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

Question

$\mathbb Z_{5}$ is the group to act on the set $\{ 1,2,3,4,5\}$. In how many ways is that possible? Now $0$ will give the identity map. $1$ will give a bijection in $5!$ ways so will the others and the number of possible bijections being $5!$ the bijections given by $1$ will coincide with those given by others. All get mixed up here.

Answer 1

In this answer, let $C_m$ denote the cyclic group of order $m$ for $m\in\mathbb{Z}_{>0}$. I also write $C_0$ for the infinite cyclic group.

There is one obvious group action: the trivial action. Now, we consider a nontrivial action. Then, there exists an element of $\{1,2,3,4,5\}$, say $1$, not fixed by $C_5$. However, the size of the orbit of $1$ would then be $5$ (since the size of the orbit divides the order of $C_5$ and it is not $1$). Hence, this action is transitive as the orbit must be the whole $\{1,2,3,4,5\}$. Therefore, this action is a permutation of $\{1,2,3,4,5\}$ on a circle (where $1\in C_5$ is rotation by $\frac{2\pi}{5}$). On the other hand, every permutation of $\{1,2,3,4,5\}$ on a circle can be made an action of $C_5$ on this set. There are $(5-1)!=24$ such permutations. Hence, in total, there are $1+24=25$ possible actions of $C_5$ on $\{1,2,3,4,5\}$.

In how many ways can the group $C_6$ act on $\{1,2,3,4,5,6\}$?

Hint: The answer is $1+15+40+45+120+15+40+120=396$.

In general, if $a^m_n$ is the number of ways the group $C_m$ can act on the set $\{1,2,\ldots,n\}$, where $m$ and $n$ are nonnegative integers, then $$\sum_{n=0}^\infty\,a^m_n\,\frac{t^n}{n!}=\exp\left(\sum_{\substack{{d\in\mathbb{Z}_{>0}}\\{d\mid m}}}\,\frac{t^d}{d}\right)\,.$$ In particular, it can be shown that $a^0_n=n!$ for every $n\in\mathbb{Z}_{\geq 0}$, or $$\sum_{n=0}^\infty\,t^n=\sum_{n=0}^\infty\,a^0_n\,\frac{t^n}{n!}=\exp\left(\sum_{\substack{{d\in\mathbb{Z}_{>0}}\\{d\mid 0}}}\,\frac{t^d}{d}\right)=\exp\left(\sum_{d=1}^\infty\,\frac{t^d}{d}\right)=\exp\Bigg(\ln\left(\frac{1}{1-t}\right)\Bigg)=\frac{1}{1-t}\,.$$

Answer 2

An equivalent way of thinking of group actions is as follows: a group $G$ acting on a (finite) set $X$ is exactly a homomorphism $$\theta:G\to S_{|X|}$$ where $S_{|X|}$ is the symmetric group on $|X|$ elements, and $\theta(g)$ is the permutation of $X$ given by $x\mapsto g(x)$. And every such homomorphism gives a group action: define $g(x)$ to be where $\theta(g)$ sends $x$.

How many homomorphisms are there from $\mathbb Z/5\mathbb Z\to S_5$?

$\mathbb Z/5\mathbb Z$ is cyclic, so any homomorphism will be completely determined by where it sends the generator $1$. But $1$ has order $5$ in $\mathbb Z/5\mathbb Z$, so it can be sent only to elements with orders dividing $5$.

So we just need to find the number of elements of orders $1$ and $5$ in $S_5$. Can you finish from here?

Answer 3

Since the stabilizers must have order either $1$ or $5$, the orbits must have accordingly size either $5$ or $1$. Therefore, the only allowed orbit equations are $5=1+1+1+1+1$ and $5=5$. The former corresponds to the trivial action (the kernel is the whole $\Bbb Z_5$). The latter corresponds to a (transitive and) faithful action, being all the stabilizers trivial; so it is equivalent to an embedding of $\Bbb Z_5$ into $S_5$. Every (cyclic) subgroup of order $5$ of $S_5$, say $H_i,$ is an image of $\Bbb Z_5$ under such an embedding and, for each $H_i(\cong C_5)$, there are $4$ distinct isomorphisms $\varphi_{ij}\colon \Bbb Z_5\to H_i$, $j=1,2,3,4$ (recall that for a prime $p$, $\operatorname{Aut}(C_p)\cong C_{p-1}$). So we are left with searching for how many $H_i$ are there; now, in general there are ${n!\over k(n-k)!}$ $k$-cycles in $S_n$, and hence $24=(5-1)!$ $5$-cycles in $S_5$; each of them lies in one and only one $H_i$, which in turn are pairwise with trivial intersection; therefore, there are six $H_i$and overall $6\cdot4=24$ nontrivial homomorphisms $\varphi\colon \Bbb Z_5\to S_5$ or, equivalently, as many nontrivial actions of $\Bbb Z_5$ on the set $\{1,2,3,4,5\}$.