Problem statement: I need to find out if $\mathrm{Aut}(D_4)$ is isomorphic to $D_4$ and explain my answer. I already know that it is isomorphic, so now all I need to do is to prove it.
I assume that first we need to look where it sends $s$ and $t$ and it must send them to elements that satisfy the same relations() And then need to show that those two are conjugation by some element in $D_4$(?)
Any help is appreciated. I was hoping for a duplicate post, but couldn't find it.
Thank you!
The shortest proof I can think of:
Since an automorphism sends a generator to a generator, and is completely determined by its image on generators, we have, at most, $10$ automorphisms $\phi \in \text{Aut}(D_4)$ since we have $2$ choices for $\phi(r)$, and $5$ choices for $\phi(s)$ (automorphisms are order-preserving).
However, since $\text{Inn}(D_4) \leq \text{Aut}(D_4)$ has order $4$ (being isomorphic to $D_4/Z(D_4)$), we see that $|\text{Aut}(D_4)| = 4$, or $8$.
Define $\rho: D_4 \to D_4$, by $\rho(r) = r, \rho(s) = sr$. Then $\rho$ is outer, as all inner automorphisms have order $2$, and $\rho$ has order $4$. One should prove that $\rho$ is indeed an automorphism (surjectivity is the main concern), but I will omit this (it should not be hard to see $\{r,rs\}$ generate $D_4$). Thus $\text{Aut}(D_4)$ has order $8$.
Let $\sigma$ be the inner automorphism induced by $s$. Note $\rho^2(r) = r$, while $\sigma(r) = r^3$, so $\langle \rho\rangle \cap \langle \sigma\rangle = 1_{D_4}$. Thus $\langle\rho\rangle\langle \sigma\rangle = \text{Aut}(D_4)$, so that $\text{Aut}(D_4) = \langle \rho,\sigma\rangle$.
Finally, note that $\rho\sigma(r) = \rho(r) = r = \sigma(r^3) = \sigma\rho^3(r)$ and:
$\rho\sigma(s) = \rho(s) = sr = \sigma(sr^3) = \sigma\rho^3(s)$, so that $\rho\sigma = \sigma\rho^3$, and:
$r \mapsto \rho, s \mapsto \sigma$ is the desired isomomorphism $D_4 \to \text{Aut}(D_4)$.
Note the two maps (of the possible $10$ above) which fail to yield automorphisms are those that send $s \mapsto r^2$, because they are not surjective.
I suppose you already know that the automorphism group of the $8$-element dihedral group $D_4$ contains $8$ elements. Here is how you can argue that it is actually isomorphic (though not canonically) to $D_4$ itself.
Given a square in the plane, its $4$ corners are also (alternatingly chosen) vertices of a unique octagon. That octagon has $4$ more vertices that also form a square; call the corners of the original square black vertices, and the remaining one white corners. The symmetry group of the octagon is a $16$-element group $D_8$. It contains your original $D_4$ as colour-preserving subgroup; it has index$~2$ and is therefore a normal subgroup. Conjugation defines a natural group morphism from $D_8$ to $\operatorname{Aut}(D_4)$. The kernel of this morphism contains the $2$-element centre $Z$ of $D_8$, and it is easy to see that it is actually equal to it: no non-central element of $D_8$ commutes with all elements of the subgroup $D_4$. So you get an injective group morphism $D_8/Z\to\operatorname{Aut}(D_4)$. Since you know that both groups have $8$ elements, this is in fact an isomorphism.
Then it remains to show that $D_8/Z\cong D_4$. For this it suffices to choose a similar presentation for $D_8$ and $D_4$ (either two adjacent reflections or a generating rotation and a reflection will do), map the generators of $D_8$ to corresponding generators of $D_4$, and check that the nontrivial element of $Z$ is mapped to the identity of $D_4$ (again using equal sizes and the fact that the morphism is clearly surjective suffice).
