Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to prove for a group $G$, that if $$a\circ b =a\circ c$$ then this is true $$b=c$$ I started with $b=b\circ e$, but this didn't help me at all.

Next I tried with this: $$(a\circ b)\circ c=a\circ (b\circ c)$$ but I don't know/understand how to go further. How can I prove this equation?

share|cite|improve this question
Divide both sides by a. – theonlygusti Feb 22 at 15:51
I think you have to make sure $a$ is not $0$. – kleineg Feb 22 at 16:33
When I was teaching, I wouldn't let my students use the word "cancel". If they did, I would have them identify if they were using the additive inverse or the multiplicative inverse (I wanted them to actually think about what they were doing and why). If this case, you're using the Distributive Property. – CharlieHorse Feb 22 at 20:41
@kleineg, what is this $0$ that you talk about? – Carsten S Feb 22 at 23:40
up vote 35 down vote accepted

Suppose $$a\cdot b = a\cdot c$$ Let $a^{-1}$ be the inverse element of $a$ in $G$ (s.t. $a^{-1}\cdot a = a\cdot a^{-1} = e$ where $e$ is the identity element), which must exist by the axioms of groups. Now consider

$$a^{-1}\cdot(a \cdot b) =a^{-1}\cdot(a\cdot c)$$

By associativity, we have

$$(a^{-1}\cdot a)\cdot b = (a^{-1}\cdot a)\cdot c$$

By the definition of inverse, we have

$$e\cdot b = e\cdot c$$

where $e$ is the identity element (s.t. $e\cdot x = x\cdot e = x$ for all $x \in G$). By the definition of the identity element,

$$b = c$$

share|cite|improve this answer
Because that's what you have to prove. – Davide F Feb 22 at 16:01
@user1717828 For two reasons. One, you're in a group, not a ring. A Rubik's cube (or rather sequences of moves on it) is an example of a group. There's no "zero." There's an identity -- doing nothing -- but more often we denote that "1" and whether you can cancel "a" has nothing to do with whether "a = 1." The other is that we're trying to prove the think you're asking "why can't we just say this." It's false in many important rings. So you need to understand when it's true and prove it. – djechlin Feb 22 at 16:33
Here's a counter-example of why you can't just "cancel on both sides": Let $A$ be a matrix. If $A \vec{v} = A \vec{w}$, does it follow that $\vec{v} = \vec{w}$? Answer: not unless $A$ has an inverse, and not all matrices have inverses. – Michael Seifert Feb 22 at 16:35
@user1717828 There is actually substance to this proof - i.e. it says that if we have inverses then we have cancellation. There are common systems without cancellation - like modular arithmetic - so we definitely shouldn't treat it as an axiom. If we start with $$3\cdot 3 \equiv 3\cdot 1 \pmod{6}$$ Then we just cancel on both sides, then we get $$3\equiv 1\pmod{6}.$$ Whoops. – Milo Brandt Feb 22 at 16:36
@user1717828 I don't think this has anything to do with being a physicist - I've always thought it was a symptom of bad (or insufficiently deep) math education. It's very common among students of all quantitative fields that they memorize the rule of cancellation without understanding why things can be cancelled, and that gets them into trouble when dealing with more sophisticated situations like matrices, modular arithmetic, differential operators, etc. – David Z Feb 22 at 21:12


If you know that $4\cdot x = 4\cdot y$, how do you prove that $x=y$?

Hint 2:

Think about inverses

share|cite|improve this answer
Because N is closed under multiplication. – djechlin Feb 22 at 21:38
I'm sorry when I said multiplication I meant multiplication, not the operation you just made up. – djechlin Feb 22 at 21:52
Because Z is an ACRU with a nonempty subset N with trichotomy that is closed under addition and multiplication. – djechlin Feb 22 at 21:57
Associative commutative ring with unit. My point was that you first hint rather begs the question. – djechlin Feb 22 at 21:59

$G$ is a group. One of the axioms of a group is that every element has an inverse. This means that $a\in G$ has an inverse $a^{-1} \in G$. This will help a lot.

share|cite|improve this answer

Ok, we know $a,b,c \in G$ $$b = e∘b = (a^{-1}∘a)∘b = a^{-1}∘(a∘b)=a^{-1}∘(a∘c) = (a^{-1}∘a)∘c = c$$

share|cite|improve this answer
Where is $e$ is neutral element in $G$ – openspace Feb 22 at 10:30
And where you use $'$ to denote the inverse (this is probably a bad idea in general). – Tobias Kildetoft Feb 22 at 10:31
@TobiasKildetoft why? We know that in $G$ $\exists a, a': a∘a' = a'∘a = e$ – openspace Feb 22 at 10:34
It is a bad idea because the inverse is denoted by $a^{-1}$. This is universal when working with groups written multiplicatively. Using ' is something done temporarily until it has been established that the inverse is unique and two-sided. This also frees up ' to denote alternative elements. – Tobias Kildetoft Feb 22 at 10:36
@TobiasKildetoft ok, I understand you – openspace Feb 22 at 10:39

By the group properties each element has an inverse. So you can just multiply your equation on the left by $a^{-1}$.

share|cite|improve this answer

Multiply both sides of the given equation $$ a\circ b=a\circ c $$ on the left by the inverse of $a$ to get the desired result.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.