped to $$()$$. So, there are overall four homomorphisms $$\Bbb Z_8\to S_3$$.