# Confusion about Actions of the Symmetric Group

I'm working on some practice questions and I am having trouble understanding actions of the symmetric group.

I have the answers, but there were no explanations as to how they were derived. I feel that there is something very fundamental that i am not understanding.

$(i)$ Let $X=\{ \{i,j,k\} \subset \{1,2,3,4\} | |\{i,j,k\}| = 3 \} =$ the set of all $3$ element subsets. What is the number of orbits of Sym$(4)$ $\curvearrowright X$?

The answer here was given as $1$.

However, I don't understand this. I have that the elements of $X$ are $\{123\},\{124\},\{234\},\{134\}$ (i.e. choosing combinations of 3 from 4)

For example, let $g_1 =$(12)(34) $\in$ Sym$(4)$ and say $x_1 = (123) \in X$

Then $$g_1 \cdot x_1 = (214) \in X$$ (Is this correct? $g_1$ sends $1 \rightarrow 2, 2 \rightarrow 1, 3 \rightarrow 4$)?

Similarly let $g_2 = (1234) \in$ Sym$(4)$, then

$$g_2 \cdot x_1 = (234) \in X$$

So from these two examples, already the orbit of $x_1$ under G are two other elements in $X$, so how can the number of orbits be $1$?

$(ii)$ Note that Sym($n$) acts on the set of all the subsets of $\{1,...,n\}$ denoted $\rho(\{1,...,n\})$. Let $X= \rho( \{1,...,4\}).$ What is the number of orbits of Sym($4$) $\curvearrowright X$?

The answer provided here is $5$. Again, I have a misunderstanding here, which is similar to above. Any insight would be greatly appreciated!

• 10-30, 10-30! The elements of $X$ are NOT 3-cycles, i.e. elements of $S_4$. They are subsets of $\{1,2,3,4\}$, so don't use the cycle notation for them. Also, the fact that you can find permutations such as $g_1$ and $g_2$ that map a given element of $X$ to any other is an indication that the number of orbits is small. If there is a permutation $g\in S_4$ such that $g(A)=g(B)$ for some $A,B\in X$, this means that $A$ and $B$ belong to the same orbit of $S_4$. – Jyrki Lahtonen Aug 24 '14 at 6:02
• In part (ii) you should notice that $A$ and $g(A)$ have the same number of elements for all $g\in S_4$ and all $A\subseteq\{1,2,3,4,\}$. – Jyrki Lahtonen Aug 24 '14 at 6:05
• thanks for your comments. I think I am getting a better understanding, just to clarify, so I have shown there are two elements of Sym($4$) that send $x_1$ to different elements of $X$ (the same orbit?), So is it true that if there is an element $g_i \in$ Sym($4$) that sends $x_j \in X$ to something $\notin X$ then this is a different orbit? Then how is 'different' orbits actually defined? Thanks again. – JackReacher Aug 24 '14 at 6:25
• If $g$ sends something in $X$ outside of $X$, then we don't have the group acting on $X$ at all. To get more orbits we need a smaller group or a bigger set. For example, if instead of all of $S_4$ we only look at the subgroup $H$ generated by $g_1=(12)(34)$, then the orbits of $H$ acting on the set $A=\{1,2,3,4\}$ are $\{1,2\}$ and $\{3,4\}$. This is because no matter how many times you apply $g_1$, you cannot map $1$ to, say, $3$. Also we might define an action of $G=S_4$ on the set $Y=\{1,2,3,4,-1,-2,-3,-4\}$ by letting $G$ act on the negative numbers by temporarily ignoring the sign. – Jyrki Lahtonen Aug 24 '14 at 6:33
• (cont'd) In that case $G$ has two orbits on $Y$. One orbit consists of the positive numbers and the other of the negative numbers. – Jyrki Lahtonen Aug 24 '14 at 6:34

You are talking about size of orbit, not the number of orbits. $X$ is the whole orbit which has $4$ elements.
For the second part, each orbit of $\rho(\{1,\dots,4\}$ contains subsets of $\{1,\dots,4\}$ of the same size. The number of orbits is equal to the number of possible sizes of the subsets which are $0,1,2,3,$ and $4$.
• sorry, I am still confused about the difference between the size of an orbit vs. the number orbits. Am i correct to say that for part $(a)$, since the size of each element resulting from Sym($4$) $\curverightarrow X$ is the same (i.e. a $3$ element subset) - then the number of orbits is $1$? Then in part $(b)$, elements of $X$ can be of sizes $0,1,2,3,4$. And when a $g \in$ Sym($4$) acts on a $x_i \in X$ it sends it to $x_j$ of the same size - so the possible sizes are as mentioned - so the answer is $5$? Is that correct? Thanks so much. – JackReacher Aug 24 '14 at 10:01