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
Start from $1$, compute its image, compute the image of this image, and so on until this goes back to $1$. Iterate starting from the smallest element not met until then.

Here $1\to5\to1$ hence your first cycle is $(15)$. Next, $2\to2$. Next, $3\to7\to9\to6\to4\to3$ hence another cycle is $(37964)$. You are left with another fixed point $8\to8$. 

Thus, $\alpha=(15)(37964)$.
$1\to5\to5\to5$

$2\to1\to1\to2$

$3\to7\to7\to7$

$4\to4\to9\to3$

$5\to3\to3\to1$

$6\to6\to4\to4$

$7\to2\to2\to9$

$8\to8\to8\to8$

$9\to9\to6\to6$

The no. after the first arrow gives the image under the first cycle, 2nd arrow after the 2nd cycle and so on,

So,ultimately we have,

$1\to5$

$2\to2$

$3\to7$

$4\to3$

$5\to1$

$6\to4$

$7\to9$

$8\to8$

$9\to6$

Now you can easily figure out the cycle,

Take $1$ figure out its image ,go to the image figure out its image and so on, and when you return back to the same no. its one cycle.Go on and get all the cycles.