Complement of a subgroup

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October 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.
October 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.
October 29th 2012, 05:45 AM
moont14263

Re: Complement of a subgroup

Your proof is clearer, thank you very much.