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.
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$\begingroup$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 $D_4/Z(D_4) = D_4/\langle r^2\rangle \cong V = C_2 \times C_2$).
$\endgroup$ 1 $\begingroup$There is no issue with what you wrote: there is no reason why an automorphism should preserve any element other than $1$. But let me see if I can lay out some ideas that might help you out.
Let $G$ be a group, and let $\sigma_g(a)=gag^{-1}$. Then $$\sigma_g \sigma h (a)=\sigma_g (hah^{-1})=g(hah^{-1})g^{-1}=(gh)a(gh)^{-1}.$$
This means that $\sigma_g \sigma_h = \sigma{gh}$, and the map $g\mapsto \sigma_g$ is in fact a homomorphism $G\to \operatorname{Aut}(G)$ whose image is $\operatorname{Inn}(G)$. This establishes the isomorphism $\operatorname{Inn}(G)\cong G/Z(G)$. However, we don't need centers and quotient groups to solve the problem in this case.
Since $D_4$ is generated by $r$ and $s$, $\operatorname{Inn}(D_4)$ will be generated by $\sigma_r$ and $\sigma_s$. Furthermore, given an automorphism $\sigma$, it is determined entirely by the values $\sigma(r)$ and $\sigma(s)$. This lets us significantly reduce the amount of computation we have to do.
It is straight forward that $\sigma_s(s)=s$ and $\sigma_r(r)=r$, and as you computed above, $\sigma_s(r)=r^{-1}=r^3$. The last bit we need to compute the inner automorphisms is $\sigma_r(s)=rsr^{-1}=rsr^{-1}ss=r^2s$.
Because $s^2=1$, we know that $\sigma_s^2=\sigma_{s^2}$ is the identity automorphism. We compute $\sigma_{r^2}(s)=\sigma_r(r^2s)=r^2(r^2s)=s$, which tells us that $\sigma_r^2$ is the identity automorphism as well. This is actually enough information to conclude that $\operatorname{Inn}(D_4)=C_2\times C_2$, but I suggest you do more calculations until you can convince yourself of this fact.
Let us label automorphisms by ordered pairs $\sigma\mapsto (\sigma(r), \sigma(s))$. We have already found 4: $$\operatorname{Id}\mapsto (r,s);\quad \sigma_r\mapsto (r,r^2s); \quad \sigma_s\mapsto (r^{-1},s); \quad \sigma_{rs}\mapsto (r^{-1},r^2s).$$
What other possibilities are there? An automorphism has to map $r$ to an element of order $4$ and $s$ to an element of order $2$. Of course, not every ordered pair $(x,y)$ with $x$ of order $4$ and $y$ of order $2$ will correspond to an automorphism. For example, $(r,r^2)$ does not. However, this gives us a starting point. We can calculate the orders of all the group elements (if I have done so correctly, everything other than $1, r, r^2$ is of order $2$). So every automorphism is of the form $(r,x)$ or $(r^{-1},x)$, and we can compose any two automorphisms of the form $(r^{-1},x)$ to get one of the form $(r,x)$, and so we can find all the automorphisms (the useful first step to finding the group structure) by finding all the automorphisms that preserve $r$. This should be a much more manageable task.
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