I have written down a proof for the proposition described in the title, but slightly different from most proofs I have found, and I feel something not correct with it, especially the part proving normality. Could anyone please verify it for me? Thanks in advance!
Proposition:
Let $G$ be a group of order $2^nm$, where $m$ is odd. If $G$ has an element of order $2^n$, then $G$ has a normal subgroup of index $2^n$.
Proof:
We use induction on $n$.
Denote by $g \in G$ the element of order $2^n$. $\langle g \rangle$ has an cyclic subgroup $\langle g^2 \rangle$ of order $2^{n-1}$, which we denote by $H$. Consider the action of $G$ on the coset space $X = \{gH|\, g \in G\}$ by left multiplication, which induces a homomorphism $\rho:\,G \to S_{2m}$.
We prove first the existence of an odd permutation in $\rho(G)$ by showing that $\rho(g)$ is such an element. Denote $\langle g \rangle$ by $P$, and consider the action of $P$ on the coset space $X$, which is the restriction of $\rho$. Then $X$ is partitioned into some $P$-orbits.
Write $\rho(g)$ as the product of disjoint cycles, each of which corresponds naturally to one and only one orbit. Hence, any cycle has length some divisor $2^j$ of $|P|$. Suppose there is some one-element cycle, say $(aH)$. Then $gaH = aH$ and $a^{-1}ga \in H$. But $a^{-1}ga$ has the same order with $g$, which is $2^{n}$, contradicting $H$ having order $2^{n-1}$. Therefore such cycles must not exist, and all cycles in the decomposition of $g$ have length a proper multiple of $2$. Thus $\rho(g)$ is an even permutation.
We have found an odd permutation in $\rho(G) = S_{2m}$, so all of the even permutations form a subgroup $T$ of order $m$. Denote its preimage by $K$. The restriction $\rho|_K:\, K \to T$ must be surjective. That is, $T = \rho(K)$. Thus, $K$ is actually a subgroup containing $\ker \rho$. Henceforth, by corresponding theorem $$[G:K] = \left[\frac{G/\ker\rho}{K/\ker\rho}\right] = [\rho(G):\rho(K)] = 2,$$ implying the normality of $K$ .
Write $|K| = 2^{n-1}m$. By the inductive hypothesis, there exists $M \lhd K$ of order $m$, which is a fortiori a subgroup of $G$ having index $2^n$.
It remains to prove normality. Suppose $M' = y^{-1}My \neq M$ for some $y \in G$. Since $K \lhd G$ and $M \leq K$, $M' \leq K$. Then $M'M \leq K$ by the normality of $M$ in $K$. $|M'M|$ is a divisor of $|M'||M| = m$, hence an odd number. But $M$ is strictly contained in $M'M$, so $M'M$ is a subgroup of $K$ having more than $m$ elements of odd order, contradicting $|K|=2^{n-1}m$. So $y^{-1}My = M$ for any $y \in G$, completing the proof.
edit:
I realize that I should have prove $H$ contain a cyclic subgroup of order $2^{n-1}$ to make the inductive hypothesis valid, thanks to Dietrich Burde who offered me Conrad's note. This is one of the severe gaps in my proof, and I try to fix it as below. Since $K \leq G$ has index $2$, every square is contained in $K$, including $g^2$. So $\langle g^2 \rangle \leq G$, and we managed to make the inductive hypothesis valid. Now I do wonder more if there are other gaps here. Could anyone help?
edit2: Gerry Myerson reminded me that the part which was supposed to give an odd permutation in $\mathbb{im}\rho$ is severely insufficient. That is to say, it is not merely a slip of the pen. Compared to the standard proof which uses regular representation, I think the problem is the cycles in the complete decomposition are of different lengths, which makes it hard to judge whether it is odd or not.
