Proving Even Order Group G Has Element a=/e Satisfying a^2=e

[fk378]

Homework Statement

If G is a group of even order, prove it has an element a=/ e satisfying a^2=e.

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?

 

[Dick]

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.

 

[Dick]

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?

 

[fk378]

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

 

[Dick]

{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.

 

[fk378]

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?

 

[fk378]

I got some hints from the TA:

Make 2 equivalence relations, one where a~a and another a~a-inverse
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.

 

[Dick]

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?

You assumed a^2=e for any a in G. That means a=a^(-1). That means {a,a^(-a)} has only one element. It's not a subgroup since if a!=e, e isn't in it. The subgroup is was referring to is {e,a}.

 

[Dick]

I got some hints from the TA:

Make 2 equivalence relations, one where a~a and another a~a-inverse
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.

I've begun to realize I'm confusing the original question with one you subsequently posed. Did you get the original question? Yes, pair a with a^(-1) for all elements in the group. e pairs with itself. If no other element pairs with itself then the order of the group must be odd. Right?

 

[fk378]

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

 

[Dick]

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.

 

[fk378]

I agree.

 

[Dick]

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

 

[fk378]

Yes, still following.

 

[Dick]

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?

 

[fk378]

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?

 

[Dick]

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.

 

[fk378]

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

 

[Dick]

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

Absolutely, then the order of G is 1 plus an even number isn't it? What could be wrong with that?

 

[fk378]

That would imply that the order of G is odd.

 

[Dick]

Right. You started by assuming the order of G is even. Can a be one?

 

[fk378]

a can be one. G can be of odd order...
But how does that relate to the question asking if G is of even order then there is an a=/e s.t. a^2=e?

 

[Dick]

IF G IS OF EVEN ORDER CAN a BE ONE? Answer me! The fact G 'could' of odd order in the world of all possible groups has nothing to do with this. We are talking about the world of groups of even order!

 

[fk378]

NO! If G is of even order, then a cannot be one.

 

[fk378]

How did you know that each set is just composed of at most 2 elements though (besides the hint I gave). You could have had a bigger set than just {a,a^-1}.

 

[HallsofIvy]

No, no one has said "G was made up of sets of 2 elements"! What Dick said before was that if a= a^-1, then {a, a^-1} contains only a single member because those (seeming) two elements are really the same. If a is NOT a^-1, then it has 2 members. Of course, any subgroup must contain the group identity, e. If either a or a^-1 is the group identity, then the other must be: a(a^-1)= e by definition of inverse. If a= e, then e(a^-1)= e or a^-1= e. That is the only case in which {a, a^-1} forms a subgroup: a= e so {a, a^-1} is the trivial subgroup, {e}. In any other case, whether a=a^-1 or not, {a, a^-1} is NOT a subgroup because it does not contain the group identity.

Now, suppose G is a set containing an even number of members. We know that the group identity, e, is a member of G and that e= e^-1. That leaves an odd number of "non-identity" members. If there were no member a\ne e such that a= a^-1, we can pair each non-identity member with its inverse: {a, a^-1}. But such a pairing partitions G-{e}, as set with an odd number of members into a collection of sets each containing two members. Do you see what's wrong with that?

Notice that this only says "G has a member a such that a= a^-1". It does not say that a is the only such member! The Klein 4 group, for example, has a= a^-1 for all its members.

G must have even order because Lagrange's theorem says that the order of any subgroup of G must divide the order of G. In particular, if a= a^-1, then {a, e} (NOT {a, a^-1}!) is a subgroup of order 2 and so the order of G must be even.

 

[fk378]

But such a pairing partitions G-{e}, as set with an odd number of members into a collection of sets each containing two members. Do you see what's wrong with that?

I'm a bit confused by the wording of this statement. Do you mean that since we now have |G|=odd, and then we multiply that number of sets with (2), then the order will actually become even?

 

[Dick]

I'm confused by your confusion. We are 27 posts into a not terribly hard problem and I'm about ready to give up. I'll recap. We've split G into disjoint subsets which have size two (if a!=a^(-1)) or size one (if a=a^(-1)). We know there is at least one set of size one, {e}. IF G IS EVEN (and don't take caps lightly) can there be only one set of size one? That's all I have to say. Finally and forever.

 

[Tobias Funke]

Hasn't everybody been confused by something that other people see as easy? To someone new to proofs, some of these responses, even if well-intentioned, may seem quite condescending. Some of them so much that it's almost certainly intentional.

 

[Dick]

Hasn't everybody been confused by something that other people see as easy? To someone new to proofs, some of these responses, even if well-intentioned, may seem quite condescending. Some of them so much that it's almost certainly intentional.

Tobias, this isn't condescension. It's frustration. Read the posts. Do you think I haven't been patient? At this point all I can think of to do is go back and repeat previous posts. And I'm tired of that because most of them are already repetitions. If you have a better idea and think you can do a better job, then please try it. Jeez. This isn't even about l'Hopital's theorem and you still find the time to criticize. Why don't you try to help? I find you condescending.

 

[HallsofIvy]

I'm a bit confused by the wording of this statement. Do you mean that since we now have |G|=odd, and then we multiply that number of sets with (2), then the order will actually become even?

If I understand what you are saying here, yes, that's the basic idea. IF G contains no member a, other than the identity e, such that a^-1= a, then we can pair all non-identity members, {a, a^-1}. If we multiply the number of such sets by 2, we get the number of non-identity members which must be even. Since there is one identity, e, the total number of members of G, |G|, must be odd.

Here's an example. Suppose we have a group with 4 members, e, x, y and z where e is the group identity. Suppose further that there is no member, other than e, such that a^-1= a. e is its own inverse. What is the inverse of x? Since x is not its own inverse, it must be either y or z. Suppose it is y: we can pair {x,y}. Now that leaves z. z is not, by hypothesis, its own inverse, but there is nothing else left to pair it with!

It occurs to me that this requires we know that the inverse of z cannot be either x or y because they are "already paired"- theorem: if a^-1= b and c^-1= b, the a= c. Do you see why that is true? Suppose a^-1= b and c^-1= b. Look at a(cb) and (ac)b.