# Multiplying both sides of an equation in proofs

I'm learning the basics of group theory, and must justify every step of a proof by referring to the basic axioms/theorems. Which axioms/theorems justify multiplying or adding an element of a group to both sides of an equation?

• It depends on how deep you want to go. It's not a group theory axiom, it's a "mathematics axiom", (in $\sf ZFC$ it would boil down to the Axiom of Extensionality). Morally, if you have "two" equal objects, whatever you do to them results in yet two equal objects. Mar 17, 2018 at 23:30
• By the way, the identities $aa^{-1}=e$ and $a^{-1}a=e$ tell us that $a$ satisfies the required properties for being the inverse of $a^{-1}$, so it is the inverse of $a^{-1}$. No manipulation is necessary. As to your question, have you got doubts when passing from $2x=4$ to $x=2$ in elementary algebra? I guess not. Well, it's the same, because you multiply both sides by $1/2$. Mar 17, 2018 at 23:47
• If $a = b$ then $a$ and $b$ are the same element (call it "fred"). A group has a binary operation and doing the operation on two elements will constantly yield the same result. So fred$\times c$ will be an elment (call it "sally") and fred$\times c$ will always be sally. So if $a = b$ and $a\times c = b\times c$ because $a$ and $b$ diferent ways of saying fred, and $a\times c$ will be fred$\times c$ and $b\times c$ is just fred$\times c$. Mar 18, 2018 at 0:52
• "As to your question, have you got doubts when passing from 2x=4 to x=2 in elementary algebra? " I imagine the OP doesn't but that the OP doesn't know how one can justify that by group axioms. In a way it's not really fair the we say "you must justify everything" and then when comes to justifying things actually exists we say "of things are what they are! We don't have to justify that, duh". And as Git Gud points out we do have existence axioms for that. But usually "equality" is considered "is the same specific thing" and whatever you do to the thing, is done to the thing is assumed. Mar 18, 2018 at 0:57
• Anyway, it's a good question. I'm sorry that my way of dealing with it isn't a good answer. Mar 18, 2018 at 0:58

Group operation is first and foremost a map, say $$\star\colon G\times G\to G$$. Therefore, if $$a,b\in G$$ are such that $$a=b$$, then $$(a,g)=(b,g)$$ for every $$g\in G$$, and hence $$\star(a,g)=\star(b,g)$$, or (in infix notation) $$a\star g=b\star g$$. So, in groups, $$a=b\Longrightarrow ag=bg$$ for every $$g$$.