Just as the title says:
Proposition. Let G be a group and suppose $a*b*c=e$ for some $a,b,c \in G$. Then $b*c*a=e$ as well.
I'm looking for a proof review, please. Here is my proof:
Proof. If $a*b*c=e$, then $a^{-1} = b*c$. Because inverses are commutative, this means that $e = a * (b*c) = a * a^{-1} = a^{-1} * a = (b*c)*a = b*c*a$, as desired.
The strange thing is, my professor has told me this proof is wrong, and that $b*c$ is "just a right inverse of $a$", so I cannot say it is the inverse of $a$. I don't know what this means though, because I thought that in groups, right inverses are the same as left inverses, i.e. they are, well, inverses. In addition, upon consulting our book at a later date, I found that my proof is identical to the proof given there. So who is correct?