# Why is this the right answer to this orbit 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.

• Do you understand how the group of permutations of $\{1,\dots,8\}$ acts on $\{1,\dots,8\}$? Also, I would rather denote the orbit $\{1,3,5,6\}$ since this is a set. – Clément Guérin Apr 3 '18 at 10:20
• I dont think i do, can youve give an example? – user531499 Apr 3 '18 at 10:21
• Ok, $(1,2)\cdot 2=1$, $(1,2)\cdot 1=2$ and $(1,2)\cdot 3=3$. – Clément Guérin Apr 3 '18 at 10:21
• To answer your question: yes, after a group acts on a set you get back elements of the set. An action of $G$ on $X$ is a map $G \times X \to X$, that sends $g$ and $x$ to the element $g\cdot x$ of $X$. – 57Jimmy Apr 3 '18 at 10:29
• @pullofthemoon Do you know what the notation $(1,2)$ represents? – Arnaud D. Apr 3 '18 at 10:46

## 1 Answer

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 $\{2\},\{4\},\{7\},\{8\}$.

• That is very helpful. I think the root of my problem still is that i don't actually understand what the group action is. For example...Why is $(1,3,5,6).2=2?$ Can you give an explanation of what it is? – user531499 Apr 3 '18 at 10:36
• Writing $(1356)$ is just a shorthand to denote the bijection that sends $1,3,5,6$ in a cycle and leaves all other elements fixed. In the same way, the permutation $(12)(58)(347)$ exchanges the places of $1$ and $2$; of $5$ and $8$; sends $3 \mapsto 4 \mapsto 7 \mapsto 4$; and leaves $6$ fixed. It is just a problem of notation, you just need to work out some examples. Have a look here. (It might be helpful to read the rest of the Wikipedia page before). – 57Jimmy Apr 3 '18 at 10:53
• That makes sense thank you! – user531499 Apr 3 '18 at 11:00