A group of order 2n conatins an element of order 2

[guildmage]

Homework Statement

Let G be a group of order 2n. Show that G contains an element of order 2. If n is odd and G is abelian, then there is only one element of order 2

Homework Equations

Theorem (Lagrange):
If H is a subgroup of a group G, then |G|=[G<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />]|H|. In particular, if G is finite, then the order of a divides G for all a \in G

The Attempt at a Solution

Case 1. G is cyclic.

This means that G \cong Z_{2n}.

We know that n + n = 0 for n \in Z_{2n}.

Thus there is an element of order 2 in G.

Case 2. G is not cyclic.

This is where I'm stuck.

I also don't know how to show the next statement when G is abelian and n is odd.

 

[guildmage]

The order of a group G, denoted |G| is the number of elements in G.
The order of an element a in G, denoted |a| is the least integer n such that a^{n}=e where e is the identity element in G.

Yes, you are correct, Z_3 has no elements of order 2 since 3 cannot be expressed as 2n for an integer n. What I want to show when G is not cyclic is that there exists an element in G of order 2. The special case when G is abelian and n is odd, on the other hand, wants uniqueness of that element.

I believe that being able to solve Case 2 (G abelian, n odd) will help me solve the special case.

I did not arrive at a relevant conclusion by assuming the contrary (that there does not exist an element of order 2 in G).

What do you suggest I do?

 

[Dick]

I just realized I was reading the question wrong, sorry for the previous post. Use the structure theorem for abelian groups. An abelian group is the direct sum of cyclic groups. In your case, only one of those cyclic groups has even order. What's it's order?

 

[guildmage]

But I think we are not allowed to use that concept yet. The only concepts available to me, as far as the problem is concerned, are the definition of groups, homomorphisms and subgroups, cyclic groups, and cosets and counting (where this problem is part of).

So what I did was to use two cases:
1. when G is cyclic -- this would mean that G is isomorphic to Z_2n which is automatically abelian. I have shown here that regardless of whether or not n is odd, G will only have one element of order 2.

2. when G is not cyclic -- I don't know where I would start

Any ideas?

 

[Dick]

Ok, if G is not cyclic. You know if G has even order, it has at least one element of order 2, right? Suppose it has two elements of order 2, 'a' and 'b'. Consider the order of the subgroup generated by {a,b} and use that G is abelian. What's the order of that subgroup?

 

[guildmage]

Ah... I see where this should be going.

I decided to scrap the idea of using two cases. Instead, I have proven that if G is of order 2n, then G has an element of order 2 by partitioning G into two -- one partite cell containing the elements whose inverse is itself; and the other, those elements that are not inverses of themselves. I then showed that the set containing the elements of G that are inverses of themselves should be of even order and thus must contain an element other than the identity.

Tell me if I got your hint correct.

Suppose that there are more than one element in G of order 2, G is abelian and n is odd.
The subgroup of G (I was able to show that it is indeed a subgroup) that contains all the elements that are inverses of themselves has order 2^{k} for k\geq2. And this should divide 2n (which is the order of G). But n is odd and so 2^{k} cannot divide the order of G. A contradiction.
So, there is only one element in G of order 2 if G is abelian and n is odd.

Is that where you are leading me?

 

[Dick]

My goals were more modest, I was just suggesting if you had two elements of order 2 in an abelian group, then you should be able to prove G has a subgroup of order 4. I don't really need the 2^k. 2^2 is enough.

 

[guildmage]

Yeah. I got that. I wanted it to be more general, though.

Thanks a lot!

 

[HallsofIvy]

Surely this is simpler than that. If G has even order, since e, the identity, is its own inverse, that leaves an odd number of non-identity members. If we tried to pair up every element with its distinct inverse we would have one left over. Therefore, there must be at least one member, a, such that a= a^-1 so a²= 1.

 

[Dick]

Surely this is simpler than that. If G has even order, since e, the identity, is its own inverse, that leaves an odd number of non-identity members. If we tried to pair up every element with its distinct inverse we would have one left over. Therefore, there must be at least one member, a, such that a= a^-1 so a²= 1.

The problem is to show G has EXACTLY one element of order 2.

 

[guildmage]

My teacher returned my assignment to me. She pointed out I made a mistake with the proof regarding my idea that the subgroup of G containing all elements of order 2 together with the identity has 2^k elements. From how I see it now, this idea could not be true. (Or is it?)

Dick has suggested that if there are two elements of order 2, then they generate a subgroup (of G) of order 4 (which does not divide 2n, n odd). This of course contradicts the theorem of Lagrange.

My question now is: Is this contradiction enough to ensure the uniqueness of the element of order 2 in G? I mean, surely there cannot be two elements of order 2 (as given by Dick's hint), but what about if there were more? Can I really just show this and state that it is done without loss of generality?

 

[Dick]

My teacher returned my assignment to me. She pointed out I made a mistake with the proof regarding my idea that the subgroup of G containing all elements of order 2 together with the identity has 2^k elements. From how I see it now, this idea could not be true. (Or is it?)

Dick has suggested that if there are two elements of order 2, then they generate a subgroup (of G) of order 4 (which does not divide 2n, n odd). This of course contradicts the theorem of Lagrange.

My question now is: Is this contradiction enough to ensure the uniqueness of the element of order 2 in G? I mean, surely there cannot be two elements of order 2 (as given by Dick's hint), but what about if there were more? Can I really just show this and state that it is done without loss of generality?

Sure. If there's more than two, pick any two and show they generate a subgroup of order 4. How would you do that?