A Group Having a Cyclic Sylow 2-Subgroup Has a Normal Subgroup. I want to prove the following:


Let $G$ be a group of order $2^nm$, where $m$ is odd, having a cyclic Sylow $2$-subgroup.
    Then $G$ has a normal subgroup of order $m$.


ATTEMPT:
We will show that $G$ has a subgroup of order $m$.
Let $\theta:G\to \text{Sym}(G)$ be the homomorphism which is defined as
$$
 \theta(g)=t_g,
 \quad
 \forall g\in G
$$
where $t_g:G\to G$ is defined as:
$$
 t_g(x)=gx,
 \quad
 \forall x\in G.
$$
Let $g$ be an element of order $2^n$ in $G$.
(There exists such an element since $G$ has a cyclic Sylow $2$-subgroup.)
The cyclce representation of $t_g$ is a product of $m$ disjoint cycles each of length $2^n$.
Therefore $t_g$ is an odd permutation.
Thus the homomorphism $\epsilon\circ \theta:G\to\{\pm 1\}$, where $\epsilon:\text{Sym}\to\{\pm 1\}$ is the sign homomorphism, is a surjection.
By the First Isomprphism Theorem, we conclude that the kernel $K$ of $\epsilon\circ \theta$ is of order $2^{n-1}m$.
If $n$ were equal to $1$ then we are done.
If $n>1$ then note that any Sylow $2$-subgroup of $K$ is also cyclic.
This is because each Sylow $2$-subgroup of $K$ is contained in a Sylow $2$-subgroup of $G$, where the latter is cyclic.
Now we can inductively show that $K$ has a subgroup of order $m$.

What I am struggling with is showing the normality.
Can anybody please help me with this.
Thanks.
 A: Claim: If a group of order $2^nm$, $m$ odd, has cyclic Sylow-2 subgroup, then $G$ has unique subgroup of order $m$. 
Proof: Induction on $n$- for $n=1$, as you showed, the kernel $K$ has order $m$, which is a normal subgroup. If there is another subgroup $H$ of order $m$, then the product $KH$ is a subgroup (since $K\trianglelefteq G$) of odd order (equal to $|H|.|K|/|H\cap K|$), and is bigger than $m$ (since $H\neq K$), which is impossible since the largest odd order dividing $|G|$ is $m$.
Suppose theorem is proved for groups of order $2m, 2^2m, \cdots, 2^{n-1}m$ (containing cyclic Sylow-$2$). Let $|G|=2^nm$, with cyclic Sylow-2. As you noted, kernel $K$ has order $2^{n-1}m$, which contains unique subgroup of order $m$ (by induction), say it is $L$. Thus $L$ is characteristic in $K$ and as $K$ is normal in $G$, it follows that $L\trianglelefteq G$. 
Again, as in previous paragraph (at the starting of proof), we can conclude that $G$ has unique subgroup of order $m$, a stronger conclusion than you expected.
(Very Simple Exercise: $H$ is characteristic in $K$ and $K\trianglelefteq G$ $\Rightarrow$ $H\trianglelefteq G$. Just apply definition otherwise see this)
A: To complete your proof (normality part), we induction on $n$.
$n =1$ case is clear. Suppose $n>1$. Let $H = \langle g\rangle$ which is Sylow $2$-subgroup of $G$. $K$ is normal in $G$ so the intersection $K\cap H$ is a Sylow $2$-subgroup of $K$ whose order is $2^{n-1}$ and cyclic. Hence by induction on $n$, there is a normal subgroup $N\triangleleft K$ with order $m$. $(\dagger)$ Such normal subgroup $N$ is unique in $K$. For each $g\in G$, $gNg^{-1}<gKg^{-1} = K$ with order $m$ so $gNg{-1} = N$. Hence, $N\triangleleft G$ with order $m$.
$(\dagger)$ Let $N'$ be any subgroup of $K$ with odd order. Consider a quotient map $K\to K/N$. Since $|K/N| = 2^{n-1}$, the image of odd order subgroup $N'$ must be trivial so $N'$ is contained in the kernel of this map which is $N$.
Note that $(\dagger)$ argument also shows $N\triangleleft G$ is unique normal subgroup of order $m$.
