Is there an action of $S_2$ on $\{1,2,3,4,5\}$ that has exactly two orbits?

Question

I am studying for an algebra midterm and I am trying to solve the following problem: Is there an action of $S_2$ on $\{1,2,3,4,5\}$ that has exactly two orbits?

So far what I have been thinking is that I know that the order of an orbit must divide the order of the group acting on the set by the orbit-stabilizer theorem. This means that, in this case, each orbit must have order $1$ or $2$. However, because the orbits are equivalence classes, they partition $\{1,2,3,4,5\}$, so the distinct orbits must have in total $5$ elements but if there are two orbits then they have a total of $2$ different elements since each has order $1$. If there are exactly two orbits, then they both must have order $1$. I don't see though how an orbit couldn't have order $2$, because $O_x = \{gx | g \in S_2\}$ and there are two elements in $S_2$.

Am I thinking about this incorrectly?

Answer 1

Am I thinking about this incorrectly?

You are on the right track, I think. Let's clarify things.

If $G$ is a group acting on a set $X$ and $x\in X$ then there is a bijection (even $G$-isomorphism) between the orbit $Gx$ and the collection of all cosets $G/G_x$ where $G_x$ is the stabilizer subgroup for $x$. This is also known as the orbit-stabilizer theorem. In particular this shows that in the finite case $|Gx|$ divides $|G|$, as a consequence of Lagrange's theorem.

And so in our case $G=S_2$ (which has exactly $2$ elements) it means that any orbit (of any action on any set) has either $1$ or $2$ elements. The case when an orbit has one element is trivial: $S_2$ acting on any set $Y$ via $(g,y)\mapsto y$. The case when an orbit has two elements is for example when $S_2$ acts on itself (or disjoint union of itselves) via $(g,h)\mapsto gh$.

And since the set $X=\{1,2,3,4,5\}$ has $5$ elements then it has to have at least $3$ orbits regardless of how $S_2$ acts on it. That's because orbits form a partition of any set.

Answer 2

In general, the orbits of an action of a group $G$ of order $2$ on a set $S$ consist of a maximum of $2$ elements (Orbit-Stabilizer Theorem: for every $s \in S$, $|O(s)||Stab(s)|=|G|=2$, thence $|O(s)| \le 2$). If $|S|=5$, being $S$ partitioned into action's orbits, there must be at least $3$ orbits. This holds in general, for any action of any group of order $2$ on any set of cardinality $5$, and this would suffice for your problem.

A different matter is if we want "to see" this all in action for a specific group (say $S_2$), a specific set (say $\{1,2,3,4,5\}$) and a specific action of the former on the latter. To this aim, we need to remind that an action of $S_2$ on $\{1,2,3,4,5\}$ is equivalent to a homomorphism from $S_2$ to $\operatorname{Sym}(\{1,2,3,4,5\})=S_5$. Now, any map $\varphi \colon S_2=\{\iota_{S_2},(12)\} \longrightarrow S_5$, defined by $\varphi(\iota_{S_2})=\iota_{S_5}$ and $\varphi((12))=\alpha$, where $\alpha$ is an element of order $2$ of $S_5$, is a homomorphism from $S_2$ to $S_5$. As an example, let's take the "most natural" such a $\varphi$, namely the one featured by $\varphi((12))=(12) \in S_5$; then, the orbit of this action by the point $i\in \{1,2,3,4,5\}$ is the set

$$O(i)=\{\varphi(\sigma)(i), \sigma \in S_2\}=\{i\} \cup \{(12)(i)\} \tag 1$$

Deploying $(1)$, we get:

$$O(1)=\{1,2\}, \quad O(2)=\{2,1\}=O(1), \quad O(3)=\{3\}, \quad O(4)=\{4\}, \quad O(5)=\{5\}$$

so, one orbit of $2$ elements and three orbits of $1$ element each.

If, instead, we take $\varphi((12))=(12)(34)$, yet another order $2$ element of $S_5$, then this other action has orbits

$$O(i)=\{\varphi(\sigma)(i), \sigma \in S_2\}=\{i\} \cup \{(12)(34)(i)\} \tag 2$$

Deploying $(2)$, we get:

$$O(1)=\{1,2\}, \quad O(2)=\{2,1\}=O(1), \quad O(3)=\{3,4\}, \quad O(4)=\{4,3\}=O(3), \quad O(5)=\{5\}$$

so, this time we have two orbits of $2$ elements each and one orbit of $1$ element.

Needless to say that both these particular cases comply with the general result gotten by the O.S.T. for all the actions of groups of order $2$ on sets of cardinality $5$, namely that at least $3$ orbits must be there.