Finding elements in a finite bijective set
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The Question:
Let $A = {1, 2, 3}$ and let $X = {f|f : A → A text{ is a bijection}}$. List the elements of $X$ and given them each names. (Hint: there are $6$.)
My Thoughts
I think I can get $3$ of them by mapping the values onto themselves, but then how would I get the other values?
functions
edited Nov 26 at 5:10
Mason
1,8811527
asked Nov 26 at 4:40
George
656
add a comment |
up vote
0
down vote
favorite
The Question:
Let $A = {1, 2, 3}$ and let $X = {f|f : A → A text{ is a bijection}}$. List the elements of $X$ and given them each names. (Hint: there are $6$.)
My Thoughts
I think I can get $3$ of them by mapping the values onto themselves, but then how would I get the other values?
functions
edited Nov 26 at 5:10
Mason
1,8811527
asked Nov 26 at 4:40
George
656
Do you know about groups and permutation groups ??
– Anik Bhowmick
Nov 26 at 4:46
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The Question:
Let $A = {1, 2, 3}$ and let $X = {f|f : A → A text{ is a bijection}}$. List the elements of $X$ and given them each names. (Hint: there are $6$.)
My Thoughts
I think I can get $3$ of them by mapping the values onto themselves, but then how would I get the other values?
functions
edited Nov 26 at 5:10
Mason
1,8811527
asked Nov 26 at 4:40
George
656
The Question:
Let $A = {1, 2, 3}$ and let $X = {f|f : A → A text{ is a bijection}}$. List the elements of $X$ and given them each names. (Hint: there are $6$.)
My Thoughts
I think I can get $3$ of them by mapping the values onto themselves, but then how would I get the other values?
functions
functions
edited Nov 26 at 5:10
Mason
1,8811527
asked Nov 26 at 4:40
George
656
edited Nov 26 at 5:10
Mason
1,8811527
asked Nov 26 at 4:40
George
656
edited Nov 26 at 5:10
Mason
1,8811527
edited Nov 26 at 5:10
Mason
1,8811527
edited Nov 26 at 5:10
Mason
1,8811527
1,8811527
asked Nov 26 at 4:40
George
656
asked Nov 26 at 4:40
George
656
asked Nov 26 at 4:40
George
656
656
Do you know about groups and permutation groups ??
– Anik Bhowmick
Nov 26 at 4:46
add a comment |
Do you know about groups and permutation groups ??
– Anik Bhowmick
Nov 26 at 4:46
Do you know about groups and permutation groups ??
– Anik Bhowmick
Nov 26 at 4:46
Do you know about groups and permutation groups ??
– Anik Bhowmick
Nov 26 at 4:46
add a comment |
3 Answers
3
active
oldest
votes
up vote
0
down vote
accepted
This is the symmetry group of three elements. You can read up on these groups here.
I like to write the set $X={Id, T_1, T_2, T_3, R, R^2 }$. I will write three elements below and leave the rest to you.
Id: Which is the identity element. It leaves all the elements where they are.
$1 to 1$
$2 to 2$
$3 to 3$
$T_1:$ Which Transposes $2$ and $3$ and leaves $1$ fixed. You can guess what is meant by $T_2$ and $T_3$ by examining this guy.
$1 to 1$
$2 to 3$
$3 to 2$
$R: $ Who rotates all the elements. $R^2$ is the result of applying $R$ twice.
$1 to 2$
$2 to 3$
$3 to 1$
answered Nov 26 at 4:47
Mason
1,8811527
add a comment |
up vote
0
down vote
If $(p_1, p_2, p_3)$ is a permutation of $(1,2,3)$.
We can let $f_p(i)=p_i$.
There are $3!$ permutations of $(1,2,3)$.
answered Nov 26 at 4:47
Siong Thye Goh
97.4k1463116
add a comment |
up vote
0
down vote
Think of what a bijective function does. Each element in the domain maps uniquely to one element of the codomain, and each element of the codomain will have a preimage.
