Exercise 4.4.5 ($\mathrm{Aut}(D_8) \simeq D_8$)

Question

Use the fact that $D_8 \unlhd D_{16}$ to prove that $\mathrm{Aut}(D_8) \simeq D_8$.

Richard Ganaye · Accepted Answer

\begin{proof} Consider the group $G = D_{16}$. We know that
$$D_{16} = \langle r,s angle,\qquad 	ext{where} \qquad r^8 = s^2 = 1,\quad rs = sr^7 = sr^{-1}.$$
Consider the subgroup
$$H = \langle r^2,sangle = \{1,r^2,r^4, r^6, 1 ,sr^2, sr^4, sr^6\}.$$
Then $H \simeq D_8$. We take $H$ as the version of $D_8$ in $D_{16}$, so that $D_8 = H \leq D_{16}$
(in the geometric version of $D_{2n}$, $D_{16}$ acts on the vertices of an octagon, and $H = D_8$ acts on the square obtained by taking one vertex out of two).

We check $H \unlhd D_{16}$:
\begin{align*}
&r r^{2} r^{-1} = r^{2} \in H,\qquad r s r^{-1} = r^{-2} = r^6 \in H,\
&s r^2 s^{-1} = r^6 \in H,\qquad s s s^{-1} = s \in H.
\end{align*}
Since $D_{16} =\langle r,s angle$ and $H = \langle r^2,s angle$, this shows that 
$$H  = D_8 \unlhd D_{16}.$$

Since $H  \unlhd D_{16}$, $D_{16}$ acts by conjugation on $H = D_8$, and by Corollary 15, $G/ C_G(H) = N_G(H)/C_G(H) $ is isomorphic to a subgroup of $\mathrm{Aut}(D_8)$.

We compute $C_G(H)$. Note that $\{1,r^4\} = Z(G) \leq C_G(H)$.

Let $g \in G$. Then 
$$g \in C_G(H) \iff r^2 g r^{-2} = g 	ext{ and } sgs^{-1} = g.$$
\begin{itemize}
\item If $g = r ^i \in C_G(H)$, where $0 \leq i < 8$, then
$$r^i = g = s g s^{-1} = sr^i s^{-1} = r^{-i},$$
thus $i \equiv  -i \pmod 8$,  $i \equiv 0 \pmod 4$, so $g \in \{1,r^4\} = Z(G)$.

\item If $g = sr^i \in C_G(H)$, where $0 \leq i < 8$, then
$$ sr^i= g= r^2 g r^{-2} = r^2 s r^i  r^{-2} = s r^{i-4},$$
therefore $r^i= r^{i-4}$, thus $r^4 = 1$, which is false, so  $g = sr^i 
ot \in C_G(H)$ for all exponents $i$. 
\end{itemize}
This shows that
$$C_G(H) = \{1, r^4\} = Z(G).$$
Hence $G/Z(G) = N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\mathrm{Aut}(D_8)$. Since $|G/Z(G)| = 16/2 = 8$, we obtain $ |\mathrm{Aut}(D_8)| \geq 8$. By Exercise 3,  $| \mathrm{Aut}(D_8) |\leq 8$, thus
$$ |\mathrm{Aut}(D_8)|= 8.$$
Therefore
$$\mathrm{Inn}(D_{16}) \simeq G/Z(G) \simeq \mathrm{Aut}(D_8),$$
i.e., every automorphism of $D_8$ is obtained by restriction to $D_8$ of a inner automorphism of $D_{16}$.

Consider the map
$$\psi
\left\{
\begin{array}{ccl}
D_{16} & 	o & D_8\
s^i r^j & \mapsto & s^i r^{2j} 
\end{array}
ight.
$$

This map is well defined:  first $s^i r^{2j} \in \langle r^2, s angle = H = D_8$, and if $s^i r^j = s^k r^l$, then $i \equiv k \pmod 2$ and $j \equiv l \pmod 8$, thus $2j  \equiv 2l \pmod {16}$, so $s^i r^{2j} = s^k r^{2l}$.

Furthermore $\psi$ is a homomorphism: if $g = s^i r^j \in D_{16}$ and $h = s^k r^l \in D_{16}$, then
\begin{align}
gh = s^i r^j s^k r^l = s^i s^k r^{(-1)^k j} r^l = s^{i+k} r^{(-1)^k j +l}.
\end{align}
and
\begin{align*}
\psi(g) \psi(h) &= s^i r^{2j} s^k r^{2l}\
&=s^is^k r^{(-1)^k 2j} r^{2l}\
&= s^{i+k} r^{2((-1)k j + l)}\
&= \psi(gh)&(	ext{by (1)}).
\end{align*}
First $\psi(r^4) = r^8 = 1$, so $\{1,r^4\} \subseteq \ker(\psi)$. If $\psi( s^ir^j) = s^i r^{2j} = 1$, then $i \equiv 0 \pmod 2$, and $2j \equiv 0 \pmod 8$, thus $j \equiv 0 \pmod 4$, so $s^i r^j \in \{1,r^4\}$. This shows that
$$\ker(\psi) = \{1,r^4\} =  Z(G).$$
Moreover, $\psi$ is surjective: if $h \in D_8 = \langle r^2, sangle$, then $h = s^i r^{2j} = \psi(s^i r^{j})$.

The First Isomorphism Theorem shows that
$$ D_8  = \psi(D_{16}) \simeq D_{16}/\ker(\psi) = G/Z(G) \simeq \mathrm{Aut}(D_8).$$
In conclusion,
$$ \mathrm{Aut}(D_8) \simeq D_8.$$
\end{proof}

Exercise 4.4.5 ($\mathrm{Aut}(D_8) \simeq D_8$)

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Exercise 4.4.5 ($\mathrm{Aut}(D_8) \simeq D_8$)

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