Question: Elements of Order 2 in Finite Abelian Group

[Nebula]

I've got a question. It pertains to a proof I'm doing. I ran into this stumbling block. If I could show this I think I could complete the proof.

G is a finite Abelian Group such that there exits more than one element of order 2 within the group.

more than one element of the form b not equal to identity
such that b^2=e

Is the product of all the elements of order 2 equal to the identity element, and why?

 

[Hurkyl]

Well, the set of everything such that b^2 = e forms a subgroup of the original group... can you figure out its decomposition into a product of cyclic groups?

 

[Nebula]

Well, the set of everything such that b^2 = e forms a subgroup of the original group...

Thats what I thought.

Now as for the cyclic group part. I suppose it has a decomposition. It's very general. I'm not dealing with any specific group. Just a subset of the group in which all elements are order 2.

...I'm not sure here... I had thought about the cyclic group. Say you had a decompositon into a product of cyclic groups. You'd have a generator it would have to generate elements of order 2... man I'm lost... nevermind...

 

[matt grime]

If x has order 2, then so does x^{-1}. So if yo'ure multiplying all the elements of order 2 together, and reorder them you get...

 

[Hurkyl]

If x has order 2, then x = x^-1, so this argument doesn't work.


Anywys, you aren't dealing with a very general group... you're dealing with groups in which every element has order 1 or 2. This is fairly unusual, you should be able to use this fact to completely determine the form of its factorization into a product of cyclic groups.

 

[matt grime]

yep, my head must be screwed up this morning, sorry.

 

[Nebula]

If all the elements have order two then doesn't each individual element generate a cyclic group? Either that or the identity paired with every element forms a group.

How does that tell us anything about the product of elements?

 

[matt grime]

any element in any group generates a cyclic subgroup.

what, i think, hurkyl was getting at was that a group where every element has order 2 (or is the identity) can only be one of a certain kind of group, namely an elementary abelian group.

Consider some examples of groups with the property you're interested in:

let H= the cyclic group with 2 elements,

HxH satisfies your condition, what is the product of all the elements there?

HxHxH also satisfies your condition, what about te product of all elements then?

 

[NateTG]

Now as for the cyclic group part. I suppose it has a decomposition. It's very general. I'm not dealing with any specific group. Just a subset of the group in which all elements are order 2.

...I'm not sure here... I had thought about the cyclic group. Say you had a decompositon into a product of cyclic groups. You'd have a generator it would have to generate elements of order 2... man I'm lost... nevermind...

What would a minimal generating subset of a group that contains only elements of order 1 or 2 look like?

 

[Hurkyl]

If you know the cyclic decomposition of your group, you can then write down every element of that group, and then the product of every element of that group, and show it's equal to 1.

For instance, in HxH, the elements are a^m b^n where a and b are the two generators, and m and n are in {0, 1}.

 

[robert Ihnot]

I don't really know this at all, BUT if we have only one element of order two, then the product e*a =a, BUT that case is omitted because the group has, as a given,at least two distinct elements of order 2.

In the case of a,b, we have a 4 group: 1,a,b,ab. ab can not be equal to a or b, since ab =a implies b=e. And if ab=e =aa, that implies that b=a, a contradiction. (This is the Klein group.) In this case, obviously: e*a*b*ab =e.

Now if we have at least three distinct elements or order 2, then by the same reasoning above we have 2^3 elements: e,a,b,c,ab,ac,bc,abc. Clearly ab not equal to a, to b, not c bcause c is distinct, not ab=e =aa, etc. Thus the product is:

e*a*b*c*ab*ac*bc*abc = e.

So proceeding in the same manner, we can handle any 4 distinct elements of order 2, etc. Since we have an Abelian group, it would seem that any other construction of n distinct elements would be isomorphic to the ones by the above construction.