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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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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,

$$b = c$$

**Hint**:

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

**Hint 2:**

Think about *inverses*

$G$ 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.

Ok, we know $a,b,c \in G$

$$b = e∘b = (a^{-1}∘a)∘b = a^{-1}∘(a∘b)=a^{-1}∘(a∘c) = (a^{-1}∘a)∘c = c$$

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

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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