Theorem: Every finite group $G$ that has a normal Sylow 2-subgroup, i.e., with $n_{2}(G) = 1$ is solvable. (let $n_{p}(G)$ be the number of Sylow p-subgroups of $G$)
I proof the following:
"Let $P$ is a Sylow 2-subgroup.
Then $|P|=2^n$ and $|G|=2^n.m$ with $(2,m)=1$.
We have $n_{2}(G) = 1$. Therefor $P \triangleleft G$ (because the condition $n_p(G) = 1$ is equivalent to saying that the Sylow p-subgroup of G is a normal subgroup.)
So, we have $1 \triangleleft P \triangleleft G$.
To prove $G$ solvable, I will prove $G/P$ and $P/1=P$ solvable.
$|G/P| = \frac{{|G|}}{{|P|}} = m$ is an odd number. Therefor $G/P$ solvable. (Because Theorem Feit-Thompson: Every odd order groups are solvable)
$|P|=2^n$ ??? ..."
How to prove "$P$ solvable"?
My problem is that $|P|=2$ or $|P|=2^n$. (If $|P|=2$ then I can prove.)
Thanks for any help.