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

[arpitm08]

Homework Statement

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

 

[sponsoredwalk]

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}?

 

[gopher_p]

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?

 

[SteveL27]

Homework Statement

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

 

[gopher_p]

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 $S$.

 

[arpitm08]

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?

 

[gopher_p]

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?

 

[arpitm08]

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.

 

[gopher_p]

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.

 

[arpitm08]

g in A of S such that g of s2 is s2

 

[arpitm08]

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

 

[gopher_p]

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 $s_2$.

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 $s_2$.

 

[gopher_p]

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 $f^{-1}\cdot f=f\cdot f^{-1}=id$

 

[arpitm08]

Okay. I understand what you're saying so far. When they say g ∈ K, what do they exactly mean with that then?

 

[gopher_p]

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.

 

[arpitm08]

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?

 

[gopher_p]

Okay, so they could've used m or n or whatever instead of g. Okay.

Good

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.

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 $H$?

 

[arpitm08]

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?

 

[gopher_p]

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, $(f^{-1}gf)(s_1)=f^{-1}\left(g\left(f(s_1)\right)\right)$.

 

[arpitm08]

thanks for the answer jerk

 

[gopher_p]

thanks for the answer jerk

okay ...

 

[arpitm08]

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