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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?

TheIntern
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    It depends on your definition of a group - with some group definitions the fact that a right inverse is always a two-sided inverse has to be proved. On the other hand some definitions require two-sided inverses as part of the definition. At the beginning of studying such structures it is good to explore how these properties work. Once this is done, the basics can be assumed. – Mark Bennet Nov 16 '16 at 08:34
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    An useful read could be: http://math.stackexchange.com/questions/65239/right-identity-and-right-inverse-implies-a-group – Shraddheya Shendre Nov 16 '16 at 09:03

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