Group with exactly one element of order 2.

[Kreizhn]

Homework Statement

Suppose G is a finite group containing precisely one element of order 2. Call this element f. Show that $h= \prod_{g \in G} g$ is actually f.

The Attempt at a Solution

Since f has order 2, it must be in the center of G, and hence commutes with all other elements. It is sufficient to show that $h^2 = e$ or equivalently $h = h^{-1}$ by uniqueness of f.

I've been playing around with this, doing things like playing with $(fhf)^2$. Haven't really been able to put 2 and 2 together though. Someone want to throw me in the right direction?

 

[jbunniii]

It is sufficient to show that $h^2 = e$ or equivalently $h = h^{-1}$ by uniqueness of f.

Not true, that shows that the order of h is either 1 or 2. You also have to exclude 1 as a possibility.

 

[Kreizhn]

Yes, very true. I can worry about that case afterwards, or is this a suggestion that showing those is not the path to take?

 

[jbunniii]

I would back up and ask a more basic question: is h well-defined? Multiplication isn't commutative, so in general the value of h depends on the order in which you multiply the elements of G.

 

[Kreizhn]

Yeah, I had been thinking that myself. Since there is no natural index on the group, h must be defined invariant on the order of the multiplication. It's not clear to me if this helps in the proof though.

 

[jbunniii]

It's an interesting question. I have been thinking about it but haven't worked it out yet. I agree that h is in the center of G and therefore you can "slide it out" of the product. Thus no matter what order you choose for the multiplication, it's possible to write

f = hg

for some element of g.

Furthermore, $f^2 = hghg = h^2g^2 = g^2$, so the problem reduces to proving that $g^2 = e$ whenever g is the product of all elements of G except h. That can happen in one of two ways: either g = h or g = e. But we can rule out the first case, for if g = h then we have f = h^2 = e, but we need f to have order 2, not 1. Thus the only possibility to make this work is g = e.

So the problem reduces to showing that if g is formed by taking the product of all elements of G except h, in any order, you get e. It's obvious that this is true if G is abelian, but I'm not seeing the trick yet for the general case. I'll keep thinking about it as time allows.

 

[micromass]

Take the quaternion group, then this has -1 has unique element of order 2, furthermore

$$1(-1)ijk(-i)(-j)(-k)=1$$

and thus is not our element of order 2.

 

[jbunniii]

Take the quaternion group, then this has -1 has unique element of order 2, furthermore

$$1(-1)ijk(-i)(-j)(-k)=1$$

and thus is not our element of order 2.

Nice, I was trying to come up with a concrete nonabelian example to check but forgot about the quaternions.

P.S. Not that it matters now, but I see that I reversed the roles of f and h in my previous post.

 

[Kreizhn]

Yeah, I saw that you reversed 'em too, but I think we all got it.

Anyway, just checked the errata of the book and found a nice little tidbit that makes a HUGE difference. The group is abelian.