It might be best to make a tree diagram or something of the sort if you're not familiar with permutation groups or such. Of the three elements you can map the element $1$ to, you pick one of them in the first branch. In each branch, you're left with two different options for which to map $2$ to (you cannot map to what you map $1$ to because it won't be injective), and in the final branch only one option to map $3$ to. Once you have that all figured out, you should be able to follow the paths in tree diagram to find your definition for a bijective $f$.
answered Nov 26 at 4:49
Eevee Trainer
2,561221
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
This is the symmetry group of three elements. You can read up on these groups here.
I like to write the set $X={Id, T_1, T_2, T_3, R, R^2 }$. I will write three elements below and leave the rest to you.
Id: Which is the identity element. It leaves all the elements where they are.
$1 to 1$
$2 to 2$
$3 to 3$
$T_1:$ Which Transposes $2$ and $3$ and leaves $1$ fixed. You can guess what is meant by $T_2$ and $T_3$ by examining this guy.
$1 to 1$
$2 to 3$
$3 to 2$
$R: $ Who rotates all the elements. $R^2$ is the result of applying $R$ twice.
$1 to 2$
$2 to 3$
$3 to 1$
answered Nov 26 at 4:47
Mason
1,8811527
add a comment |
up vote
0
down vote
accepted
This is the symmetry group of three elements. You can read up on these groups here.
I like to write the set $X={Id, T_1, T_2, T_3, R, R^2 }$. I will write three elements below and leave the rest to you.
Id: Which is the identity element. It leaves all the elements where they are.
$1 to 1$
$2 to 2$
$3 to 3$
$T_1:$ Which Transposes $2$ and $3$ and leaves $1$ fixed. You can guess what is meant by $T_2$ and $T_3$ by examining this guy.
$1 to 1$
$2 to 3$
$3 to 2$
$R: $ Who rotates all the elements. $R^2$ is the result of applying $R$ twice.
$1 to 2$
$2 to 3$
$3 to 1$
answered Nov 26 at 4:47
Mason
1,8811527
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
This is the symmetry group of three elements. You can read up on these groups here.
I like to write the set $X={Id, T_1, T_2, T_3, R, R^2 }$. I will write three elements below and leave the rest to you.
Id: Which is the identity element. It leaves all the elements where they are.
$1 to 1$
$2 to 2$
$3 to 3$
$T_1:$ Which Transposes $2$ and $3$ and leaves $1$ fixed. You can guess what is meant by $T_2$ and $T_3$ by examining this guy.
$1 to 1$
$2 to 3$
$3 to 2$
$R: $ Who rotates all the elements. $R^2$ is the result of applying $R$ twice.
$1 to 2$
$2 to 3$
$3 to 1$
answered Nov 26 at 4:47
Mason
1,8811527
This is the symmetry group of three elements. You can read up on these groups here.
I like to write the set $X={Id, T_1, T_2, T_3, R, R^2 }$. I will write three elements below and leave the rest to you.
Id: Which is the identity element. It leaves all the elements where they are.
$1 to 1$
$2 to 2$
$3 to 3$
$T_1:$ Which Transposes $2$ and $3$ and leaves $1$ fixed. You can guess what is meant by $T_2$ and $T_3$ by examining this guy.
$1 to 1$
$2 to 3$
$3 to 2$
$R: $ Who rotates all the elements. $R^2$ is the result of applying $R$ twice.
$1 to 2$
$2 to 3$
$3 to 1$
answered Nov 26 at 4:47
Mason
1,8811527
answered Nov 26 at 4:47
Mason
1,8811527
answered Nov 26 at 4:47
Mason
1,8811527
answered Nov 26 at 4:47
Mason
1,8811527
1,8811527
add a comment |
add a comment |
up vote
0
down vote
If $(p_1, p_2, p_3)$ is a permutation of $(1,2,3)$.
We can let $f_p(i)=p_i$.
There are $3!$ permutations of $(1,2,3)$.
answered Nov 26 at 4:47
Siong Thye Goh
97.4k1463116
add a comment |
up vote
0
down vote
If $(p_1, p_2, p_3)$ is a permutation of $(1,2,3)$.
