Proof Problem: Show f(x-1)=f(x) for All x ∈ E

  • Context: Graduate 
  • Thread starter Thread starter Cyannaca
  • Start date Start date
  • Tags Tags
    Proof
Click For Summary

Discussion Overview

The discussion revolves around a proof problem involving a group homomorphism and the properties of elements in a group. Participants are tasked with showing that for all elements x in a set E, the equation f(x-1) = f(x) holds, given that x*x is in the kernel of the homomorphism f. The scope includes mathematical reasoning and proof strategies related to group theory.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant requests assistance in starting the proof and questions whether the kernel of f can be assumed to be the identity element.
  • Another participant clarifies that if x*x is in the kernel of f, then f(x*x) equals the identity element of H, suggesting that properties of homomorphisms can be utilized to show f(x*y) = f(x)#f(y) and f(x^{-1}) = (f(x))^{-1}.
  • A different participant points out that the notation used for inverses should be consistent and that one cannot assume the kernel of f is the identity, as this would imply f is injective, which is not established.
  • There is a clarification regarding the notation, with one participant suggesting that f(x-1) is intended to mean f(x^(-1)), which simplifies the problem.
  • Another participant notes that if f(x*x) equals the identity, then x must be its own inverse, emphasizing the uniqueness of inverses in group theory.

Areas of Agreement / Disagreement

Participants express differing views on the notation and assumptions regarding the kernel of f. While there is some agreement on the properties of homomorphisms, the discussion remains unresolved regarding the initial assumptions and the correct interpretation of the notation.

Contextual Notes

There are limitations in the clarity of notation and assumptions about the kernel of f, which could affect the proof's approach. The discussion also highlights the need for precise definitions in group theory.

Cyannaca
Messages
22
Reaction score
0
I would really appreciate if anyone could help me with this problem.

F is a group homomorphism from G= (E, *) to H= (F,#).
If , for all x e E , x*x e ker(f).
Show that for all x e E, f(x-1)=f(x)

Now I don't know how I should start the proof. Also, I would like to know if I can assume that Ker(f) is equal to the identity element.
 
Physics news on Phys.org
If [itex]x*x \in Ker(f)[/itex] then this means precisely that [itex]f(x*x) = 1_F[/itex] where [itex]1_F[/itex] is the identity element of H. You want to show something like:

[tex]f(x*y) = f(x)#f(y)[/tex]

and

[tex]f(x^{-1}) = (f(x))^{-1}[/tex]

In fact, you probably already have these proved somewhere. If you do, then use them and the rest should be very simple. Actually, I don't remember the definitions, but I have a feeling that the first thing (that f(x*y) = f(x)#f(y)) might just be a property of homomorphisms. The second property follows from the first at least in the case where x*x is in Ker(f), and probably in general too.
 
You've got F as a homomorphism, and the underlying set of H (at least that's what I presume H=(F,#) means. Usually there is no reason to distinguish the group from its elements like that. Also, if you're not going to tex it then it's common to indicate inverses by putting the minus 1 in an exponent, since x-1 doesn't really make sense for a group. You may not assume ker(f) is the identity element, since that is saying f is an injection which is not given.

From the definition y in ker f iff f(y)=e conclude something about f(xx), and use AKG's stuff above which is the (usual) definitoin of what a homomorphism is.

And remember showing that y is the inverse of x is showing exactly that xy=yx=e.
 
Matt, I believe he/she meant f (lower case) to be the homomorphism, and F to be the set underlying group H.
 
Is that what the guy is talking about, F(x-1) is actually F(x^(-1))? Just written differently? No wonder the problem look mysterious! This is a completely simple problem now that it is written so it can be read! If F(x*x) goes to 1, then x is its own inverse, since by group property there can be only a unique inverse.

You got to be careful about how you write the inverse if you use a calculator, you know!
 
Last edited:

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 26 ·
Replies
26
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
48
Views
7K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 14 ·
Replies
14
Views
4K