# 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.

• The left and right cancellation properties follow from the fact that every element in a group has an inverse, one of the defining properties of a group. So you can't really ask for something more basic than that. – Ethan Alwaise May 4 '16 at 1:39
• @EthanAlwaise : The question isn't about proving cancellation. The question is about justifying simultaneous left- or right-multiplication of a pre-existing equation. – Eric Towers May 4 '16 at 1:42
• Adding to Eric Towers' comment: if $a=b$, that means that $a$ and $b$ are "the same" (without getting into philosophical discussions), so you do something to it (namely, multiply by $c$ on the right or on the left) and you end up with the same result. That is a first principle from equality. – Luiz Cordeiro May 4 '16 at 1:43
• The definition of a group requires a set $G$ and a binary operation $\cdot$ on $G$ — i.e., a function $f\colon G \times G \to G$, $(a,b) \mapsto f(a,b)$, which we denote by $a \cdot b$ or $ab$. By definition, a binary operation is a well defined function, so if $a = a'$, then $f(a,b) = f(a',b)$ (or $ab = a'b$). This is a matter of definition. You can't "prove" it from the axioms. – M. Vinay May 4 '16 at 1:57
• @M.Vinay Thank you, this is the level of scrutiny I was looking for. – Oiler May 4 '16 at 1:59

## 3 Answers

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 $ac=bc$.

• The ordered pair comes to the rescue once again! For a much less trivial example see this answer for the role that the ordered pair played in rigorizing the idea of complex numbers. – Bill Dubuque May 4 '16 at 21:24
• This is what I was looking for and actually how I convinced myself after M. Vinay's comment. It might be worth mentioning that this came up when working through a model theory exercise, which is why I was looking for such a high level of scrutiny. – Oiler May 4 '16 at 22:05

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.

• This assumes that $\phi_{c}$ is well-defined, which is what we are trying to prove. – Oiler May 4 '16 at 1:49
• Incorrect. That's a group axiom. Every row and every column of the multiplication table is a permutation of the elements of the group. This is a consequence of inverses existing. I.e. $\phi_{c^{-1}}$ is well defined by axiom, but every $c$ is someone else's $c^{-1}$. – Eric Towers May 4 '16 at 2:06
• You merely asserted that $\phi_{c}$ is a function, and all functions are well-defined by definition. However based on the basic group axioms (identity, associativity, closure, and inverses), I fail to see how every row and every column of the multiplication table is a permutation of the elements of the group (though I know this to be true). The comments of the original post is more what I was looking for. – Oiler May 4 '16 at 2:13
• @Oiler a function doesn't have to be a permutation. The cancellation property tells you that multiplication by an element does permute the group, but that's not necessary to manipulate equations. You can do that whether multiplication is invertible or not. All it requires is that it be defined at all. If it's not a function, then it doesn't even make sense, so there isn't really a situation where what you're asking could ever fail. It's practically tautological. +1, I think the downvote is unjustified. – Matt Samuel May 4 '16 at 3:05
• @Oiler : It occurs to me that you are incorrect for an even more fundamental reason. A group is a set with a binary relation (called multiplication). Consequently, $\phi_c$ is well defined by definition. – Eric Towers May 5 '16 at 1:18

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.