Exercise 4.5.49 ( If $|G| = 2^n m, 2
mid m$, where $G$ has a cyclic Sylow $2$-subgroup, then $G$ has a normal subgroup of order $m$)

Question

Prove that if $|G| = 2^n m$ where $m$ is odd and $G$ has a cyclic Sylow $2$-subgroup then $G$ has a normal subgroup of order $m$. [Use induction and Exercises 11 and 12 in Section 2.]

Richard Ganaye · Accepted Answer

This is a generalization of Exercise 4.2.13. 
\begin{proof} 

Let $m$ be a fixed odd positive integer. We suppose that $|G| = 2^n m$ for some integer $n\geq 0$.

If $n = 0$, then $|G| = m$, and $G$ has a normal subgroup of order $m$, the whole group $G$. 

 Reasoning by induction on $n\geq 1$, we assume that every group of order $2^{n-1} m$, where $m$ is odd, which has a cyclic Sylow $2$-subgroup, has a normal subgroup of order $m$.

\bigskip
Consider now a group $G$ of order $2^n m$, where $m$ is odd, such that $G$ has a cyclic Sylow $2$-subgroup $P$. 

Let $x$ be a generator of $P$, so that $|x| = 2^{n}$ is even and $\frac{|G|}{|x|} = m$ is odd. By Exercise 4.2.11, where $\pi : G 	o S_G$ is the regular representation, $\pi(x)$ is an odd permutation, and by Exercise 4.2.12, $G$ has a normal subgroup $H$ of index $2$ in $G$, of order $2^{n-1}m$.


Since $H \unlhd G$, $PH$ is a subgroup of $G$, which contains $H$. Moreover $|G:H| = 2$, so $H$ is a maximal subgroup of $G$, so $PH = G$ or $PH = H$. If $PH = H$, then $P \leq H$: this is impossible, because $|P| = 2^n$  does not divide $|H| = 2^{n-1} m$, otherwise $2 \mid m$. This shows that
$$PH = G.$$
Consider the subgroup $P \cap H$. The Second Isomorphism Theorem (see figure) gives
$$P/P\cap H \simeq G/H \simeq Z_2.$$
Therefore
$$|P \cap H |= 2^{n-1},$$
so $P \cap H$ is a Sylow $2$-subgroup of $H$. Moreover $P \cap H$ is a subgroup of a cyclic group, so $P \cap H$ is cyclic. The induction hypothesis shows that there is some normal subgroup $N$ of order $m$ of $H$.

\bigskip
\begin{center}
\begin{tikzpicture}

ode (c1) at (4,1) {$ \{1\}$};

ode (c3) at (5.5,2.5) {$N$};

ode (c2) at (3,2.25) {$P\cap H\ \ $};

ode (c4) at (4.5,3.75) {$H$};

ode (c5) at (2,3.5) {$ P$};

ode (c6) at (3.5,5) {$G$};

\draw (c1) edge node [auto = left]{ $2^{n-1}$} (c2);
\draw (c1) edge  node [auto = right]{$ m$} (c3);
\draw (c3) edge  node [auto = right]{$2^{n-1}$}(c4);
\draw (c2) edge  node [auto = right]{ $m$}(c4);

\draw (c2) edge node [auto = right]{ $2$} (c5);
\draw (c5) edge node [auto = left]{ $m$} (c6);
\draw (c4) edge node [auto = right]{ $2$} (c6);
\end{tikzpicture}
\end{center}
\bigskip
It remains to show that $N$ is normal in $G$ (*).

Let $M$ be another subgroup of $H$ of order $m$. Since $N \unlhd G$, $NM$ is a subgroup of $H$, and
$$|NM| =\frac{|N| |M|}{|N \cap M|} = \frac{m^2}{k},$$
where $k = |N \cap M|$ is a divisor of $m = |N|$. Moreover $NM$ is a subgroup of $H$. By Lagrange's Theorem, $|NM| = \frac{m^2}{k} \mid 2^{n-1}m = |H|$, thus $m \mid 2^{n-1}k$, where $\mathrm{g.c.d}(m, 2) = 1$, therefore $m \mid k$. Since $m\mid k$ and $k \mid m$, we obtain $m= k$. So $|NM| = m = |N| = |M|$. This shows that 
$$ N = NM = M.$$
There is a unique subgroup $N$ of $H$ of order $m$. Hence $N$ is a characteristic subgroup of $H$, and $H$ is a normal subgroup of $G$, therefore $N$ is normal in $G$:
$$(N \unlhd^{\mathrm{char}} H, \ H \unlhd G) \Rightarrow N \unlhd G.$$

The induction is done. 

 If $|G| = 2^n m$ where $m$ is odd and $G$ has a cyclic Sylow $2$-subgroup then $G$ has a normal subgroup of order $m$.

\end{proof}

Note: I got the idea  for the proof of $N \unlhd G$ from Aryaman Maithani:

https://aryamanmaithani.github.io/alg/groups/sylow-exercises/

Exercise 4.5.49 ( If $|G| = 2^n m, 2 \nmid m$, where $G$ has a cyclic Sylow $2$-subgroup, then $G$ has a normal subgroup of order $m$)

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Exercise 4.5.49 ( If $|G| = 2^n m, 2 \nmid m$, where $G$ has a cyclic Sylow $2$-subgroup, then $G$ has a normal subgroup of order $m$)

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