$\mathrm{Aut}(D_4)$ is isomorphic to $D_4$

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!

Solution 1:

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.

Solution 2:

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).