Let $D_8$ be the dihedral group of order $8$. Using the generators and relations, we have $D_8=⟨r,s∣r^4=s^2=1,sr=r^{−1}s⟩$.
(a) Let $A$ be the subgroup of $D_8$ generated by $r$, that is, $A=\{ 1,r,r^2,r^3\}$. Prove that the centralizer $C_{D_8}(A)=A$.
Proof,
Since $C_{D_8}(A) \leq D_8$ , By lagrange's theorem order of $C_{D_8}(A)$ might be $1,2,4$ or $8$
Since $A$ is abelian group, It is clear that for any $g \in A$, $gag^{-1}=a \space \forall a \in {A}$.
That is $A \subseteq C_{D_8}(A) $ this follows $|C_{D_8}(A)| \geq 4$
Note that $srs^{-1}=r^3 \not= r$. Therefore $s \not \in C_{D_8}(A)$. This follows $|C_{D_8}(A)| \not= 8$.
Hence $|C_{D_8}(A)|=4$ and $C_{D_8}(A)$ must be $A$
Can anyone verify my answer?