How does $ab= a^2ba$ for the Dihedral group $D_8 = \langle a, b \rangle$ and a $\mathbb{C}D_8$-Algebra?

Question

I'm trying to show for $z = b + a^2b \in \mathbb{C}D_8$, $az = za$.

I have the presentation $D_8 = \langle a, b \ : \ a^4 = 1, b^2 = 1, b^{-1}ab = a^{-1} \rangle$. So I want to show $az = ab + a^3b = ba + a^2ba = za$.

I have shown $a^3b = ba$ by $b = b^{-1}$, $a^{-1} = a^3$, and $bab^{-1} = a^{-1}$ to get $ba = a^{-1}b = a^3b$.

But now I have to how $ab = a^2ba$. The furthest I can get is $ab = ba^{-1}$. I've been thinking about this for a long time and cannot think of how to get this equality. I appreciate any help.

Answer 1

But this is easy: $ab=ba^{-1}\implies b=aba\implies ab=a(aba)=a^2ba$.