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?