If $A\ast B = B$ and if $B\ast C = A$ then compute $B^2$ and $C^2$.

Question

Let $G = \{A,B,C\}$ denote a group of order three with respect to a binary operation, which we denote by $\ast$.

If $A\ast B = B$ and if $B\ast C = A$ then compute $A^2,B^2,C^2,B\ast A,C\ast B, A \ast C, C \ast A$

We can tell from the given information that $A$ is acting as the identity and $B$ and $C$ are inverses of each other.

Therefore I computed

$A^2=A$

$B\ast A=B$

$C\ast B=A$

$A \ast C=C$

$C \ast A=C$

but I cannot figure out $B^2$ and $C^2$.

Answer 1

You have $B^3=A$ because the group is of order $3$. Therefore $$ \begin{align} B^2 &= B^2 * A\\ & = B^2 * B * C\\ & =B^3*C\\ &=A*C\\ & = C. \end{align}$$

You can obtain similarly $C^2$.

Another way to see it is to say that the order of $B$ is either $3$ or $1$ (the order of $B$ should divide $3$, order to the group). Since it is not $1$, it is $3$. And therefore $B^2$ is neither $B$ nor $A$, and consequently it is $C$.

Finally, an interesting fact is that the only finite abelian group of order $3$ is $(\mathbb Z/3\mathbb Z, +)$ where $1$ and $2$ are opposite and $1+1=2$ and $2+2=1 [3]$. Therefore you can identify $B$ to $1$ (or $2$) and $C$ to the other one and conclude that $B^2=C$ and $C^2=B$.

Answer 2

Hint $$A*B=B \Rightarrow A=e \\ B*C =e \Rightarrow C=B^{-1}=B^2$$

Answer 3

Since $A\ast B=B$, we can multiply on the right by $B^{-1}$ to get that $A$ is the identity in $(G, \ast)$.

Now $B\ast C=A$ implies $B=C^{-1}=C^2$ by uniqueness of inverses and the fact that $C^{|G|}=C^3=A$.

But then $$\begin{align} B^2&=C^2\ast C^2 \\ &=C^4 \\ &=C^3\ast C\\ &=A\ast C\\ &=C. \end{align}$$