Does every element of order 2 in a finite group have a complement in the group?

[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=<x^{2}> K and < x^{2}> $\cap$K=1 for some subgroup K of G. Clearly, G=< x> K and < x>$\cap$ K=1$, but |G|=|< x^{2}>||K|<|< x >||K|=|G|, a contradiction. Therefore G has no element of order 4.

Is above true? Thanks in advance.

 

[Vargo]

It looks right to me, but to understand your last inequality I had to look up and verify the fact that the order of a product is the product of the orders divided by the order of the intersection. So I'd recommend putting that in there. My algebra prof recommended writing proofs so that someone 3 weeks behind could understand.

 

[moont14263]

Notice that $|HK|=\frac{|H||K|}{|H\cap K|}$, for any subgroups $H$ and $K$ of $G$. Sorry, I could not add it there, but thank you very much.