MHB Solving the Mystery of the Table: Explaining $g = (123)$

  • Thread starter Thread starter Guest2
  • Start date Start date
  • Tags Tags
    Mystery Table
Guest2
Messages
192
Reaction score
0
View attachment 5399

Could someone please explain how they're getting the answers in the table, for example $g = (123)$.
 

Attachments

  • Screen Shot 2016-03-14 at 13.53.08.png
    Screen Shot 2016-03-14 at 13.53.08.png
    23.3 KB · Views: 99
Physics news on Phys.org
$$(123)(1)$$

$(123)$ represents the function that maps $1$ to $2$, $2$ to $3$ and $3$ to $1$.
$(1)$ represents the identity function, i.e. the function that maps $1$ to $1$, $2$ to $2$ and $3$ to $3$.

So to compute $(123)(1)$ we do the following:

From the right cycle we have that $1$ is mapped to $1$, and at the first cycle $1$ is mapped to $2$, therefore we get that $1$ is mapped to $2$.
From the right cycle we have that $2$ is mapped to $2$, and at the first cycle $2$ is mapped to $3$, therefore we get that $2$ is mapped to $3$.
From the right cycle we have that $3$ is mapped to $3$, and at the first cycle $3$ is mapped to $1$, therefore we get that $3$ is mapped to $1$.

So, we get $(123)(1)=(123)$.
$$(123)(12)$$

$(123)$ represents the function that maps $1$ to $2$, $2$ to $3$ and $3$ to $1$.
$(12)$ represents the function that maps $1$ to $2$, $2$ to $1$ and $3$ to $3$.

So to compute $(123)(12)$ we do the following:

From the right cycle we have that $1$ is mapped to $2$, and at the first cycle $2$ is mapped to $3$, therefore we get that $1$ is mapped to $3$.
From the right cycle we have that $2$ is mapped to $1$, and at the first cycle $1$ is mapped to $2$, therefore we get that $2$ is mapped to $2$.
From the right cycle we have that $3$ is mapped to $3$, and at the first cycle $3$ is mapped to $1$, therefore we get that $3$ is mapped to $1$.

So, we get $(123)(12)=(13)$.
 
mathmari said:
$$(123)(1)$$

$(123)$ represents the function that maps $1$ to $2$, $2$ to $3$ and $3$ to $1$.
$(1)$ represents the identity function, i.e. the function that maps $1$ to $1$, $2$ to $2$ and $3$ to $3$.

So to compute $(123)(1)$ we do the following:

From the right cycle we have that $1$ is mapped to $1$, and at the first cycle $1$ is mapped to $2$, therefore we get that $1$ is mapped to $2$.
From the right cycle we have that $2$ is mapped to $2$, and at the first cycle $2$ is mapped to $3$, therefore we get that $2$ is mapped to $3$.
From the right cycle we have that $3$ is mapped to $3$, and at the first cycle $3$ is mapped to $1$, therefore we get that $3$ is mapped to $1$.

So, we get $(123)(1)=(123)$.
$$(123)(12)$$

$(123)$ represents the function that maps $1$ to $2$, $2$ to $3$ and $3$ to $1$.
$(12)$ represents the function that maps $1$ to $2$, $2$ to $1$ and $3$ to $3$.

So to compute $(123)(12)$ we do the following:

From the right cycle we have that $1$ is mapped to $2$, and at the first cycle $2$ is mapped to $3$, therefore we get that $1$ is mapped to $3$.
From the right cycle we have that $2$ is mapped to $1$, and at the first cycle $1$ is mapped to $2$, therefore we get that $2$ is mapped to $2$.
From the right cycle we have that $3$ is mapped to $3$, and at the first cycle $3$ is mapped to $1$, therefore we get that $3$ is mapped to $1$.

So, we get $(123)(12)=(13)$.
Wonderful explanations, thanks!
 
mathmari said:
$$(123)(1)$$

$(123)$ represents the function that maps $1$ to $2$, $2$ to $3$ and $3$ to $1$.
$(1)$ represents the identity function, i.e. the function that maps $1$ to $1$, $2$ to $2$ and $3$ to $3$.

So to compute $(123)(1)$ we do the following:

From the right cycle we have that $1$ is mapped to $1$, and at the first cycle $1$ is mapped to $2$, therefore we get that $1$ is mapped to $2$.
From the right cycle we have that $2$ is mapped to $2$, and at the first cycle $2$ is mapped to $3$, therefore we get that $2$ is mapped to $3$.
From the right cycle we have that $3$ is mapped to $3$, and at the first cycle $3$ is mapped to $1$, therefore we get that $3$ is mapped to $1$.

So, we get $(123)(1)=(123)$.

$$(123)(12)$$

$(123)$ represents the function that maps $1$ to $2$, $2$ to $3$ and $3$ to $1$.
$(12)$ represents the function that maps $1$ to $2$, $2$ to $1$ and $3$ to $3$.

So to compute $(123)(12)$ we do the following:

From the right cycle we have that $1$ is mapped to $2$, and at the first cycle $2$ is mapped to $3$, therefore we get that $1$ is mapped to $3$.
From the right cycle we have that $2$ is mapped to $1$, and at the first cycle $1$ is mapped to $2$, therefore we get that $2$ is mapped to $2$.
From the right cycle we have that $3$ is mapped to $3$, and at the first cycle $3$ is mapped to $1$, therefore we get that $3$ is mapped to $1$.

So, we get $(123)(12)=(13)$.

Spoken like a pro.
 

Thread 'Trouble understanding an online solution to an exercise in Dummit & Foote'

##\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...
View full post »

Thread 'Questions about non existence of GCDs vs (coimages, cokernels)'

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...
View full post »

Thread 'Decomposition into irreps of compact Lie group'

When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
View full post »

Similar threads

Insights Groups, The Path from a Simple Concept to Mysterious Results
Replies
17
Views
8K
MHB Finding a Cayley Table for a Groupoid: an Example
Replies
3
Views
2K
I Decomposing arbitrary representations
Replies
2
Views
1K
Insights How Quantum Information Theory Solves “the only mystery” of Quantum Mechanics
Replies
1
Views
4K
How to solve a circuit mystery box
Replies
8
Views
2K
Downdraft table -blocking off part of table surface to increase flow?
Replies
10
Views
3K
Why body A has greater acceleration?
Replies
13
Views
1K
MHB Example on Triangular Rings - Lam, Example 1.14
Replies
4
Views
1K
MHB How Do You Manage Database Tables in SQL?
Replies
8
Views
943
A About Schur's lemma in lie algebra
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top