How to find all the orbits of a permutation defined by a function?

I have a function, where T sends Z-->Z, where T(x)=x+5

The only thing I could think of for defining the orbits were orbits like cyclic subgroups under the operation of T (like for instance <1>={1, 6, 11 ...}

but not only would those orbits be infinite, I don't think they are right.

so does anyone have some sort of help to offer on finding orbits of a function with no parameters?

-jackie

Re: How to find all the orbits of a permutation defined by a function?

Quote:

Originally Posted by

**jackie213** I have a function, where T sends Z-->Z, where T(x)=x+5

The only thing I could think of for defining the orbits were orbits like cyclic subgroups under the operation of T (like for instance <1>={1, 6, 11 ...}

but not only would those orbits be infinite, I don't think they are right.

so does anyone have some sort of help to offer on finding orbits of a function with no parameters?

-jackie

Orbits in what sense of the word? There isn't an action mentioned. If you mean though the orbit in the sense of say dynamics (i.e. iteratives of the function) then it's pretty easy to see that $\displaystyle T(x)=T(y)$ if and only if $\displaystyle 5\mid x-y$ if and only if $\displaystyle [x]_5=[y]_5$ where $\displaystyle [\cdot]_5$ is the residue class modulo five. Does that help?

Re: How to find all the orbits of a permutation defined by a function?

That seems to make sense I am just wary because I have never seen it with not being strictly numerical values.

And just to make sure we're on the same page, I mean orbits in terms of cycles/equivalence classes in abstract.

And thank you very much for responding!

Re: How to find all the orbits of a permutation defined by a function?

Quote:

Originally Posted by

**jackie213** That seems to make sense I am just wary because I have never seen it with not being strictly numerical values.

And just to make sure we're on the same page, I mean orbits in terms of cycles/equivalence classes in abstract.

And thank you very much for responding!

Yes, I think we're on the same wave length.