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.

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

So we can also say that there is y such that , right?

The trivial group is fine. e^2 = e.

Return to “Mathematics GRE Forum: The GRE Subject Test in Mathematics”

Users browsing this forum: Bing [Bot] and 9 guests