How to prove $b=c$ if $ab=ac$ (cancellation law in groups)? 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?
 A: Hint:
If you know that $4\cdot x = 4\cdot y$, how do you prove that $x=y$?
Hint 2:
Think about inverses
A: Ok, we know $a,b,c \in G$
$$b = e∘b = (a^{-1}∘a)∘b = a^{-1}∘(a∘b)=a^{-1}∘(a∘c) = (a^{-1}∘a)∘c = c$$
A: $G$ 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.
A: By the group properties each element has an inverse. So you can just multiply your equation on the left by $a^{-1}$.
A: 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,
$$b = c$$
A: 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.
