Need Help with Abstract Algebra Please

[Redhead711]

My question is regarding abstract algebra.

Suppost that B is a 10-cycle.
For which integers i between 2 and 10 is
B^(i) also a 10-cycle?

I know that the answer is 3, 7, and 9 I just don't know
how you arrive at these numbers. If someone could explain the
process clearly to me I would greatly appreciate that.
Thank you sooo much. :-)

[NateTG]

The i are values that are coprime (have no common factors other than 1) with 10.

Here's a general proof:

Let's say we have some element g with order n. Then g^i has order \frac{n}{<n,i>} where <n,i> is the greatest common factor of n and i.

Proof:
({g^{i}})^{\frac{n}{<n,i>}}=g^{i \times \frac{n}{<n,i>}}
but
i = k \times <n,i>
for some k so
g^{i \times \frac{n}{<n,i>}}=g^{k \times <n,i> \times \frac{n}{<n,i>}}=g^{k \times n}=g^{n\times k}=(g^{n})^{k}=e^k=e
so the order ok g{i} is at most \frac{n}{<n,i>}

Now, let's say we have some j so that
e=({g^{i}})^j=g^{ij}
then, since the order of g is n
so n | ij (n divides ij)
so \frac{n}{<n,j>} | j \Rightarrow j \geq \frac{n}{<n,i>}

So the order of g^i is \frac{n}{<n,i>}.

In this particular case, you have \frac{n}{<n,i>} = n so <n,i>=1.

[Redhead711]

Thank you so much I understand much better now. I am very grateful for all your help.

[Huyen Nguyen]

I think that g has order 10 does not guarantee that g is a 10-cycle. If g is a product of two disjoint cycles of order 2 and 5, it can still have order 10. Is that right?