Help with proof involving the set of 1-1 mappings of S onto itself.

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Okay. So we are saying then that g is an element of K but can we use that to prove that the inverse of f times g times f is in H?Okay. So we are saying then that g is an element of K but can we use that to prove that the inverse of f times g times f is in H?The elements of ##K## are automorphisms of ##S## that have the additional property that they fix ##s_2##. I've told you this three times now. You don't prove that the inverse of ##f\cdot g\cdot f^{-1}## is in ##H##. You prove that ##f\cdot g\cdot f^{-1}## is in ##H
  • #1
arpitm08
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Homework Statement



Suppose that s1≠s2 are in S and f(s1)=s2, where f[itex]\in[/itex]A(S). Then if H = (f[itex]\in[/itex]A(S)|f(s1)=s1)) and K = (g[itex]\in[/itex]A(S)|g(s2)=s2) show that:
a) If g[itex]\in[/itex]K, then f^-1 *g *f [itex]\in[/itex] H

Homework Equations



I don't know.

The Attempt at a Solution



I don't really know where to go with this problem. I don't understand what it means that g[itex]\in[/itex]K if in the definition of K involves g like that, shouldn't that be implied? Does it just mean that g(s2)=s2?

Then how do I figure out the inverse of f? Since f(s1)=s2, would the inverse just be f^-1(s2)=s1? Then how would I got about multiplying inverse of f to g and f and proving they are in H??

Thanks in advance for any help.
 
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  • #2
Wait, f is a function such that f(s₁) = s₂ but f is also a function such that f(s₁) = s₁?
Maybe H is supposed to be {(f∈A(S)|f(s₁) = s₂}?
 
  • #3
Restatement of problem using (perhaps) less confusing notation:

Let ##s_1,s_2\in S## with ##s_1\neq s_2##, and put ##H=\{a\in A(S):a(s_1)=s_1\}## and ##K=\{a\in A(S):a(s_2)=s_2\}##. Let ##f\in A(S)## be such that ##f(s_1)=s_2## and ##g\in K##. Show that ##f^{-1}gf\in H##.

Now. What questions do you have?
 
  • #4
arpitm08 said:

Homework Statement



Suppose that s1≠s2 are in S and f(s1)=s2, where f[itex]\in[/itex]A(S). Then if H = (f[itex]\in[/itex]A(S)|f(s1)=s1)) and K = (g[itex]\in[/itex]A(S)|g(s2)=s2) show that:
a) If g[itex]\in[/itex]K, then f^-1 *g *f [itex]\in[/itex] H


Homework Equations



I don't know.

The Attempt at a Solution



I don't really know where to go with this problem. I don't understand what it means that g[itex]\in[/itex]K if in the definition of K involves g like that, shouldn't that be implied? Does it just mean that g(s2)=s2?

Then how do I figure out the inverse of f? Since f(s1)=s2, would the inverse just be f^-1(s2)=s1? Then how would I got about multiplying inverse of f to g and f and proving they are in H??

Thanks in advance for any help.

I didn't look at this in detail but I'm confused by your notation. Your title talks about the set of all 1-1 mappings; but A(S) is typically the alternating group on S. Did you mean S(S), the symmetric group on S?
 
  • #5
SteveL27 said:
I didn't look at this in detail but I'm confused by your notation. Your title talks about the set of all 1-1 mappings; but A(S) is typically the alternating group on S. Did you mean S(S), the symmetric group on S?

I'm guessing he means the automorphism group of ##S##.
 
  • #6
Sorry about the confusion guys. The chapter is about the set of 1-1 Mappings of S onto itself. They had listed the function H in another problem and referenced it in this one, but they seem to be the same I guess. Is that how I go about solving this problem? Is H = k?
 
  • #7
What is the name of the class that this problem comes from?

I don't mean to be rude, but I get the impression that you may not understand some of the notation. Do you understand the problem statement? Do you understand it after I rewrote it above?
 
  • #8
It's an Abstract Algebra class. I somewhat understand the problem. I posted it here because I wasn't sure what it really was asking or how to go about doing it.
 
  • #9
Do you understand the notation? Do you know, for instance, what ##K=\{g\in A(S)|g(s_2)=s_2\}## means? Can you translate the symbols into english (or your native language) words?

Sorry again if these questions seem silly or if it seems as if I'm talking down. I just need a jumping-off point.
 
  • #10
g in A of S such that g of s2 is s2
 
  • #11
Would K and H be equal to each other or does it make a difference when it is s1 and s2?
 
  • #12
arpitm08 said:
g in A of S such that g of s2 is s2

I would call this a literal translation. I would maybe say "The set of automorphisms of ##S## which map ##s_2## to itself" or "The set of automorphisms of ##S## which fix ##s_2##". It really doesn't matter as long as you understand it for what it is: a set of functions that have certain properties related to being automorphisms of ##S## along with the additional property that they all fix ##s_2##.

arpitm08 said:
Would K and H be equal to each other or does it make a difference when it is s1 and s2?

