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? 

Your calculations look wrong. Keep in mind that $a = (1\ 3\ 8\ 2\ 5\ 10)(4\ 7\ 6\ 9)$ is the map $a:\{1,2,\dots,10\}\to\{1,2,\dots,10\}$ given by
$$
a : 1\mapsto 3, 3\mapsto 8, 8 \mapsto 2, 2\mapsto5, 5\mapsto 10, 10\mapsto 1, 4\mapsto 7, 7\mapsto 6, 6 \mapsto 9, 9\mapsto 4.
$$
Applying this map twice yields
$$
a^2 : 1\mapsto 8, 8\mapsto 5, 5\mapsto 1, 3\mapsto 2, 2\mapsto 10, 10\mapsto 3, 4\mapsto 6, 6\mapsto 4, 7\mapsto 9, 9 \mapsto 7,
$$
which is written in cycle notation $a^2 = (1\ 8\ 5)(3\ 2\ 10)(4\ 6)(7\ 9)$.
For $a^3$ you have to apply $a$ three times and get $a^3 = (1\ 2)(3\ 5)(8\ 10)(4\ 9\ 6\ 7)$.
