Re: Complement of a subgroup

Quote:

Originally Posted by

**moont14263** Let $\displaystyle G$ be a finite group. Suppose that every element of order $\displaystyle 2$ of $\displaystyle G$ has a complement in $\displaystyle G$, then $\displaystyle G$ has no element of order $\displaystyle 4$.

Proof. Let $\displaystyle x$ be an element of $\displaystyle G$ of order $\displaystyle 4$. By hypothesis, $\displaystyle G=\langle x^{2} \rangle K$ and $\displaystyle \langle x^{2} \rangle \cap K=1$ for some subgroup $\displaystyle K$ of $\displaystyle G$. Since $\displaystyle [G:K]=2$, then $\displaystyle K$ is normal in $\displaystyle G$. Clearly, $\displaystyle G=\langle x \rangle K$ and $\displaystyle \langle x\rangle \cap K=1$

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}.

Quote:

, but $\displaystyle |G|=|\langle x^{2} \rangle||K|<|\langle x \rangle ||K|=|G|$, a contradiction. Therefore $\displaystyle G$ has no element of order $\displaystyle 4$.

Is above true? Thanks in advance.

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.