cyclic subgroup elements I'm having hard time finding elements of the cyclic subgroup $\langle a\rangle$ in $S_{10}$, where $a = (1\ 3\ 8\ 2\ 5\ 10)(4\ 7\ 6\ 9)$
This is my attempt:
\begin{align}
a^2 &= (1\ 8\ 5\ 10)(4\ 6\ 9) \\
a^3 &= (1\ 3\ 5\ 10)(4\ 7\ 9\ 6) \\
a^4 &= (1\ 5\ 10)(4\ 9\ 7) \\
a^5 &= (1\ 3\ 8\ 2\ 10)(7\ 6) \\
a^6 &= (1\ 8\ 10)(4\ 6\ 9) \\
a^7 &= (1\ 3\ 10)(4\ 7\ 9\ 6) \\
a^8 &= (1\ 10)(4\ 9\ 7) \\
a^9 &= (1\ 3\ 8\ 2\ 5\ 10)(7\ 6) \\
a^{10} &= (1\ 8\ 5)(4\ 6\ 9) \\
a^{11} &= (1\ 3\ 5\ 10)(4\ 7\ 9) \\
a^{12} &= (1\ 5)(4\ 9\ 7\ 6)
\end{align}
I suspect I already went wrong somewhere. I understand I need to get to $e = (1)$ at some point. Is there a way to check and make sure there are no mistakes when you calculate this? 

If you have $$a = (1\,3\,8\,2\,5\,10)(4\,7\,6\,9)$$
that means that $a$ is the permutation that takes 1 to 3, 3 to 8, 8 to 2, and so on.
The permutation $a^2$ is obtained by applying $a$ twice.  Since $a$ takes 1 to 3, and then 3 to 8, $a^2$ takes 1 to 8.  $$\begin{array}{ccc}
a^0 & a^1 & a^2 \\ \hline
1 & 3  & 8 \\
2 & 5 & 10 \\
3 & 8 & 2  \\
4 & 7 & 6  \\
5 & 10 & 1 \\
6 & 9 & 4 \\
7 & 6 & 9 \\
8 & 2 & 5 \\
9 & 4 & 7 \\
10 & 1 & 3 
\end{array}$$
Reading off the first and last column of the first row, we have that $a^2$ takes 1 to 8, so it begins $a^2 = (1\,8\ldots)\ldots$.  Reading the first and last column of the 8th row, we see that $a^2$ takes 8 to 5, so $a^2 = (1\,8\,5\ldots)\ldots$.  Reading off the rest of the rows similarly, we get: $$a^2 = (1\,8\,5)(2\,10\,3)(4\,6)(7\,9).$$
Perhaps you can take it from here.
