MHB Do disjoint cycles commute under exponentiation? (Curious)

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Induction
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

1. Let $1\leq n\in \mathbb{N}$ and $\pi\in \text{Sym}(n)$. For $1\leq k\in \mathbb{N}$ we define $\pi^{-k}:=\left (\pi^n\right )^{-1}$.

Show for all $k,\ell\in \mathbb{Z}$ the equation $\pi^k\circ \pi^{\ell}=\pi^{k+\ell}$. 2. Let $1\leq n\in \mathbb{N}$. Show that $\pi^{n!}=\text{id}$ for all $\pi\in \text{Sym}(n)$.
Do we show both statements using induction? (Wondering)
 
Physics news on Phys.org
mathmari said:
Hey! :o

1. Let $1\leq n\in \mathbb{N}$ and $\pi\in \text{Sym}(n)$. For $1\leq k\in \mathbb{N}$ we define $\pi^{-k}:=\left (\pi^n\right )^{-1}$.

Hey mathmari!

Is that a typo? Shouldn't it be $\pi^{-k}:=\left (\pi^k\right )^{-1}$? (Wondering)

mathmari said:
Show for all $k,\ell\in \mathbb{Z}$ the equation $\pi^k\circ \pi^{\ell}=\pi^{k+\ell}$. 2. Let $1\leq n\in \mathbb{N}$. Show that $\pi^{n!}=\text{id}$ for all $\pi\in \text{Sym}(n)$.
Do we show both statements using induction?

I don't think we need induction for 1.
Instead I think we need to distinguish cases.
If k and l are positive, the relation follows from the definition of repeated application of a function.
However, if either is negative or zero, we still need to see what happens. (Thinking)

For question 2, I think we need group theory. Specifically that the order of an element divides the size of a finite group. Then we don't need induction. (Thinking)
 
I noticed now that the hint for induction is for question 1.

There is the following hint:

$\pi\in \text{Sym}(n)$

$\pi=\pi_1\circ \pi_2\circ \ldots \circ \pi_m$, $m\leq n$

$\pi$ has maximum length $n$

We are looking for $x\in \mathbb{N}$ such that $\pi^x=(\pi_1\circ \pi_2\circ \ldots \circ \pi_m)^x\ \overset{\text{Induction}}{ =} \ \pi_1^x\circ \pi_2^x\circ \ldots \circ \pi_m^x=\text{id}$

For $1\leq i\leq m$ the $\pi_i$ is a cycle of length $m_i$.

It holds that $(\pi_i)^{m_i}=\text{id}$ so $(\pi_i)^{n!}=\text{id}$ .

Then $\pi^{n!}=\pi_1^{n!}\circ \pi_2^{n!}\circ \ldots \circ \pi_m^{n!}=\text{id}\circ \text{id}\circ \ldots \circ \text{id}=\text{id}$.

(Wondering)
 
mathmari said:
I noticed now that the hint for induction is for question 1.

There is the following hint:

$\pi\in \text{Sym}(n)$

$\pi=\pi_1\circ \pi_2\circ \ldots \circ \pi_m$, $m\leq n$

$\pi$ has maximum length $n$

What do you mean by maximum length? (Wondering)

mathmari said:
We are looking for $x\in \mathbb{N}$ such that $\pi^x=(\pi_1\circ \pi_2\circ \ldots \circ \pi_m)^x\ \overset{\text{Induction}}{ =} \ \pi_1^x\circ \pi_2^x\circ \ldots \circ \pi_m^x=\text{id}$

For $1\leq i\leq m$ the $\pi_i$ is a cycle of length $m_i$.

I don't think this is true in general.
Consider for instance $(1) \circ (1\,2) \circ (1\,2\,3) = (2\,3)$ and $x=3$.
Then:
$$\big((1) \circ (1\,2) \circ (1\,2\,3)\big)^3 = (2\,3)^3=(2\,3)
\ne (1\,2)= (1)^3 \circ (1\,2)^3 \circ (1\,2\,3)^3 $$
isn't it? (Worried)

Is it perhaps supposed to be a disjoint decomposition in cycles? (Wondering)

mathmari said:
It holds that $(\pi_i)^{m_i}=\text{id}$ so $(\pi_i)^{n!}=\text{id}$ .

Then $\pi^{n!}=\pi_1^{n!}\circ \pi_2^{n!}\circ \ldots \circ \pi_m^{n!}=\text{id}\circ \text{id}\circ \ldots \circ \text{id}=\text{id}$.
 
Klaas van Aarsen said:
What do you mean by maximum length? (Wondering)

I forgot the index. It should be: The cycle $\pi_i$ has maximum length $n$.
Klaas van Aarsen said:
I don't think this is true in general.
Consider for instance $(1) \circ (1\,2) \circ (1\,2\,3) = (2\,3)$ and $x=3$.
Then:
$$\big((1) \circ (1\,2) \circ (1\,2\,3)\big)^3 = (2\,3)^3=(2\,3)
\ne (1\,2)= (1)^3 \circ (1\,2)^3 \circ (1\,2\,3)^3 $$
isn't it? (Worried)

Is it perhaps supposed to be a disjoint decomposition in cycles? (Wondering)

Ahh ok! So if we suppose that $\pi_i\neq \pi_j$ for $i\neq j$, does the above hold? (Wondering)
 
mathmari said:
I forgot the index. It should be: The cycle $\pi_i$ has maximum length $n$.

Ahh ok! So if we suppose that $\pi_i\neq \pi_j$ for $i\neq j$, does the above hold?

That is not enough, is it?
If any pair $\pi_i$ and $\pi_j$ are cycles that overlap, the result won't hold. (Worried)

However, if the cycles are disjoint, the result does hold.
So if for instance $\pi=(1\,2\,3)(5\,6)$ with $\pi_1=(1\,2\,3)$ and $\pi_2=(5\,6)$, we have for any $x\in\mathbb Z$:
$$\pi^x = \big((1\,2\,3)\circ (5\,6)\big)^x = (1\,2\,3)^x \circ (5\,6)^x$$
Then we have $m=2$ and $m_1=3,\,m_2=2$. (Thinking)
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...

Similar threads

Replies
9
Views
2K
Replies
7
Views
2K
Replies
52
Views
3K
Replies
10
Views
2K
Replies
10
Views
1K
Replies
4
Views
1K
Replies
5
Views
2K
Back
Top