Abstract Algebra

Abstract Algebra Proof, I am not sure my proof is legit.
Suppose that [math]$a$ and [math]$b$ belong to a group,[math] $a$ has odd order, and [math]aba^{-1}=b^{-1}. Prove [math]b^{2}=e.

So my proof is

We know [math] b\inC(b) ( b is in C(b) and that C(b) is a subgroup of the group) Since C(b) is a subgroup and b is in it then [math]b^{-1} is also in C(b). Since [math]aba^{-1}=b^{-1} this also means a is in C(b). ( this is what I am questioning) Therefore (b commutes with a) and [math]aa^{-1}b=b^{-1} which implies [math]b=b^{-1} which implies [math]b^{2}=e.

[math]
Since b^{-1}\inC(b)

and aba{-1}=b{-1}

thus that implies a is in C(b)

I have no idea why ab(a^(-1)) = b^(-1) would imply a is in the centralizer of b.  I'm pretty sure that's false but I'm too lazy to come up with a counterexample.

The problem is easy though.  Take ab(a^(-1)) = b^(-1) and multiply both sides on the left by b and on the right by a to get bab = a.
Take ab(a^(-1)) = b^(-1) and multiply both sides on the right by a(b^(-1)) to get a = b^(-1)a(b^(-1)).
For some odd natural i, a^i = e.  If i = 1 the result is obvious, otherwise e = a^i = (aa)^((i - 1)/2)a = ((bab)(b^(-1)a(b^(-1))))^((i - 1)/2)(bab) = (baa(b^(-1)))^((i - 1)/2)bab = b(a^(i - 1))(b^(-1))bab = b(a^i)b = beb = b^2..