To prove that $G$ is the group the condition is not necessary $\forall a,b, c \in G(ba=ca\to b=c)$.? $1.$ Let $G$ be a finite semigroup such that  
$\forall a,b, c \in G(ab=ac\to b=c)$.
Then $G$ is Group. ?
I know the following result : If $G$ be a finite semigroup such that 
$\forall a,b, c \in G(ab=ac\to b=c)$ and $\forall a,b, c \in G(ba=ca\to b=c)$
Then $G$ is Group.
To prove that $G$ is the group the condition is not necessary $\forall a,b, c \in G(ba=ca\to b=c)$. ?
There exists an contraexample for $1.$
 A: Let $S$ be any finite set, and define $a\star b = b$ for all $a,b\in S$. Show this is a semigroup, and has your property. Show it is not a group if $|S|>1$.
If a finite semigroup $S$ has the above property plus the property that if $a\neq b$ then there is a $c$ so that $ac\neq bc$, then you can prove it is a group. (This second condition is true, for example, when $S$ has a right-identity, or if cancellation works the other way as well: $ba=ca\implies b=c$.)
This is because the map $\phi_a:b\mapsto a\cdot b$ is always a permutation of $S$ if $S$ is finite and has your condition, and this second condition says that $a\to\phi_a$ is one-to-one - if $\phi_a=\phi_b$ then $a=b$. We can show this is a homomorphism of $S$ to $\Sigma(S)$, the group of permutations of $S$, it is $1-1$, and therefore makes the $S$ isomorphic to the image of $S$, which is easily seen to be a subgroup of $\Sigma(S)$.
In my example with $a\star b=b$ for all elements, the map $S\to \Sigma(S)$ is not one-to-one. Every element of $S$ is sent to the identity permutation. 
