Write the following permutation as product of disjoint cycles
$$(12)(13)(14)(15)$$
Could someone explain how to proceed with this question ? I have four more similar, so I just want somebody to solve this one so that I can have a basic example.
Thank you in advance
$\endgroup$2 Answers
$\begingroup$We write this permutation on its standard form $$\sigma=\left(\begin{array}\\ 1&2&3&4&5\\ 5&1&2&3&4 \end{array}\right)$$ and this a cycle since $$1\overset \sigma\rightarrow5\overset \sigma\rightarrow4\overset \sigma\rightarrow3\overset \sigma\rightarrow2\overset \sigma\rightarrow1$$ so $$\sigma=(1\ 5\ 4\ 3\ 2)$$
$\endgroup$ 13 $\begingroup$Assuming that the transpositions are applied from left to right, this permutation first takes $1$ to $2$, and $2$ is unaffected by the remaining three transpositions. It takes $2$ to $1$, which the second transposition then takes to $3$; $3$ is unaffected by the last two transpositions, so the permutation takes $2$ to $3$. Now what does it do to $3$? The first transposition has no effect on $3$, the second takes it to $1$, and the third then takes this $1$ to $4$; since the last transposition does not affect the $4$, the net effect is to take $3$ to $4$. Similar reasoning shows that $4$ is unaffected by the first two transpositions and sent to $1$ by the third; this $1$ is then sent to $5$ by the last transposition, so the permutation ends up taking $4$ to $5$. Finally, $5$ is affected only by the last transposition, which takes it to $1$; there are no further transpositions to be applied at that point, so the permutation takes $5$ to $1$. The overall effect is therefore
$$1\mapsto 2\mapsto 3\mapsto 4\mapsto 5\mapsto 1\;,$$
which is represented by the single cycle $(12345)$. That is, this permutation is a cycle.
With another permutation we might initially have found that $1\mapsto 3\mapsto 4\mapsto 1$. In that case we’d then look to see what the permutation does to the first number missing from this cycle, namely, $2$. In this particular case we’d then find one of two things: either it takes $2$ to itself and $5$ to itself, or it takes $2$ to $5$ and $5$ to $2$. In the second case we have the permutation $(134)(25)$; in the first we have $(134)(2)(5)$, though the $1$-cycles are often omitted in practice.
If the permutation is $\pi$, the general idea is to find $\pi(1)$, $\pi\big(\pi(1)\big)$, and so on, until you close a cycle. Then take the first number not in that cycle and track its orbit under repeated applications of $\pi$. Keep doing this until all elements of the domain have been exhausted. These orbits never intersect, so you get the decomposition of $\pi$ into a product of pairwise disjoint cycles.
$\endgroup$
