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Group theory

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  • #1
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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 thats 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.
 

Answers and Replies

  • #2
Dick
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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
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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
Dick
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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
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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
Dick
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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
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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
Dick
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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
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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
Dick
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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
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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 dont have inverses in the new set, so I don't see how they can be grouped.
 
  • #12
Dick
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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
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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 realise 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
Dick
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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
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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
Dick
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If g is it's own inverse, what is g^2? Sigh.
 
  • #17
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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
Dick
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No, it's not g, g^2=e, isn't it??? That's what you meant, right?
 
  • #19
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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
Dick
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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.
 

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