Expression as a product of disjoint cycles Let $\alpha = (9312)(496)(37215) \in S_n, n \ge 9$. Express $\alpha$ as a product of disjoint cycles.
I know this is probably a really easy question, but my professor didn't elaborate on how to exactly do this and neither does my assigned text. If anyone could elaborate on the algorithm of going about this I would really appreciate it.
Thanks you 

Forgive me if the answer is a little sloppy, this is my first answer on stack exchange.
We begin by writing in the following format, 
(37215)=\begin{pmatrix}
1&2&3&4&5&6&7&8&9\\
5&1&7&4&3&6&2&8&9
\end{pmatrix}
Now, if we take the cycle $(496)$ composed of this, we get
(496)o (37215)=\begin{pmatrix}
1&2&3&4&5&6&7&8&9\\
5&1&7&9&3&4&2&8&6
\end{pmatrix}
and then one more time
(9312)o ((496)o(37215))=\begin{pmatrix}
 1&2&3&4&5&6&7&8&9\\
 5&2&7&3&1&4&9&8&6
 \end{pmatrix}
Now, we can conveniently put this in one giant matrix(and breach some mathematical etiquette while we're at it) and write
\begin{pmatrix}
1&2&3&4&5&6&7&8&9\\
\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow \\
5&1&7&4&3&6&2&8&9\\
\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow \\
5&1&7&9&3&4&2&8&6\\
\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow \\
5&2&7&3&1&4&9&8&6
\end{pmatrix}
This is not necessarily the quickest or most elegant approach, but serves as a good substitute while familiarity with cycle notation grows.  In cycle notation, the answer is (15)(37964), because, (from the top row of the bottom matrix to the bottom row)1 goes to 5, which goes back to 1, and 3 goes to 7 which goes to 9 and so on.
