Exercise 4.4.3 ($|\mathrm{Aut}(D_8)| \leq 8$)

Question

Prove that under any automorphism of $D_8$, $r$ has at most $2$ possible images and $s$ has at most $4$ possible images ($r$ and $s$ are the usual generators --- cf. Section 1.2). Deduce that $|\mathrm{Aut}(D_8)| \leq 8$.

Richard Ganaye · Accepted Answer

\begin{proof} 
The order of $r$ is $4$. Any automorphism $\sigma$ preserves the order, so $|\sigma(r)| = 4$. The only elements of order $4$ in $D_8$ are $r$ and $r^3 = r^{-1}$ (see Exercise 1.2.1). Therefore
$$\sigma(r) \in \{r, r^{3}\}.$$
Similarly, $|\sigma(s)|= 2$, thus
$$\sigma(s) \in \{r^2\} \cup \{s,rs,r^2s, r^3s\}.$$
If $\sigma(s) = r^2$, then 
$$\sigma(G)  = \sigma(\langle r,s angle) = \langle \sigma(r), \sigma(s) angle  \leq \langle r^{\pm1} , r^2 angle \subseteq \langle r angle.$$ 
This is impossible, because $|\sigma(G) | = 8$ and $| \langle r angle| = 4$. Therefore
$$\sigma(s) \in  \{s,rs,r^2s, r^3s\}.$$

Since $G = \langle r, s angle$, every automorphism of $G$ is characterized by $\sigma(r)$ and $\sigma(s)$, i.e.,  for every $(i,j) \in \{1,3\} 	imes \{0,1,2,3\}$, there is at most one automorphism such that $\sigma(r) = r^{i}$ and $\sigma(s) = r^j s$. Therefore there are at most $8$ automorphism of $D_8$:
$$|\mathrm{Aut}(D_8)| \leq 8.$$
\end{proof}

Exercise 4.4.3 ($|\mathrm{Aut}(D_8)| \leq 8$)

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Exercise 4.4.3 ($|\mathrm{Aut}(D_8)| \leq 8$)

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