Proving Even Cardinality of a Set w/o Inverse Elements in G

In summary: I realize that there has to be at least one g in G-t(G) that isn't equal to e (as it's even), but it could be of any...order?
  • #1
Firepanda
430
0
Show that the Cardinality of a set which doesn't include inverse elements from a group G is always even.

So the set with all g in G, which includes no inverse elements from G. (g=!g^-1)

I can get this in every example I've done, checking mainly with dihedral groups, it's always been true but I can't find a pattern.

I know that the neutral element can't be in the set, so that's one down, then I thought maybe halfing it, which is wrong I know, as I kept finding cases where the element itself was it's own inverse.

I've really no idea where to go from here.
 
Physics news on Phys.org
  • #2
If you mean the set H of all g in G such that g!=g^(-1), just think about grouping the elements in H in pairs, like {h,h^(-1)}.
 
  • #3
Dick said:
If you mean the set H of all g in G such that g!=g^(-1), just think about grouping the elements in H in pairs, like {h,h^(-1)}.

Yeah that's what I mean.

But how do I group an element in a pair if the inverse of that element is itself?
 
  • #4
Firepanda said:
Yeah that's what I mean.

But how do I group an element in a pair if the inverse of that element is itself?

H doesn't contain any elements that are inverses of themselves, does it?
 
  • #5
Dick said:
H doesn't contain any elements that are inverses of themselves, does it?

Well H doesn't, but G does, and those are the elements (like the neutral element) that I'm trying to exclude.

Like the dihedral group D6 of a triangle, the symmetry element is it's own inverse.. I think.

In effect I'm trying to find a way to count all the elements in H after I've excluded all the inverse elements from G, and see if it's divisible by 2.
 
  • #6
You are confusing me. You are trying to count H, right? Not G. The definition of H excludes all elements of G that are their own inverse. Why are you worried about them?
 
  • #7
Dick said:
You are confusing me. You are trying to count H, right? Not G. The definition of H excludes all elements of G that are their own inverse. Why are you worried about them?

Well the question is this (part b hint)

t0q642.jpg


If I don't exclude them somehow from G and then count I don't see how I can prove it's even.
 
  • #8
You know t(G) has even order, right? If G has even order then G-t(G) also has even order, right? You know e is in G-t(G), right? That means G-t(G) must contain an element g that is not equal to e. What's the order of g?
 
  • #9
Dick said:
You know t(G) has even order, right? If G has even order then G-t(G) also has even order, right? You know e is in G-t(G), right? That means G-t(G) must contain an element g that is not equal to e. What's the order of g?

Hmm. How did you get the order of t(G) was even? I'm not familiar with the order of a set, I know the order of a group is its cardianlity, and I know the definition of the order of an element, but not that lol.

By G-t(G) do you mean literally G minus t(G)?
 
  • #10
I know t(G) is even because I grouped the elements in pairs, like we were discussing earlier. G-t(G) just means the set of elements in G that aren't in t(G). By order of a set, I mean it's cardinality.
 
  • #11
Dick said:
I know t(G) is even because I grouped the elements in pairs, like we were discussing earlier. G-t(G) just means the set of elements in G that aren't in t(G). By order of a set, I mean it's cardinality.

That's what I'm not getting, you grouped the elements of t(G) as (h,h^-1), but I don't see any h^-1 in the set, as I thought that because we excluded all the inverses then the elements don't have inverses in the new set, so I don't see how they can be grouped.
 
  • #12
t(G) is the set of all elements that AREN'T their own inverses. If h isn't it's own inverse then h^(-1) isn't it's own inverse either. If h is in t(G), so is h^(-1).
 
  • #13
Dick said:
t(G) is the set of all elements that AREN'T their own inverses. If h isn't it's own inverse then h^(-1) isn't it's own inverse either. If h is in t(G), so is h^(-1).

Oooh now I get it, I thought it meant there were no inverses, not specifically only the inverses that were the same element!

I get this line now:

'You know t(G) has even order, right? If G has even order then G-t(G) also has even order, right? You know e is in G-t(G), right?'

As for this:

'That means G-t(G) must contain an element g that is not equal to e. What's the order of g?'

I realize that there has to be at least one g in G-t(G) that isn't equal to e (as it's even), but it could be of any order surely, depending on the group.
 
  • #14
If g is in G-t(G) and isn't e, then it is it's own inverse. How could it be of any order? Repeat, it is it's own inverse.
 
  • #15
Dick said:
If g is in G-t(G) and isn't e, then it is it's own inverse. How could it be of any order? Repeat, it is it's own inverse.

I know you're probably hinting the order of g is 2 lol, but still..

I can see how it applys to this dihedral group D6, as I have s.r (rotation then symmetry) as my element of order 2, but I can't see why that order is special.
 
  • #16
If g is it's own inverse, what is g^2? Sigh.
 
  • #17
Dick said:
If g is it's own inverse, what is g^2? Sigh.

Oh then it's g again!

Sorry.. no excuses

Thanks for all the help!
 
  • #18
No, it's not g, g^2=e, isn't it? That's what you meant, right?
 
  • #19
Dick said:
No, it's not g, g^2=e, isn't it? That's what you meant, right?

Yes that's what i meant!

Sorry it's half past midnight here :D
 
  • #20
Firepanda said:
Yes that's what i meant!

Sorry it's half past midnight here :D

I guess that explains a certain slowness. Good job on checking the dihedral groups though. It's always a good idea to check concrete examples before trying to prove something.
 

What is the definition of "Proving Even Cardinality of a Set w/o Inverse Elements in G"?

"Proving Even Cardinality of a Set w/o Inverse Elements in G" refers to the mathematical process of determining the number of elements in a set, where the set does not contain any inverse elements, within a given mathematical structure or group G.

Why is proving even cardinality important in mathematics?

Proving even cardinality is important in mathematics because it allows us to understand and analyze the structure and properties of a given set or group. It helps us determine the symmetry and balance of the set, which can be useful in solving various mathematical problems.

What are some methods for proving even cardinality of a set without inverse elements in G?

There are several methods for proving even cardinality of a set without inverse elements in G, including counting the elements in the set, using mathematical induction, and using properties of even numbers such as divisibility by 2.

What are some real-world applications of proving even cardinality?

Proving even cardinality has various real-world applications, such as in computer science for analyzing algorithms and data structures, in economics for analyzing market trends and supply and demand, and in physics for understanding symmetries and conservation laws.

Is it possible to prove even cardinality without inverse elements in any type of set or group?

No, it is not possible to prove even cardinality without inverse elements in every type of set or group. This method of proof is specific to sets or groups where inverse elements do not exist, and may not be applicable to other types of mathematical structures.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
492
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
23
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
981
Back
Top