Group with Cyclic Sylow 2-subgroup

Proposition

If a group of order $2^{n}m$ with odd $m$ has cyclic Sylow $2$-subgroup, then $G$ has the unique normal subgroup of order $m$ and so $G$ is solvable.

\begin{proof} When $n=1$, there exists a $g\in G$ with $|g|=2$. Then $\check{g}\in \mathrm{Sym}(G)$, which is defined here, is a product of $2$-cycles. Since $m$ is odd, we have that $\check{g}$ is a product of $m$‘s $2$-cycles and so $\check{g}$ is an odd permutation. Define $\phi :\check{G}\to {\mathbb{Z}}_2$ such that $\ker \phi =\check{G}\cap A_{|G|}$. Then $\ker \phi ⊲\check{G}$ is a group of order $m$ and so we get a normal subgroup $H$ of $G$ with order $m$. Furthermore, if there is another normal subgroup $K⊲G$ of order $m$, then the product $KH$ is a subgroup as $K⊲G$. Note that $|KH|=|K||H|/|K\cap H|>m$ is odd, which is a contradiction. Therefore, it has the unique normal subgroup of order $m$.

Now suppose that it holds for groups of order $2m,2^{2}m,\cdots ,2^{n-1}m$, and suppose that $|G|=2^{n}m$. With the similar argument, there is a $L⊲G$ such that $|L|=2^{n-1}m$. As each Sylow $2$-subgroup of $L$ is contained in a Sylow $2$-subgroup of $G$, each Sylow $2$-subgroup of $L$ is also cyclic. Then by induction hypothesis, group $L$ has the unique normal subgroup of order $m$, denoted by $L_0$. Since $L_0$ is a characteristic subgroup of $L$, there is $L_0⊲G$. Additionally, with the same argument in the previous paragraph, we can prove that $L_0$ is unique.

By Feit-Thompson theorem, a group $G$ with cyclic Sylow $2$-subgroup is solvable. \end{proof}

Remark. Also see Group with Cyclic Sylow Subgroups.