Permutation as disjoint cycles

Question

I want to write the permutation $(123)(45)(16789)(15)$ as disjoint cycles.

This is what I did:

In standard notation, I believe it would be

$$\sigma=\begin{pmatrix} 1 & 2 &3 & 4& 5&6 &7 &8 &9\\ 4&3&1&5&6&7&8&9&1 \end{pmatrix}$$

But how do I write the disjoint cycles? $(1456789)$ is a cycle but these leaves out $2\to3, 3\to 1$ that do not cycle.

Edit:

$1\to 5, 5\to5, 5\to4, 4\to4$

$2\to2, 2\to2,2\to2, 2\to3$

$3\to3, 3\to3,3\to3, 3\to 1$

$4\to4, 4\to4,4\to5, 5\to 5$

$5\to1, 1\to6,6\to6, 6\to 6$

$7\to7, 7\to8,8\to8, 8\to 8$

$8\to8, 8\to9,9\to9, 9\to 9$

$9\to9, 9\to1,1\to1, 1\to 1$

Answer 1

Given the permutation $(1\,\,2\,\,3)(4\,\,5)(1\,\,6\,\,7\,\,8\,\,9)(1\,\,5)$, you can find the disjoint union of cycles by considering what this permutation does to each element. Observe that, from left to right, we have that $1 \mapsto 2$ by the first cycle, and then $2$ remains fixed by the following three cycles, so that $1 \mapsto 2$ in total. Now consider $2$. It maps to $3$ in the first cycle, and then remains fixed in the other three, so $2 \mapsto 3$ in total. What does $3$ map to? Well $3 \mapsto 1$ in the first cycles, and then $1 \mapsto 6$ in the third cycle, so that $3 \mapsto 6$ in total. We have determined that the permutation starts with $(1\,\,2\,\,3\,\,6\,\ldots)$. By continuing this, you should obtain the permutation $$ (1\,\,2\,\,3\,\,6\,\,7\,\,8\,\,9\,\,5\,\,4) $$ since $4 \mapsto 1$. Check that you agree.

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Following your update, I see that you write permutations right to left so that your answer will be different to mine. You are correct everywhere up to your final step --- you should have that $9 \mapsto 2$ since $1 \mapsto 2$ in the final cycle.

Answer 2

You just apply iteratively the given permutations, starting with $1$ until you get back to $1$, i.e. until you obtain a first cycle, then start from the next number that is not in the first cycle, and so on. I apply the right-to-left convention.

For instance, $1\to 5 = 5\to 4 = 4$. So $1\to 4$.

Next, $4=4=4\to 5=5$. So $1\to 4\to 5$.

Next, $5\to 1\to 6=6=6$. So $1\to 4 \to5\to 6$.

Next, $6 =6\to 7=7=7 $. So $1\to 4 \to 5\to 6\to 7$.

Next $\,7=7\to8=8=8$, $8=8\to9=9=9$, $9=9\to 1\to 2=2$, $2=2=2=2\to 3$, $3=3=3\to1$

We finally obtain a single $9$-cycle: $$(1\: 4\: 5\: 6\:5\:7\:8\:9\: 2\: 3).$$