Let $D_8\cong H\leq D_{16}=G$, where $$D_{16}=\{1, a, a^2, a^3, a^4, a^5, a^6, a^7, b, ba, ba^2, ba^3, ba^4, ba^5, ba^6, ba^7\}$$ and $$H=\{1, a^2, a^4, a^6, b, ba^2, ba^4, ba^6\}.$$ Let $G$ act on $H$ by conjugation. That is, $g\cdot h=ghg^{-1}$. Since $H$ is normal in $G$, this action is well-defined.
Consider the permutation representation $\theta:G\to S_H$. Recall that $\ker{\theta}=C_G(H)$. In this case, $\theta(g)$ is a group homomorphism on $H$, the image of $\theta$ is contained in $\text{Aut }{H}$. Then $$G/\ker{\theta}\cong \text{Im }{\theta}\leq \text{Aut }{H}.$$
It is easy to show that $\ker{\theta}=C_G(H)=Z(G)$. Hence, $G/\ker{\theta}=G/Z(G)\cong D_8$. Now we show that $\theta$ is onto. Then $D_8\cong \text{Aut }{D_8}$ follows from the First Isomorphism Theorem.
For any $\varphi\in \text{Aut }{H}$, we want to find $g\in G$ such that $\theta(g)=\varphi$. Note that $a^2$ and $b$ are independent generator of $H$. $\varphi$ is completely determined by $\varphi(a^2)$ and $\varphi(b)$. Since $\varphi$ is an automorphism, it preserve the order of the element. Thus, $|\varphi(a^2)|=|a^2|=4$ and $|\varphi(b)|=|b|=2$. Comparing the order, we have $\varphi(a^2)\in \{a^2, a^6\}$ and $\varphi(b)\in \{b, ba^2, ba^4, ba^6\}$. ($\varphi(b)\neq a^4$, otherwise, $\theta$ is not onto.) We have the following $8$ possibilities.
$$ \left\{\begin{array}{l} \varphi(a^2)=a^2 \\ \varphi(b)=b \end{array}\right. \Rightarrow \varphi=\theta(a^4)|_{H} \text{ or }\theta(a^8)|_{H} ~~,~~ \left\{\begin{array}{l} \varphi(a^2)=a^6 \\ \varphi(b)=b \end{array}\right. \Rightarrow \varphi=\theta(ba^4)|_{H} \text{ or }\theta(b)|_{H} $$ $$ \left\{\begin{array}{l} \varphi(a^2)=a^2 \\ \varphi(b)=ba^2 \end{array}\right. \Rightarrow \varphi=\theta(a^3)|_{H} \text{ or }\theta(a^7)|_{H} ~~,~~ \left\{\begin{array}{l} \varphi(a^2)=a^6 \\ \varphi(b)=ba^2 \end{array}\right. \Rightarrow \varphi=\theta(ba^5)|_{H} \text{ or }\theta(ba)|_{H} $$ $$ \left\{\begin{array}{l} \varphi(a^2)=a^2 \\ \varphi(b)=ba^4 \end{array}\right. \Rightarrow \varphi=\theta(a^2)|_{H} \text{ or }\theta(a^6)|_{H} ~~,~~ \left\{\begin{array}{l} \varphi(a^2)=a^6 \\ \varphi(b)=ba^4 \end{array}\right. \Rightarrow \varphi=\theta(ba^6)|_{H} \text{ or }\theta(ba^2)|_{H} $$ $$ \left\{\begin{array}{l} \varphi(a^2)=a^2 \\ \varphi(b)=ba^6 \end{array}\right. \Rightarrow \varphi=\theta(a)|_{H} \text{ or }\theta(a^5)|_{H} ~~,~~ \left\{\begin{array}{l} \varphi(a^2)=a^6 \\ \varphi(b)=ba^6 \end{array}\right. \Rightarrow \varphi=\theta(ba^7)|_{H} \text{ or }\theta(ba^3)|_{H} $$ Therefore, $\theta$ is onto. (For any $\varphi\in \text{Aut }{H}$, if $\varphi(a^2)=a^{2+4i}$ and $\varphi(b)=ba^{2j}\in G=D_{16}$, where $i\in \{0,1\}$ and $j\in \{0,1,2,3\}$, then choose $g=a^{4-j}b^i$. We have $\theta(g)=\varphi$.)
