odd order
Re: odd order
x^{2n+1}=e
Then
(x^{n+1})^2=x^{2n+2}=x.
y^2=x, where y=x^{n+1}.
Answer: Yes.
Then
(x^{n+1})^2=x^{2n+2}=x.
y^2=x, where y=x^{n+1}.
Answer: Yes.
Re: odd order
Say S = { e, x, x^2,..............x^n-1 } be a subgroup, where x^n = e and n is odd.
say y = x ^ (n+1)/2 ; y^2 = x^(n+1) = x
say y = x ^ (n+1)/2 ; y^2 = x^(n+1) = x
Re: odd order
So we can also say that there is y such that $$x^2=y$$, right?
Re: odd order
The trivial group is fine. e^2 = e.