Here is another problem I struggled with about finite groups. Thanks!
1. Let G be a finite group. Show that the number of elements, x, in G such that x^3=1 is odd.
Thanks!
Here is another problem I struggled with about finite groups. Thanks!
1. Let G be a finite group. Show that the number of elements, x, in G such that x^3=1 is odd.
Thanks!
The only things that this will be true for are things with order dividing 3. That is order 1 and 3. The identity is the only thing with order 1.
I don't know what you guys have proved, but all of these are elementary to prove, so I will leave it to you.
Inverses are unique.
The order of an element is the same as the order of it's inverse.
Anything that is it's own inverse has order 2.
The point is, for every element in G with order 3, it has a unique counterpart (it's inverse) that also has order 3. You don't have to worry about double counting something because an element cannot be it's own inverse and still have order 3. So that means there must be an even number of things with order 3. Then you add in the identity which has order 1 and will still satisfy the requirement. So this is an odd number of elements in the group which satisfy $\displaystyle x^3=1$.
Let me know if any of those 3 observations are not clear to you, but I think you can check those things by just following your nose.