I Clarification of a specific orbit example

[mk9898]

Hello,

In my professor's lecture notes she gives this example and I have a couple of questions regarding it:Let M = \mathbb Z/6\mathbb Z and f: M \rightarrow M, x \rightarrow x+1 the cyclical permutation of the elements from M. Then is G := \{id_M, f, f^2,f^3,f^4,f^5\} a subset from S_M. Just like the symmetric group, G operates on M through applicaton. M has exactly one orbit namely M = G*0. We can have G operate also on the 2-element subset from M. |Pot_2(M)| = 15 and G has on Pot_2(M) exactly 3 orbits:G \cdot \{0,1\} = \{\{i,i+1\} | i \in M \}, of the length 6,

G \cdot \{0,2\} = \{\{i,i+2\} | i \in M \}, of the length 6

G \cdot \{0,3\} = \{\{0,3\},\{1,4\},\{2,5\} | i \in M \}, of the length 3.Questions:

1. How can one quickly calculate the cardinality of the power set Pot_2 without writing out all of the possibilities? The cardinality of the power set is 2^n but in this case it is 15 which confuses me.

2. The definition of an orbit is: Gm:= \{gm| g \in G\} \subseteq M and there are 6 elements from M. Why is G*0 the only orbit? Any help/insight is appreciated.

 

[andrewkirk]

It looks like she is defining $Pot_2(M)$ to be the set of two-distinct-element subsets of $M$, and has defined an action of $G$ on that set such that for $g\in G$ and $(m,n)\in Pot_2(M)$ we have $g((m_1,m_2))=(gm_1,gm_2)$. This has nothing to do with power sets, hence the power set cardinality formula $2^{|M|}$ is not relevant. Instead use permutations or combinations to find the cardinality of $Pot_2(M)$. Which should it be (perms or combs?) given that the subsets are not ordered pairs?

$G*0$ is the only orbit because the orbits form a partition of the set. In particular they are mutually exclusive. Since $G*0$ covers all six elements of $M$, that mutual exclusivity implies that no other orbit can have any elements. So $G*0$ is the only orbit.

 

[mk9898]

Hallo andrewkirk,

Thanks for the response. The G*0 makes a lot of sense now. That is simply the all of the g's in G acting on M and all of them are on the same orbit and orbits are disjunct. Regarding the cardinality of the set |Pot_2(M)|. The 15 is the total pairs of the tuples given the mapping of M (I believe). That means that if we were to write out all possibilities of pairs disregarding the order and remove all of the {i,i} i = (1,2,3,4,5) then we would have 15. I.e.: {0,1},{0,2},{0,3},{0,4},{0,5},{1,2},{1,3},{1,4},{1,5},{2,3},{2,4},{2,5},{3,4},{3,5},{4,5}. I THINK that it was she meant. She expected us to know the answer within 2 seconds so I am wondering if there is a formula to know this answer?

 

[mk9898]

I got it. It's just the binomial coefficient of 6 choose 2. Writing it out helped me realize it. Thanks!