Permutation in $S_{35}$ find $s^{12}$ and $s^{27}$ I was assigned the following Permutation in $S_{35}$
$$s=(1 2 3 4 5 6 7 8 ... 22)$$
I have to find the following

*

*$s^2$


*$s^3$


*$s^{12}$ and


*$s^{27}$
So I went ahead and calculated the first 2 problems simply by hand but I'im struggling to find how to do the other 2.
Thoughts on the third one are non existent
For the last one $s^{27}$ I know that the order of $s$ is $22$ since its length is $22$ so I can write $s^{27} = s^{22} \times s^3 \times s^2=s^3 \times s^2$.
Is this the solution to it?
 A: Your idea for $s^{27}$ is nice and works fine.

Let's consider $s^{12}$. Since $s$ fixes every element of $\{23,\cdots,35\}$, all its powers do to, so for cycle notation we don't need to think about them. Think of the other elements as the integers mod $22$, except we write $22$ instead of $0$ as one of our numbers. Then $s$ is just the function $x\mapsto x+1$. (If you're not familiar with modular arithmetic, that means $s(x)=x+1$ but with wrap-around if numbers go past $22$.)
If we iterate $s$ multiple times, we are in effect adding more than just $1$ to $x$.  Concretely,
$$ s^2(x)=(x+1)+1=x+2 $$
and similarly $s^3(x)=x+3$, etc. Thus, $s^{12}(x)=x+12$. That means $s^{12}$ sends $1\mapsto13$ and $2\mapsto14$ and so on, up until $10\mapsto22$ followed by $11\mapsto1$ all the way up to $21\mapsto11$ and $22\mapsto12$.

More generally, if you have a cycle $(a_1~a_2~\cdots~a_n)$ which you want to take to a power $k$, you cycle $a_1$ a total of $k$ units forward to $a_{1+k}$ (with indices interpreted mod $n$, so they wrap around again). That is, $(a_1~a_2~\cdots~a_n)^k=(a_{1+k}~a_{2+k}~\cdots~a_{n+k})$. Since disjoint cycles commute, this lets us take any permutation given in cycle notation to any power.
A: Are you free to choose how to represent the resulting permutation? As a function it should be simple. In two-line notation, straightforward but requires a bit of writing.
Or does it have to be in cycle notation? Then just start from $1$, see where $s^{12}$ takes it ($13$), then where that goes ($3$), so at least one cycle starts like $(1,13,3,\ldots)$. Continue until the cycle is complete. If some element was not encountered yet, but is moved by the permutation, start a new cycle etc.
