# Complement of a subgroup

• Oct 28th 2012, 02:05 PM
moont14263
Complement of a subgroup
Let $G$ be a finite group. Suppose that every element of order $2$ of $G$ has a complement in $G$, then $G$ has no element of order $4$.

Proof. Let $x$ be an element of $G$ of order $4$. By hypothesis, $G=\langle x^{2} \rangle K$ and $\langle x^{2} \rangle \cap K=1$ for some subgroup $K$ of $G$. Since $[G:K]=2$, then $K$ is normal in $G$. Clearly, $G=\langle x \rangle K$ and $\langle x\rangle \cap K=1$, but $|G|=|\langle x^{2} \rangle||K|<|\langle x \rangle ||K|=|G|$, a contradiction. Therefore $G$ has no element of order $4$.

Is above true? Thanks in advance.
• Oct 29th 2012, 12:04 AM
Deveno
Re: Complement of a subgroup
Quote:

Originally Posted by moont14263
Let $G$ be a finite group. Suppose that every element of order $2$ of $G$ has a complement in $G$, then $G$ has no element of order $4$.

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

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

of course, if it is, then x2 is in K as well (by closure, since x*x must lie in K), contradicting <x2>∩K = {1}.

Quote:

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

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 = x2k, for some k in K (we ruled out x = k above).

thus x-1 = x3 = (x2)x = (x2)(x2k) = k, so x = k-1 which is in K, contradiction.
• Oct 29th 2012, 05:45 AM
moont14263
Re: Complement of a subgroup
Your proof is clearer, thank you very much.