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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?

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    $\begingroup$ See also: Cancellation property in groups and Proof of the right and left cancellation laws for Groups $\endgroup$ – Martin Sleziak Feb 22 '16 at 12:27
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    $\begingroup$ Divide both sides by a. $\endgroup$ – theonlygusti Feb 22 '16 at 15:51
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    $\begingroup$ I think you have to make sure $a$ is not $0$. $\endgroup$ – kleineg Feb 22 '16 at 16:33
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    $\begingroup$ 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. $\endgroup$ – CharlieHorse Feb 22 '16 at 20:41
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    $\begingroup$ @kleineg, what is this $0$ that you talk about? $\endgroup$ – Carsten S Feb 22 '16 at 23:40
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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,

$$b = c$$

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    $\begingroup$ Because that's what you have to prove. $\endgroup$ – yellowquark Feb 22 '16 at 16:01
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    $\begingroup$ 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. $\endgroup$ – Michael Seifert Feb 22 '16 at 16:35
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    $\begingroup$ @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. $\endgroup$ – Milo Brandt Feb 22 '16 at 16:36
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    $\begingroup$ @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. $\endgroup$ – David Z Feb 22 '16 at 21:12
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    $\begingroup$ @DarrelHoffman: but the question concerns a general group with operator $\circ$. We don't usually introduce the $/$ notation for anything other than multiplication, and it'd be wrong to say $a^{-1}$ is equal to $1/a$ in the case where $\circ$ is addition. $\endgroup$ – Steve Jessop Feb 22 '16 at 22:15
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Hint:

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

Hint 2:

Think about inverses

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  • $\begingroup$ Because N is closed under multiplication. $\endgroup$ – djechlin Feb 22 '16 at 21:38
  • $\begingroup$ I'm sorry when I said multiplication I meant multiplication, not the operation you just made up. $\endgroup$ – djechlin Feb 22 '16 at 21:52
  • $\begingroup$ Because Z is an ACRU with a nonempty subset N with trichotomy that is closed under addition and multiplication. $\endgroup$ – djechlin Feb 22 '16 at 21:57
  • $\begingroup$ Associative commutative ring with unit. My point was that you first hint rather begs the question. $\endgroup$ – djechlin Feb 22 '16 at 21:59
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$G$ 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.

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

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  • $\begingroup$ Where is $e$ is neutral element in $G$ $\endgroup$ – openspace Feb 22 '16 at 10:30
  • $\begingroup$ And where you use $'$ to denote the inverse (this is probably a bad idea in general). $\endgroup$ – Tobias Kildetoft Feb 22 '16 at 10:31
  • $\begingroup$ @TobiasKildetoft why? We know that in $G$ $\exists a, a': a∘a' = a'∘a = e$ $\endgroup$ – openspace Feb 22 '16 at 10:34
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    $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Feb 22 '16 at 10:36
  • $\begingroup$ @TobiasKildetoft ok, I understand you $\endgroup$ – openspace Feb 22 '16 at 10:39
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By the group properties each element has an inverse. So you can just multiply your equation on the left by $a^{-1}$.

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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.

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