MHB Understanding Disjoint Cycles: $k_1$ and $k_2$

[mathmari]

Hey!

$$k_1=(a_1 \ \ \ a_2 \ \ \ \dots \ \ \ a_m)$$
$$k_2=(b_1 \ \ \ b_2 \ \ \ \dots \ \ \ b_n)$$

These cycles are disjoint if each $a_i$ is $\neq$ from each $b_j$.
We remark that $k_1(a_i) \neq a_i$ and since $a_i$ isn't any of the $b_j$ that means that $k_2(a_i)=a_i$.
Respectively: $k_2(b_j) \neq b_j \ \ \ \forall j \ \ \ \text{ and } \ \ \ k_1(b_j)=b_j$.What I understand from that is the following:
These cycles are disjoint if all the elements are different.
It stands that $k_1(a_i) \neq a_i$ because if $k_1(a_i)=a_i$ then we wouldn't write this element in the cycle but since it exists there it must be $k_1(a_i) \neq a_i$. And since the $a_i$'s and $b_j$'s are all different, at the cycle $k_2$ there is no $a_i$, that means that it must be $k_2(a_i)=a_i$.Could you tell me if I have understood it right?? (Wasntme)

 

[I like Serena]

Hey!

$$k_1=(a_1 \ \ \ a_2 \ \ \ \dots \ \ \ a_m)$$
$$k_2=(b_1 \ \ \ b_2 \ \ \ \dots \ \ \ b_n)$$

These cycles are disjoint if each $a_i$ is $\neq$ from each $b_j$.
We remark that $k_1(a_i) \neq a_i$ and since $a_i$ isn't any of the $b_j$ that means that $k_2(a_i)=a_i$.
Respectively: $k_2(b_j) \neq b_j \ \ \ \forall j \ \ \ \text{ and } \ \ \ k_1(b_j)=b_j$.What I understand from that is the following:
These cycles are disjoint if all the elements are different.
It stands that $k_1(a_i) \neq a_i$ because if $k_1(a_i)=a_i$ then we wouldn't write this element in the cycle but since it exists there it must be $k_1(a_i) \neq a_i$. And since the $a_i$'s and $b_j$'s are all different, at the cycle $k_2$ there is no $a_i$, that means that it must be $k_2(a_i)=a_i$.Could you tell me if I have understood it right?? (Wasntme)

Yep. You've understood it right! (Smile)But... suppose we pick $m=1$.
Then $k_1 = (a_1)$.
But... but... that would mean that $k_1(a_1) = a_1$.
How can that be? (Wondering)

 

[Deveno]

There are some "conventions" that get around that difficulty:

1. We do not define 1-cycles at all, or
2. We note ALL 1-cycles are the identity, and define disjoint cycles for non-identity cycles only, or
3. We amend the statement of the original post to stipulate that $m,n > 1$

The decomposition of a permutation into disjoint cycles isn't "unique" much like factorization of integers isn't unique if we allow factors of units, or different ordering of primes. We can get around this by:

a) Omitting all 1-cycles
b) Ordering each cycle $(a_1\ a_2\ \dots\ a_n)$ so that $a_1 = \min(a_1,a_2,\dots,a_n)$
c) Ordering each product of disjoint cycles:

$(a_1\ a_2\ \dots\ a_{n_1})(b_1\ b_2\ \dots\ b_{n_2})\cdots(k_1\ a_2\ \dots\ k_{n_r})$

so that: $a_1 < b_1 < \cdots < k_1$.

These "technical details" are often glossed over, just as factorization of primes usually doesn't insist the primes are listed in increasing order, but we tend to do it anyway.

So, for example, instead of writing (5 4 6)(3 2), we would write (2 3)(4 6 5), even though both refer to the same permutation of $S_6$.

 

[mathmari]

Yep. You've understood it right! (Smile)But... suppose we pick $m=1$.
Then $k_1 = (a_1)$.
But... but... that would mean that $k_1(a_1) = a_1$.
How can that be? (Wondering)

Maybe we have to suppose that $m>1$... (Wasntme)

For example when we have the following permutation:
$$\sigma=\begin{pmatrix}
1 &2 &3 &4 &5 &6 &7 &8 &9 &10 \\
2 &4 &1 &5 &9 &6 &8 &7 &10 &3
\end{pmatrix}$$
the analysis into disjoint cycles is:
$$\sigma=\bigl(\begin{smallmatrix}
1 & 2 & 4 & 5 & 9 &10 &3
\end{smallmatrix}\bigr)\bigl(\begin{smallmatrix}
7 & 8
\end{smallmatrix}\bigr)$$

$6$ does not appear at the cycles since $\displaystyle{\sigma(6)=6}$What do you mean?? (Thinking)

 

[I like Serena]

Maybe we have to suppose that $m>1$... (Wasntme)

Yep.
We'll need to do something like that as Deveno explained.
Otherwise those propositions are simply incorrect.

For example when we have the following permutation:
$$\sigma=\begin{pmatrix}
1 &2 &3 &4 &5 &6 &7 &8 &9 &10 \\
2 &4 &1 &5 &9 &6 &8 &7 &10 &3
\end{pmatrix}$$
the analysis into disjoint cycles is:
$$\sigma=\bigl(\begin{smallmatrix}
1 & 2 & 4 & 5 & 9 &10 &3
\end{smallmatrix}\bigr)\bigl(\begin{smallmatrix}
7 & 8
\end{smallmatrix}\bigr)$$

$6$ does not appear at the cycles since $\displaystyle{\sigma(6)=6}$What do you mean?? (Thinking)

You have a free choice to include $(6)$ in the cycles or not.
You could also write it as:
$$\sigma=\bigl(\begin{smallmatrix}
1 & 2 & 4 & 5 & 9 &10 &3
\end{smallmatrix}\bigr) \bigl(\begin{smallmatrix}
6
\end{smallmatrix}\bigr) \bigl(\begin{smallmatrix}
7 & 8
\end{smallmatrix}\bigr)$$
Usually $(6)$ is left out, since its presence is not relevant in this context. (Mmm)