I was asked to proof the right and left cancellation laws for groups, i.e.
If $a,b,c \in G$ where $G$ is a group, show that $ba = ca \implies b=c $ and $ab = ac \implies b = c$
For the first part, I went about saying $$ba = ca \iff a = b^{-1}ca \iff b^{-1}c = e \iff (b^{-1})^{-1} = c \iff b = c$$
Similar proof for the second part.
However, I am afraid that I am thinking in circles here. Is this a valid proof?
$\endgroup$ 12 Answers
$\begingroup$Much simpler if you take $a^{-1}$ right away. As $G$ is a group and $a \in G$ then $a^{-1} \in G$, thus
$$ba = ca \implies (ba)a^{-1} = (ca)a^{-1} \implies b (aa^{-1}) = c (aa^{-1}) $$
using associativity and the result follows.
$\endgroup$ $\begingroup$Your proof is valid but there is a slightly shorter way, namely $$ba=ca\implies baa^{-1}=caa^{-1}\implies b=c$$
$\endgroup$
