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, $\alpha=(15)(37964)$.
From the given cycle structure, you read
1 -> 5
5 -> 3 -> 1 (remember, multiplication is done from right to left)
so you are back at 1, and the first cycle (1 5) is complete.
Continue with a number which is not yet in a cycle:
2 -> 1 -> 2 (so the next cycle is (2), which is a fixed point)
3 -> 7
7 -> 2 -> 9
9 -> 6
6 -> 4
4 -> 9 -> 3
This gives the cycle (3 7 9 6 4)
Now all numbers in the given permutation are done, and the result is (1 5)(2)(3 7 9 6 4) = (1 5)(3 7 9 6 4).
Most probably you guys multiply cycles from right to left, as most authors do nowadays, so begin with $\,1\,$ and follow its "cycle" (again, from right to left):
$$1\to 5$$
Now do the same for $\,5\,$
$$5\to 3\to 1$$
and we've closed the first cycle: $\,(15)\,$
Now with the smallest number not included already above:
$$2\to 1\to 2\Longrightarrow 2\to 2$$
and this time we write nothing
$$4\to 3\to 7\to 9\to 6\to 4$$
and we have another cycle: $\,(43796)\,$ , etc.
This does not really add much that has not already been said other than give a slightly modified way of writing down the algorithm the long way. Using $(37215)(496)(9312)$ and going from left to right the first row below reads $1$ goes to $5$ in the first cycle. Since there is no $5$ in the second cycle write $5$ goes to $5$ and no $5$ in the third cycle write $5$ goes to $5$. This summarizes to $1$ goes to $5$ at the end. Repeat this for $2$ through $9$ then write the multiplication based on the last two columns.
$$1\to 5\ 5\to5\ 5\to5\ 1\to 5$$ $$2\to 1\ 1\to1\ 1\to2\ 2\to 2$$ $$3\to 7\ 7\to7\ 7\to7\ 3\to 7$$ $$4\to 4\ 4\to9\ 9\to3\ 4\to 3$$ $$5\to 3\ 3\to3\ 3\to1\ 5\to 1$$ $$6\to 6\ 6\to4\ 4\to4\ 6\to 4$$ $$7\to 2\ 2\to2\ 2\to9\ 7\to 9$$ $$8\to 8\ 8\to8\ 8\to8\ 8\to 8$$ $$9\to 9\ 9\to6\ 6\to6\ 9\to 6$$
Summarizing the last two columns $(15)(37964)$
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.
$1\to5\to5\to5$
$2\to1\to1\to2$
$3\to7\to7\to7$
$4\to4\to9\to3$
$5\to3\to3\to1$
$6\to6\to4\to4$
$7\to2\to2\to9$
$8\to8\to8\to8$
$9\to9\to6\to6$
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.
