Re: Complement of a subgroup

Quote:

Originally Posted by

**moont14263**

to me, reading this, this is NOT so clear. how do we KNOW x is not in K?

of course, if it is, then x^{2} is in K as well (by closure, since x*x must lie in K), contradicting <x^{2}>∩K = {1}.

rather than argue that G = <x>K (which still requires further proof), i would use the fact that x is in G, and G = <x^2>K, to assert that x = x^{2}k, for some k in K (we ruled out x = k above).

thus x^{-1} = x^{3} = (x^{2})x = (x^{2})(x^{2}k) = k, so x = k^{-1} which is in K, contradiction.

Re: Complement of a subgroup

Your proof is clearer, thank you very much.