Exercise 4.5.12 (Sylow $2$-subgroups of $D_{2n}$)

Question

Let $2n = 2^a k$ where $k$ is odd. Prove that the number of Sylow $2$-subgroups of $D_{2n}$ is $k$. [Prove that if $P \in Syl_2(D_{2n})$ then $N_{D_{2n}}(P) = P$.]

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $2n = 2^a k$ where $k$ is odd. 
We show first that if $P \in Syl_2(D_{2n})$ then $N_{D_{2n}}(P) = P$.

Note that $|r| = n = 2^{a-1} k$ thus
$$|r^k| = 2^{a-1}.$$
Consider the subgroup $H = \langle s, r^k angle$, and $S = \{1,r^k,r^{2k}, \ldots, r^{(2^{a-1} - 1) k}, s, sr^k,sr^{2k},\ldots, sr^{(2^{a-1} - 1) k} \}$.  
Note first that since $|r^k| = 2^{a-1}$, $\langle r^k angle = \{1,r^k,r^{2k}, \ldots, r^{(2^{a-1} - 1) k}\}$, thus $S = \langle r^k angle \cup s \langle r^k angle$.

We show that $H = S$:
\begin{itemize}
\item Since $s \in H$ and $r^k \in H$, then $r^{ik} \in H$ and $sr^{ik} \in H$ for all integers $i$, thus $S \subseteq H$.
\item Conversely, S is a subgroup of $G$, because for all integers $i,j$, 
\begin{align*}
r^{ik} r^{jk} &= r^{(i+j)k} \in S,\
r^{ik} (s r^{jk}) &= s r^{(j-i) k} \in S,\
(s r^{ik})  r^{jk} &= s r^{(i+j)k} \in S,\ 
(s r^{ik}) (s r^{jk}) &= r^{(j-i)k} \in S,
\end{align*}
and $S$ is finite, so $S$ is a subgroup of $G$, which contains $r^k$ and $s$, thus $S \supseteq H$.

\end{itemize}
Hence
\begin{align*}
H &= \langle s, r^k angle\
&= \{1,r^k,r^{2k}, \ldots, r^{(2^{a-1} - 1) k}, s, sr^k,sr^{2k},\ldots, sr^{(2^{a-1} - 1) k} \}.
\end{align*}
Therefore $|H| = 2^a$, so $H$ is a particular Sylow $2$-subgroup of $D_{2n}$ (the others Sylow $2$-subgroups are conjugate of $H$).

We know that $H \subseteq N_{D_{2n}}(H)$. Conversely, let $g \in  N_{D_{2n}}(H)$.
\begin{itemize}
\item If $g = r^i$ for some integer $i$, then $gs g^{-1} = r^i s r^{-i}  = sr^{-2i} \in H$, thus $k \mid 2i$, where $k$ is odd, so $k \mid i$. Then $g= r^i \in H$.

\item If $g = sr^i$ for some integer $i$, then $gs g^{-1} = sr^i s sr^i = sr^{2i} \in H$, thus $k \mid 2i$, where $k$ is odd, so $k \mid i$. Then $g= s r^i \in H$.

\end{itemize}
This shows that $N_{D_{2n}}(H) \subseteq H$, so
$$N_{D_{2n}}(H) = H.$$
Now take any Sylow $2$ subgroup $P$ of $D_{2n}$. By Sylow's Theorem, $P$ and $H$ are conjugate, so there is some $g \in G$ such that $P = gHg^{-1}$. Then for all $u \in G = D_{2n}$,
\begin{align*}
u \in N_G(P) & \iff u P u^{-1} = P\
&\iff u gH g^{-1} u^{-1} = gH g^{-1}\
&\iff(g^{-1} u g) H (g^{-1} u g)^{-1}\
&\iff g^{-1} u g \in N_G(H)\
&\iff u \in gN_G(H) g^{-1},
\end{align*}
thus
$$N_G(P) = gN_G(H)g^{-1}.$$
Here $N_G(H) = H$, therefore $N_G(P) = gN_G(H)g^{-1} = gHg^{-1} = P$.

For all Sylow $p$-subgroups of $D_{2n}$,
$$N_{D_{2n}}(P) = P.$$

By the third part of Sylow's Theorem,
$$n_2(D_{2n}) = |D_{2n} : N_{D_{2n}}(P) | = |D_{2n} : P| = \frac{2n}{2^a} = k.$$
This proves
$$n_2(D_{2n}) = k.$$
If $2n = 2^a k$ ($k$  odd), then the number of Sylow $2$-subgroups of $D_{2n}$ is $k$.
\end{proof}

Exercise 4.5.12 (Sylow $2$-subgroups of $D_{2n}$)

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Exercise 4.5.12 (Sylow $2$-subgroups of $D_{2n}$)

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