I want to find $Aut(D_4)$ and $Inn(D_4)$. The group $D_4 = \{1,r,r^2,r^3, s,rs,r^2s,r^3s\}$.
An automorphism of a group $G$ is an isomorphism from the group to itself.
An inner automorphism of a group is a automorphism of the form $f_x: G \to G$ s.t. $\forall g\in G,\; f_x(g) = xgx^{-1}$ (given by $x$ where $x$ is some element in the group).
Does anyone have some advice for this? I'm quite confused on this.
Best Answer
Here is a "geometrical" way of looking at $\text{Aut}(D_4)$.
In this view, it should be clear that a generating rotation has to map to another generating rotation, and a generating reflection has to do the same. This gives us $8$ (at most) possible automorphisms:
$r \mapsto r,r^3$
$s\mapsto s,rs,r^2s,r^3s$
(just "mix and match").
Now, geometrically, sending $r \to r^3$ "reverses" the direction of the rotation, as if we had reflected about the $x$-axis (for example). That is, there is a natural association of $\sigma_s$ with $s$ (using Aaron's notation).
What is the geometrical result of sending $s \to rs$ (and leaving $r$ fixed)? It changes the axis of reflection, that is, it corresponds to the rotation $r$. If we call this automorphism (first, you might try proving it IS one) $\rho$, it isn't hard to see that $\rho$, like $r$, is of order $4$.
Now all that remains to be done is show that $\text{Aut}(D_4) = \langle \rho,\sigma_s\rangle$, that $\sigma_s\rho = \rho^{-1}\sigma_s$, and that the $8$ automorphisms obtained this way are all distinct.
(There's nothing I can add to Aaron's elucidation of $\text{Inn}(D_4)$, except to remark that $D_4/Z(D_4) = D_4/\langle r^2\rangle \cong V = C_2 \times C_2$).