# In how many ways can $\mathbb{Z_5}$ act on $\{1,2,3,4,5\}$

In how many ways can $$\mathbb{Z_5}$$ act on $$\{1,2,3,4,5\}$$?

I could figure out if $$\mathbb{Z_5}$$ acts on a set $$\{1,2,3,4,5\}$$ the orbit $$\theta_i$$ is either $$1$$ (the tribial action) or $$5$$.

Now we see that if the action is transitive then the shouldnt it correspond to $$5$$ cycles of the form $$(12345)^i$$ where $$1 \le i \le 4$$?

So since the homomorphism $$\mathbb{Z_5} \to S_5$$ is completely determined by $$1_5$$, it can send the first symbol to any one of $$4$$ choices the second symbol to any one of the $$3$$ choices and so on. So there are $$4!$$ choices.

Theres an answer to the question in the website whoch I was not able to understand. Is this ok?

Also I think we will get the same answer if we replace $$\mathbb{Z}_m$$ by any other cyclic group, say $$\mathbb{Z}$$.

A group action $$\phi$$ is homomorphism from a group $$G$$ to the symmetric group of a set $$X$$ i.e. $$\phi: G \rightarrow s y m(X)$$ is called group action of group $$G$$ to the set $$X$$. We have $$G=\mathbb{Z}_{5}$$ and $$X=\{1,2,3,4,5\}$$, Sym $$(X)=S_{5}$$ So, we have to find number of homomorphisms from $$\mathbb{Z}_{5}$$ to $$S_{5}$$. One is trivial homomorphism in which all elements goes to identity. For others, $$\mathbb{Z}_{5}=\{0,1,2,3,4\} ; O(1)=O(2)=O(3)=O(4)=5$$ We have a result if $$\phi: G \rightarrow G_{1}$$ is a homomorphism, then $$O(\phi(x))$$ will divide $$O(x)$$ So $$O(\phi(1))|O(1) \Rightarrow O(\phi(1))| 5$$ Now, all 5 cycles in $$S_{5}$$ have order 5 only. Number of 5 cycles in $$S_{5}$$ is 24 So $$\phi$$ (1) has 24 choices. $$\Rightarrow 24$$ are the only choices and one is trivial. $$\Rightarrow$$ Total 25 number of homomorphisms are possible.
You can't send $$1$$ to any element, you can only send it to elements whose oreder divide $$5$$ as you have already noted, because if $$\varphi$$ is a homomorphism to $$S_5$$ then $$\varphi(1+1+1+1+1) = \varphi(1)^5 = \varphi(0) = e$$
Now if we want the action to be transitive it is enough for it to be non trivial. So the orbit will be any $$5$$ cycle, WLOG we can assume the cycle representation to start from $$1$$, thus $$\varphi(1) = (1 a_1 a_2 a_3 a_4)$$ now $$a_i$$ can be arrarnged in any way so there will be $$4!$$ ways
• $\theta(1_5)$ will be mapped to an element of order $5$ in $S_5$ and also we want the action to e transitive.i am not able to understand how are we doing it? May 6, 2022 at 4:52
• There are $4$ elements of order $5$ in $S_5$. The action will be transitive if we send it to an element of the form $(12345)^i$? How is the answer $24$? May 6, 2022 at 4:53
• I think the conjugates of $(12345)$ are the elements of order $5$ not only its powers, and there are $4!$ conjugates May 6, 2022 at 4:55