The centralizer $C_{D_8}(A)$ is a subgroup of $D_8$ whose elements commute with $A$.
That is $C_{D_8}(A)=\{ g\in D_8 \mid gxg^{-1}=x \text{ for all } x\in A\}$.

The normalizer $N_{D_8}(A)$ is a subgroup of $D_8$ defined as
$N_{D_8}(A)=\{ g\in D_8 \mid gxg^{-1} \in A \text{ for any } x \in A\}$.

The center $Z(D_8)$ is a subgroup of $D_8$ whose elements commute with all elements of $D_8$.
That is, $Z(D_8)=\{g \in D_8 \mid gxg^{-1}=x \text{ for all } x\in D_8\}$.

Proof.

(a) The centralizer $C_{D_8}(A)=A$

Since any power of $r$ commutes with each other we have $A < C_{D_8}(A)$.
Since $sr=r^{-1}s$ and $r^{-1}s\neq rs$ (otherwise $r^2=1$), we see that $s\not \in C_{D_8}(A)$.

This also implies that any element of the form $r^as \in D_8$ is not in $C_{D_8}(A)$. In fact, if $r^as\in C_{D_8}(A)$, then $s=(r^{-a})\cdot (r^a s)$ is also in $C_{D_8}(A)$ because $r^{-a}\in C_{D_8}(A)$ and $C_{D_8}(A)$ is a group.

This is a contradiction. Therefore $C_{D_8}(A)=A$.

(b) The normalizer $N_{D_8}(A)=D_8$

In general, the centralizer of a subset is contained in the normalizer of the subset. From this fact we have $A=C_{D_8}(A) < N_{D_8}(A)$.
Thus it suffices to show that the other generator $s \in D_8$ belongs to $N_{D_8}(A)$.

We have $sr^as^{-1}=r^{-1}ss^{-1}=r^{-1}\in A$ using the relation $sr=r^{-1}s$.
Thus $s \in N_{D_8}(A)$ as required.

(c) The center $Z(D_8)=\langle r^2 \rangle=\{1,r^2\}$

The center $Z(D_8)$ is contained in the centralizer $C_{D_8}(A)=A$.
Since $rsr^{-1}=sr^2 \neq s$ and $r^3s(r^3)^{-1}=r^{-1}sr=sr^2\neq s$, the elements $r$ and $r^3$ are not in the center $Z(D_8)$.

On the other hand, we have $sr^2=r^{-1}sr=r^{-2}s=r^2s$. Thus $r^2s(r^2)^{-1}=s$ and $r^2 \in Z(D_8)$.
Therefore we have $Z(D_8)=\{1, r^2\}$.

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[…] know the center is $Z(G)={1,r^2}$ by Problem Centralizer, normalizer, and center of the dihedral group D8. Thus $K_1={1}$ and […]