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Even order of Gby fk378
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#1
Sep1608, 08:50 PM

P: 367

1. The problem statement, all variables and given/known data
If G is a group of even order, prove it has an element a=/ e satisfying a^2=e. 3. The attempt at a solution I showed that a=a^1, ie a is its own inverse. So, can't every element in G be its own inverse? Why does G have to be even ordered? 


#2
Sep1608, 10:36 PM

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The problem says "IF G is group of even order", that's why G has even order. The question of whether there is a nontrivial group of odd order such that a^2=e for all a is a completely different question. Do you want to try and solve it? You don't have to since you already solved the original question. Didn't you? Though you didn't really say how.



#3
Sep1608, 11:22 PM

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If you want to know, then if a^2=e for a not equal to e, the G has a subgroup of order 2. Doesn't it? What does that mean?



#4
Sep1608, 11:32 PM

P: 367

Even order of G
Subgroup of order 2, then the subgroup has 2 elements. a and ainverse. But can't there exist a b with a binverse in the subgroup as well? Why are there only 2 elements?



#5
Sep1708, 08:31 AM

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{a,a^(1)} isn't a subgroup. And it doesn't even have two elements. It's certainly not what I was considering. I don't think you are really thinking about this.



#6
Sep1708, 09:28 PM

P: 367

I'm sorry, I'm new to proofs. Can you explain your last post? Why are there not 2 elements in {a, a^1} and why is this not a subgroup?



#7
Sep1708, 09:33 PM

P: 367

I got some hints from the TA:
Make 2 equivalence relations, one where a~a and another a~ainverse Show that every class size is 2 or 1 Show that the sum of the sizes of the equivalence classes=the size of the group If what the question is asking is false, prove it is equivalent to "for all a =/ 0, the size of a equals 2." This gives a contradiction to the size of G being even. 


#8
Sep1708, 10:16 PM

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#9
Sep1708, 10:29 PM

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#10
Sep1708, 10:46 PM

P: 367

Not quite following...in fact I don't even know how to use the hints the TA gave me.



#11
Sep1708, 10:52 PM

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Ok, let's go back to the original question. I thought you said you had it and you don't. Forget all of the stuff in between, please? Think of all of the sets {a,a^(1)} for a in G. Every element is in ONE of those sets, yes? And G is the union of all of those sets, also ok? If you don't agree, say so now.



#12
Sep1708, 11:09 PM

P: 367

I agree.



#13
Sep1708, 11:18 PM

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I hope so. Now if a=a^(1), then the set {a,a^(1)} has one element, otherwise it has two. Still with me?



#14
Sep1708, 11:19 PM

P: 367

Yes, still following.



#15
Sep1708, 11:26 PM

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Great! So G is the union of a bunch of sets. Some have two elements, some have one. Now {e} is one of those sets, since e=e^(1). And {e} only has one element. Suppose n1 is the number of sets that have one element and n2 is the number that have two elements. Then the number of elements in G is n1*1+n2*2. Given that the number of elements in G is even, is it possible n1=1?



#16
Sep1708, 11:29 PM

P: 367

Oh, I'm lost here. I thought G was made up of sets of 2 elements, not two and one. I understand that e is one of the sets of one element, though, you implied that there are more.
Where did you get n2*2 from? How do you know there are 2 sets with 2 elements in them? 


#17
Sep1708, 11:35 PM

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Sorry. I meant n1 and n2 to be different numbers. Try this "Suppose a is the number of sets that have one element and b is the number that have two elements. Then the number of members of G is a*1+b*2. Given that the number of members of G is even, is it possible a=1?" If G has an member such that g=g^(1), then {g,g^(1)} has only one element. There may be other members that belong to one element sets.



#18
Sep1808, 12:02 AM

P: 367

it is possible for a=1, right? e could be the only element which is its own multiplicative inverse.



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