We can let $f_p(i)=p_i$.
There are $3!$ permutations of $(1,2,3)$.
answered Nov 26 at 4:47
Siong Thye Goh
97.4k1463116
add a comment |
up vote
0
down vote
up vote
0
down vote
If $(p_1, p_2, p_3)$ is a permutation of $(1,2,3)$.
We can let $f_p(i)=p_i$.
There are $3!$ permutations of $(1,2,3)$.
answered Nov 26 at 4:47
Siong Thye Goh
97.4k1463116
If $(p_1, p_2, p_3)$ is a permutation of $(1,2,3)$.
We can let $f_p(i)=p_i$.
There are $3!$ permutations of $(1,2,3)$.
answered Nov 26 at 4:47
Siong Thye Goh
97.4k1463116
answered Nov 26 at 4:47
Siong Thye Goh
97.4k1463116
answered Nov 26 at 4:47
Siong Thye Goh
97.4k1463116
answered Nov 26 at 4:47
Siong Thye Goh
97.4k1463116
97.4k1463116
add a comment |
add a comment |
up vote
0
down vote
Think of what a bijective function does. Each element in the domain maps uniquely to one element of the codomain, and each element of the codomain will have a preimage.
It might be best to make a tree diagram or something of the sort if you're not familiar with permutation groups or such. Of the three elements you can map the element $1$ to, you pick one of them in the first branch. In each branch, you're left with two different options for which to map $2$ to (you cannot map to what you map $1$ to because it won't be injective), and in the final branch only one option to map $3$ to. Once you have that all figured out, you should be able to follow the paths in tree diagram to find your definition for a bijective $f$.
answered Nov 26 at 4:49
Eevee Trainer
2,561221
add a comment |
up vote
0
down vote
Think of what a bijective function does. Each element in the domain maps uniquely to one element of the codomain, and each element of the codomain will have a preimage.
It might be best to make a tree diagram or something of the sort if you're not familiar with permutation groups or such. Of the three elements you can map the element $1$ to, you pick one of them in the first branch. In each branch, you're left with two different options for which to map $2$ to (you cannot map to what you map $1$ to because it won't be injective), and in the final branch only one option to map $3$ to. Once you have that all figured out, you should be able to follow the paths in tree diagram to find your definition for a bijective $f$.
answered Nov 26 at 4:49
Eevee Trainer
2,561221
add a comment |
up vote
0
down vote
up vote
0
down vote
Think of what a bijective function does. Each element in the domain maps uniquely to one element of the codomain, and each element of the codomain will have a preimage.
It might be best to make a tree diagram or something of the sort if you're not familiar with permutation groups or such. Of the three elements you can map the element $1$ to, you pick one of them in the first branch. In each branch, you're left with two different options for which to map $2$ to (you cannot map to what you map $1$ to because it won't be injective), and in the final branch only one option to map $3$ to. Once you have that all figured out, you should be able to follow the paths in tree diagram to find your definition for a bijective $f$.
answered Nov 26 at 4:49
Eevee Trainer
2,561221
Think of what a bijective function does. Each element in the domain maps uniquely to one element of the codomain, and each element of the codomain will have a preimage.
It might be best to make a tree diagram or something of the sort if you're not familiar with permutation groups or such. Of the three elements you can map the element $1$ to, you pick one of them in the first branch. In each branch, you're left with two different options for which to map $2$ to (you cannot map to what you map $1$ to because it won't be injective), and in the final branch only one option to map $3$ to. Once you have that all figured out, you should be able to follow the paths in tree diagram to find your definition for a bijective $f$.
answered Nov 26 at 4:49
Eevee Trainer
2,561221
answered Nov 26 at 4:49
Eevee Trainer
2,561221
answered Nov 26 at 4:49
Eevee Trainer
2,561221
answered Nov 26 at 4:49
Eevee Trainer
2,561221
2,561221
add a comment |
add a comment |
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Do you know about groups and permutation groups ??
– Anik Bhowmick
Nov 26 at 4:46