One of the most common manipulations performed when working with group equations is left or right multiplication, i.e. if you have a group $G$ with $a,b,c \in G$ and you have something of the form $a = b$, we can then say that $ac = bc$ and $ca = cb$. How can one prove this statement from the basic axioms of group theory? I have proved it under the assumption of left and right cancellation properties, but the only proofs I have seen of those properties relies on this fact.
[Math] Group operation is well defined
abstract-algebragroup-theory
Best Answer
I'm not sure what part of other answers are not satisfying, so let me break this down to basic set theoretic concepts and see if that helps.
Formally, a group is defined to be a set $G$ together with a function $\phi : G \times G \to G$ (the group operation) satisfying some axioms (which won't be used for this problem). We usually use the shorter "binary operation notation" $$ab = \phi(a,b), \quad a,b \in G $$ but let me stick with the function notation for the moment.
Suppose we are given $a,b,c \in G$.
If $a=b$ then we have equality of the ordered pairs $(a,c)=(b,c) \in G \times G$ (this is one of the basic properties of ordered pairs, part of set theory).
Since $\phi$ is a function, we therefore have equality of the following values of $\phi$, namely $\phi(a,c)=\phi(b,c) \in G$ (this is one of the basic properties of a function, also part of set theory).
Returning now to binary operation notation, it follows that $ac=bc$.