A permutation $\sigma= (15673)(8246)(357)$ written as a product of conjoined cycles.

Question

On $A=\{1,2,3,4,5,6,7,8\}$ the following product of cycles is given

$$\sigma = (15673)(8246)(357)$$

The cycles above are conjoined.

1) Does this mean it's just one cycle?

2) Can $\sigma$ be written as a product of disjoint cycles. If it can, how should one do it?

PS This is actually not a problem but a question I have.

Answer 1

In this context, conjoint probably means that the cycles are not disjoint, i.e., some of their elements overlap.

Any permutation can be written as a product of disjoint cycles. To get the product you are after, you can first write it as a product of $2$-cycles.

First $3$ is sent to $5$ in the rightmost cycle, and later, in the leftmost cycle, $5$ is sent to $6$. This means that $3$ is sent to $6$, so you get the first $2$-cycle $(3~6)$. Then you look at what happens to $6$, and find it is sent to $8$, so you get $(6~8)$. You now continue with $8$.

After having done all, you get exactly eight $2$-cycles like this:

$$ (3~6) (6~8) (8~2) (2~4) (4~7) (7~1) (1~5) (5~3) = (36824715) $$

This is indeed a product of disjoint cycles. The number of cycles happens to be $1$.

Answer 2

Every permutation can be written as a product of disjoint cycles.

$\sigma = (15673)(8246)(357)$ is not expressed as a product of disjoint cycles, since $3, 5, 6, $ and $7$ are elements in more than one cycle.

When two cycles are juxtaposed, they are composed. There are different conventions as to which permutation is applied first. Here I will take the convention that the rightmost permutation acts first.

The rightmost cycle $(357)$ does not affect $1$, nor does the middle cycle $(8246)$, but the leftmost cycle $(15673)$ maps $1\mapsto5$, so $\sigma:1\mapsto5$. The rightmost cycle maps $5\mapsto7$ and the leftmost cycle maps $7\mapsto3$, so $\sigma:5\mapsto3$. The rightmost cycle maps $3\mapsto5$ and the leftmost cycle maps $5\mapsto6$, so $\sigma:3\mapsto6$. The rightmost cycle does not affect $6$, but the middle cycle maps it to $8$, so $\sigma:6\mapsto8$. The rightmost cycle does not affect $8$ either, but the middle cycle maps $8\mapsto2$, so $\sigma:8\mapsto2$. Similarly, $\sigma:2\mapsto4$, $\sigma:4\mapsto7$, and $\sigma:7\to1$.

Thus, $\sigma$ can be written as $(15368247)$ in disjoint cycle notation.

Answer 3

Reading the permutation left-to-right, let $f=(15673), g=(8246), h=(357)$. Then $\sigma=fgh$.

We have that $\sigma$ is given by the following calculations:

$$\begin{align} 1&\stackrel{f}{\mapsto} 5\stackrel{g}{\mapsto}5\stackrel{h}{\mapsto} 7 \\ 7&\stackrel{f}{\mapsto} 3\stackrel{g}{\mapsto}3\stackrel{h}{\mapsto} 5 \\ 5&\stackrel{f}{\mapsto} 6\stackrel{g}{\mapsto}8\stackrel{h}{\mapsto} 8 \\ 8&\stackrel{f}{\mapsto} 8\stackrel{g}{\mapsto}2\stackrel{h}{\mapsto} 2 \\ 2&\stackrel{f}{\mapsto} 2\stackrel{g}{\mapsto}4\stackrel{h}{\mapsto} 4 \\ 4&\stackrel{f}{\mapsto} 4\stackrel{g}{\mapsto}6\stackrel{h}{\mapsto} 6 \\ 6&\stackrel{f}{\mapsto} 7\stackrel{g}{\mapsto}7\stackrel{h}{\mapsto} 3 \\ 3&\stackrel{f}{\mapsto} 1\stackrel{g}{\mapsto}1\stackrel{h}{\mapsto} 1 \\ \end{align}$$

so $\sigma=(17582463)$ (but note, again, that this is reading left-to-right; the right-to-left case is handled in other answers and is, for this $\sigma$, different).