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

1. May 21, 2012

### arpitm08

1. The problem statement, all variables and given/known data

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

2. Relevant equations

I don't know.

3. 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$\in$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.

2. May 21, 2012

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. May 21, 2012

### 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?

4. May 21, 2012

### SteveL27

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. May 21, 2012

### gopher_p

I'm guessing he means the automorphism group of $S$.

6. May 22, 2012

### 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?

7. May 22, 2012

### 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?

8. May 22, 2012

### 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.

9. May 22, 2012

### 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.

10. May 22, 2012

### arpitm08

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

11. May 22, 2012

### arpitm08

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

12. May 22, 2012

### gopher_p

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 $s_2$.

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. May 22, 2012

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

14. May 22, 2012

### arpitm08

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

15. May 22, 2012

### gopher_p

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. May 22, 2012

### 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?

17. May 22, 2012

### gopher_p

Good

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.

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. May 22, 2012

### 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?

19. May 22, 2012

### gopher_p

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. May 23, 2012