Group Theory using binary operation Let $G$ be a group, and suppose that $G=\{e,a,b,c\}$ (different elements).
There exists $e \in G$ such that  $e \ast a = a \ast e = a$ for all $a \in G$.
Prove that $a \ast b \neq c \ast a$.
I have a very hard time proving this. I think the only way to prove is to suppose that $a \ast b = c \ast a$. the problem is that no matter what rule of group I tried, I was always get back to $a \ast b = c \ast a$ again and again.
 A: Assume (toward a contradiction) that $a*b=c*a$, and consider the options for $a*b$:


*

*$a*b \neq a$, since b is not the neutral element $e$.

*$a*b \neq b$, since a is not the neutral element $e$.

*$a*b \neq e$, since $a*b=e$ would mean that $b=a^{-1}$ and $c=a^{-1}$ (since $c*a=a*b=1$).

*$a*b \neq c$, since otherwise, $a*b=c*a=a*b*a$, which would mean that $a$ is the neutral element.


Hence, $a*b \notin G$, which is a contradiction. Hence, $a*b \neq c*a$.
A: I will write $xy$ for $x \ast y$.
Suppose that $ab = ca$.
If $ab = e$, then $b=a^{-1}$, and by the assumption, $ca = ab = e$, so $c = a^{-1}$ as well. Hence $c = b$, a contradiction.
Thus, $ab$ has to be one of $a,b$ and $c$. If $ab=a$ then $b=e$, contradiction, and if $ab = b$ then $a = e$, contradiction. Thus $ab = c$. Hence, $ca = c$, so $a=e$, a contradiction.
Therefore, $ab \neq ca$.
A: Every group $G$ with $|G| = 4$ is commutative: by Lagrange all the elements have order 1,2 or 4 and if...


*

*$\exists$ some element of order 4 $\implies$ $G$ cyclic $\implies$ $G$ commutative.

*The order of all the elements $\ne e$ is 2 $\implies$ $\forall x,y\in G:\ xyxy = (xy)^2 = e$  $\implies$ $\forall x,y\in G:\ xy = x(xyxy)y = x^2yxy^2 = yx$.
Now, $b\ne c\implies ab\ne ac = ca$.
