TITLE: In how many ways can the group $\mathbb Z_5$ act on the set $\{1,2,3,4,5\}$ $?$
QUESTION [4 upvotes]: $\mathbb Z_{5}$ 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.

REPLY [0 votes]: 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 $\{1,2,3,4,5\}$.
{"set_name": "stack_exchange", "score": 4, "question_id": 1383251}