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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?

greedsin
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6 Answers6

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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,

$$b = c$$

Yiyuan Lee
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    This is why, as a physicist, I never understood math proofs. Since $a\neq0$, why not just say a cancels on both sides or whathaveyou? – user1717828 Feb 22 '16 at 15:55
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    Because that's what you have to prove. – yellon Feb 22 '16 at 16:01
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    @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 '16 at 16:33
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    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 '16 at 16:35
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    @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 '16 at 16:36
  • Nice; for complete clarity I might start with the reflexive statement $a^{-1} \cdot (a \cdot b) = a^{-1} \cdot (a \cdot b)$? – Daniel R. Collins Feb 22 '16 at 17:45
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    @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 '16 at 21:12
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    Should note that $a^{-1}$ is equal to $1/a$, which explains why there's no proof of equality when $a = 0$. You can't multiply a number by something undefined and get a defined result, so that covers every situation. – Darrel Hoffman Feb 22 '16 at 21:35
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    @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. – Steve Jessop Feb 22 '16 at 22:15
  • Let's hope that the existence of left inverses is among the axioms ;) – Carsten S Feb 22 '16 at 23:38
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    @CarstenS: well, technically it doesn't actually need to be among the axioms. You can state the group axioms with only right inverses and a right identity, and prove from that the existence of left inverses and a left identity, and that the lefts are equal to the rights. You do this using everyone's favourite elementary group lemma, $z\circ z = z \Rightarrow z = e$, applied to $a^{-1}a$ (where $a^{-1}$ is the right inverse) to prove it's also a left inverse, and then $e\circ a = a \circ a^{-1} \circ a = a \circ e = a$ shows $e$ is a left identity. – Steve Jessop Feb 23 '16 at 01:45
  • @user1717828 In physics, why do we need water and a solid object to show Archimedes' Principle? Why can't we just see that it's true because we know the object displaces an amount of water equivalent to its own mass? – user253751 Feb 23 '16 at 02:00
  • @SteveJessop: A 'shorter' way is $a'∘a = a'∘a∘e = a'∘a∘a'∘a'' = a'∘e∘a'' = a'∘a'' = e$ and then $e∘a = a∘a'∘a = a∘e = a$ (where $a'$ is the right-inverse of $a$ and $e$ is the right-identity). – user21820 Feb 23 '16 at 09:05
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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

5xum
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Ok, we know $a,b,c \in G$ $$b = e∘b = (a^{-1}∘a)∘b = a^{-1}∘(a∘b)=a^{-1}∘(a∘c) = (a^{-1}∘a)∘c = c$$

openspace
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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.

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

CAT0
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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.

Joe
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