Group operation is well defined 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. 
 A: 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 $ac=bc$.
A: Let $G$ be a group, and for all $c \in G$ define $\phi_c$to  be the function $\phi_c:G \rightarrow G:a \mapsto ca$.  Starting with the equation $a=b$, we apply $\phi_c$ to both sides and equality still holds: $ca = cb$, because functions are well-defined: given a specific input, the output is fixed.  When we say $a=b$, we guarantee that we provide the same input to both copies of $\phi_c$:  \begin{align}
    a &= b & &\text{given} \\
    \phi_c(a) &= \phi_c(b) & &\text{definition of function} \\
    ca &= cb & &\text{definition of $\phi_c$}  \text{.}
\end{align}
A similar argument applies to right-multiplication.
In short, we define "function" so that equal inputs give equal outputs.
A: The $=$ generally implies that the two sides are in fact equal. In that case there's nothing to justify; since the two sides are identical, you can do whatever you want to one side as long as you do the same to the other side.
You can come up with a "proof" by saying that left multiplication by $c$ is a function from the group to itself, and "when two elements of the domain are equal then the value of the function is the same on each of them". But in that case "two" does not count the number of unique elements, since there really is only one.
It's possible to learn so much math you forget that some things are obvious. As you progress this will pass.
Edit: This concept is not really obvious in the broadest sense of the word. It was one of the most brilliant breakthroughs in mathematics and it is the basis of all algebra. But anyone studying group theory has most likely been familiar with this concept for years.