No. ##K## and ##H## are not the same set. They are similar, and they do share some elements. But elements of ##H## fix ##s_1## while elements of ##K## fix ##s_2##.
 
  • #13
OK. Now the group operation in ##A(S)## is composition. For ##f,g\in A(S)##, the function ##f\cdot g## is given by ##(f\cdot g)(s)=f\left(g(s)\right)##. In the algebra classes, we use ##\cdot## to designate a lot of group operations which aren't necessarily multiplication. Sometimes it's convenient, other times it's just there (I'm convinced) to confuse as many people as possible. You need to be aware at all times what the group operation is. Also ##f^{-1}## denotes the compositional inverse of ##f##, i.e. the inverse function, i.e ##f^{-1}\cdot f=f\cdot f^{-1}=id##
 
  • #14
Okay. I understand what you're saying so far. When they say g ∈ K, what do they exactly mean with that then?
 
  • #15
arpitm08 said:
Okay. I understand what you're saying so far. When they say g ∈ K, what do they exactly mean with that then?

It means that ##g## is an automorphism of ##S## with the additional property that ##g(s_2)=s_2##.

And this is the potentially confusing part: It would make sense (and is in fact true) to write ##g\in\{g\in A(S)|g(s_2)=s_2\}## where the ##g## on the left means something different than the ##g## in the set description. The ##g## in the set description is used to denote an arbitrary automorphism of ##S## while the ##g## on the left denotes a specific automorphism in ##K##. I know, it's confusing if you're not used to it. That's why I say "automorphisms of ##S##" instead of "##g## in ##A(S)##", and that's why I changed up the notation several posts back.
 
  • #16
Okay, so they could've used m or n or whatever instead of g. Okay. So my next questions is about the inverse of f. Let's assume the inverse is h. Then h(s2)=s1, I think. But how do I go about finding out what f−1gf is?
 
  • #17
arpitm08 said:
Okay, so they could've used m or n or whatever instead of g. Okay.

Good

arpitm08 said:
So my next questions is about the inverse of f. Let's assume the inverse is h. Then h(s2)=s1, I think.

Yes. But why not just call it ##f^{-1}##? If you want to see how I'm getting my fancy writing, just quote the post and take a peek. It's not too hard.

arpitm08 said:
But how do I go about finding out what f−1gf is?

Well we don't have enough information to know what ##f^{-1}gf## is. But we're not being asked to find what the function is. We're being asked to show that it is an element of ##H##. Presumably you know that ##f^{-1}gf\in A(S)##? So we kind of get that one "for free". What else do we need to know about ##f^{-1}gf## in order to say that it is an element of ##H##?
 
  • #18
I understand that ##f^{-1}##gf∈A(S), because g, f ∈ A(S).

We need to determine that it sets s1. How would we go about that though?
 
  • #19
arpitm08 said:
I understand that ##f^{-1}##gf∈A(S), because g, f ∈ A(S).

We need to determine that it sets s1. How would we go about that though?

Well, we could maybe try to compute ##(f^{-1}gf)(s_1)##?

Remember, ##(f^{-1}gf)(s_1)=f^{-1}\left(g\left(f(s_1)\right)\right)##.
 
  • #20
thanks for the answer jerk
 
  • #21
arpitm08 said:
thanks for the answer jerk

okay ...
 
  • #22
Sorry, my friend is an idiot sometimes. I really appreciate your help on everything. It helped a ton. Sorry about my friend again.
 

1. What is a 1-1 mapping?

A 1-1 mapping, also known as a one-to-one mapping or a bijection, is a function that maps each element of one set to a unique element in another set. In other words, for every input, there is only one corresponding output.

2. What is the set of 1-1 mappings of S onto itself?

The set of 1-1 mappings of S onto itself, denoted as 1-1(S), is the set of all functions that map the elements of set S to themselves in a one-to-one manner. This means that each element in S has a unique image in S.

3. How do you prove that a mapping is 1-1?

To prove that a mapping is 1-1, you need to show that for every two distinct elements in the domain, the corresponding elements in the range are also distinct. This can be done by using a proof by contradiction or by showing that the inverse of the function exists.

4. Can a set have multiple 1-1 mappings onto itself?

Yes, a set can have multiple 1-1 mappings onto itself. For example, the set of real numbers can have multiple 1-1 mappings onto itself, such as the identity function, the square function, and the cube function.

5. How is the set of 1-1 mappings of S onto itself related to the concept of symmetry?

The set of 1-1 mappings of S onto itself is closely related to the concept of symmetry. A function is 1-1 if and only if it preserves symmetry. This means that if a set has a symmetry property, then there exists a 1-1 mapping that preserves that symmetry.

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