Question

What am I supposed to do?
Let n be an integer, \(n\geq 2\)
Let f \(\in S_n\)
Let X be a nonempty subset of {1,2,....,n}.
Say that f fixes X if F(X) =X,
and say that f acts irreducibly provided the only nonempty subset of {1,2,...,n} fixed by f is {1,2,...,n}itself.
Prove: f acts irreducibly if and only if f is an n-cycle.
Please, help

Answer 1

I know if f in S_n , then f is bijective.

Answer 2

idk these stuff,
Is \(S_n\) a symmetric group ?

Answer 3

but f(X) =X, is it not that f is identity?

Answer 4

Yes,

Answer 5

To me, everything is set up, f fixed X (f(X) =X is given)
the condition to say f irreducible is given also,
they ask me to prove in n-cycle. Does it mean in other cycle < n , f is not irreducible?

Answer 6

yeah biconditional works both directions right
whats the definition of `irreducible` and `cycle` ?

Answer 7

\[ \large \text{f acts irreducibly} \iff \text{f is an n-cycle}\]

Answer 8

\(\left(\begin{matrix}1&2&3\\2&3&1\end{matrix}\right)\)
from top to bottom, 1goes to 2, 2 goes to 3, 3 goes to 1
so that (1--> 2-->3-->1) = (1,2,3) is a cycle.

Answer 9

Oh and this a cycle of length 3 ?

Answer 10

yes