Why is this the right answer to this orbit question?

Question

Here is the question i am curious about:

Exercise 3.

Find the number of orbits in the set $ \{ 1, 2, 3, 4, 5, 6, 7, 8 \} $ under the action of the cyclic subgroup $ \langle (1356) \rangle $ of $ S_8 $. [Recall that $ S_n $ acts on $ [ n ] := \{ 1, 2, \dots, n \} $ by permutation: Each element $ \sigma \in S_n $ is a permutation of $ [ n ] $, i.e. a bijective function $ \sigma : [ n ] \to [ n ] $. This induces the action $ S_n \times [ n ] \to [ n ], (\sigma, k) \mapsto \sigma(k) $.]

I have that the answer is $5$ orbits, one being $\{1,3,5,6\}$ and the rest are singleton sets that are 'fixed'.

Im not sure i even completely understand what the group action actually is. Would someone be able to write it out for me and how we get what the orbits actually are?

I know that the orbit is defined as $Gx=\{g.x|g \in G\}.$ After a group acts on a set do we get an element of the set again?

I know that the orbits must partition the set.

Answer 1

The action of $S_n$ is by permutations. So is the action of a subgroup. $(1356)$ is the permutation that send $1 \mapsto 3, 3\mapsto 5, 5 \mapsto 6, 6 \mapsto 1$ and leaves all other numbers fixed. $\langle(1356)\rangle$ is just the subgroup of $S_n$ generated by this element, which consists of all its powers:

$$ \langle(1356)\rangle = \{ (1356)^n \mid n \in \mathbb{N}_0\} = \{ \text{id},(1356), (15)(36), (1653)\}.$$

You should make yourself comfortable with these notations and facts (they are certainly explained somewhere in your reference).

Now, clearly $\{1,3,5,6\}$ is an orbit under the action of $\langle(1356)\rangle$, because you can send each element to each other element and they never get sent to something outside this set. All other numbers are fixed by all the permutations in $\langle(1356)\rangle$, hence their orbits are just the singletons $\{2\},\{4\},\{7\},\{8\}$.

Answer 2

Let's set: $$G:=\langle(1356)\rangle=\{(1356),(15)(36),(1653),()\}\tag1$$ By definition of orbit under the action of $G$ as group of permutations ("natural"), and $(1)$: \begin{alignat}{1} O(1) &= \{\sigma(1), \sigma\in G\}=\{3,5,6,1\} \\ O(2) &= \{\sigma(2), \sigma\in G\}=\{2\} \\ O(3) &= \{\sigma(3), \sigma\in G\}=\{5,6,1,3\}=O(1) \\ O(4) &= \{\sigma(4), \sigma\in G\}=\{4\} \\ O(5) &= \{\sigma(5), \sigma\in G\}=\{6,1,3,5\}=O(1) \\ O(6) &= \{\sigma(6), \sigma\in G\}=\{1,3,5,6\}=O(1) \\ O(7) &= \{\sigma(7), \sigma\in G\}=\{7\} \\ O(8) &= \{\sigma(8), \sigma\in G\}=\{8\} \\ \end{alignat} Therefore, five orbits overall.