$|G|=2^{k}m$ has a normal subgroup of index $2^k$ I am being involved with the following problem:

A group $G$ of order $2^{k}m$ wherein $m$ is odd has a cyclic subgroup of order $2^k$. Prove that $G$ has a normal subgroup of index $2^k$.

Honestly, I have one solution of this problem which is based on induction on $k$ and permutation groups. I am just asking if someone knows any other approaches. Thans
 A: Consider the regular representation $\rho_{G}$ of $G$. Let $S$ be a Sylow $2$-subgroup of $G,$ and suppose that $S = \langle s \rangle$ is cyclic. Then $\rho_{G}$ restricts to $S$ as a drect sum of $m$ copies of $\rho_{S}.$ Hence the eigenvalues of $\rho_{G}(s)$ all occur with multiplicity $m,$ and each $2^{k}$-th root of unity occurs as such an eigenvalue. Now the product of all complex $2^{k}$th roots of unity is $-1$, because all such roots except $1$ and $-1$ occur in complex conjugate pairs. Hence $\rho_{G}(s)$ has determinant $(-1)^{m} = -1$ as $m$ is odd. Now $g \to {\rm det}(\rho_{G}(g))$ is a $1$-deimensional representation of $G.$ If $g$ has odd order, then ${\rm det}(\rho_{G}(g)) =1,$ because all $|g|$-th rooots of unity occur as eigenvalues with equal multiplicity, and all except 1 occur in complex conjugate pairs. Hence the image of this $1$-dimensional representation is $\{1,-1\}$ (I have skipped a few details here), and its kernel has index $2$. By induction, the kernel has a normal subgroup of index $2^{k-1}$ which is characteristic, so a normal subgroup of index $2^{k}$ of $G.$
