If $xa=xb$ then $a=b$ We just defined the axioms of a group in our lecture notes on algebra, but silently assumed that the properties of the $=$ relation are known. A few lemmas after the definition of a group we prove that (if $G$ is a group):
$\forall x \in G, \forall a,b \in G: \ xa=xb \implies a=b$
by multiplying both sides by $x^{-1}$.
Now I wonder why we are allowed to do so. Is it a general property of equivalence relationships that if $a=b,c=d \implies ac=bd$ , or is it a group property?  
 A: 
Now I wonder why we are allowed to do so. Is it a general property of equivalence relationships that if $a=b,c=d \Rightarrow ac=bd$ , or is it a group property? 

It is property of binary operations. Binary operation on a set $G$ is defined as a function $G\times G\to G$. 
This means that the result $ac$ is uniquely determined by $a$ and $c$. Which is just rephrasing of the implication you wrote: $a=b \land c=d \Rightarrow ac=bd$.
(For functions we have $x=y$ $\Rightarrow$ $f(x)=f(y)$.)
A: Assume you have an equivalence relation $\sim$ on a Group $G$. The property you will require is that $\sim$ respects the group structure, i.e. $a\sim b$ implies $xa\sim xb$ and $ax\sim bx$ for all $x\in G$. Denote by $[a]$ the equivalence class of an element $a\in G$ under $\sim$. 
Then, $\sim$ respects the group structure iff $G/\sim$ becomes a group by multiplying representatives, i.e. $[a][b]:=[ab]$ is well-defined.
Indeed, if $G/\sim$ is a group in this way, then $a\sim b$ implies $[a]=[b]$ and hence, $[xa]=[x][a]=[x][b]=[xb]$, so $xa\sim xb$. In the same way, we get $ax\sim bx$.
On the other hand, if $a\sim b$ implies $xa\sim xb$ for all $x\in G$, then let $[a]=[b]$ and $[c]=[d]$. We want to show that $[ac]=[bd]$ to have that the multiplication on $G/\sim$ is well-defined. But indeed, $c\sim d$ implies $ac\sim ad$. Furthermore, $a\sim b$ implies $ad\sim bd$ and so we have $ac\sim ad\sim bd$, i.e. $[ac]=[bd]$.
If $\tilde G:= G/\sim$ is a group in this way, then $a\mapsto [a]$ is a surjective group homomorphism $G\to \tilde G$ which has a kernel $H$ and $\tilde G=G/H$. Hence, $\sim$ respects the group structure iff $G/\sim$ is a quotient of $G$ by a subgroup $H$. In other words, this works precisely for those equivalence relations $\sim$ which are of the form $a\sim b :\Leftrightarrow ab^{-1}\in H$ for some subgroup $H\le G$. 
The equality equivalence relation is the case $H=\{1\}$.
A: The element of the group $xa$ is the same element as $xb$. Therefore, the element $x^{-1}(xa)$ is the same element as $x^{-1} (xb)$, since both these are the product of the same elements.
From that, you can assume that $a$ is the same element as $b$, since $x^{-1}(xa) = a$ and $x^{-1}(xb) = b$.
Yes, in any set $S$ in which multiplication is defined, you can, from $a=b$ and $c=d$, always assume that $ac = bd$, because $m:(x,y)\mapsto xy$ is a function from $S\times S$ to $S$, and if $(x_1,y_1) = (x_2,y_2)$, then $m(x_1,y_1) = m(x_2,y_2)$
A: $a=b$, so
$$
ac=bc
$$
If $a$ and $b$ are the same object, then you get the same value when you multiply by $c$.  They're just the same.
Now $c=d$, so
$$
bc=bd
$$
and therefore
$$
ac=bd
$$
A: Be: $ g * h = g * k$ ; then there is a $s$ in $ G $, so that: $g*s = e $; 
$s * (g * h) = (s*g) * h = e * h = h;  $
$s * (g * k) = (s*g) * k = e * k = k;$
And this gets you:  $ h = k; $ 
You can show in very similar manner that $ h = k $, if $ h * g = k * g; $   
