
A1. A set $A$ is endowed with a binary operation $*$ satisfying the following four conditions:

(1) If $a, b, c$ are elements of $A$, then $a *(b * c)=(a * b) * c$;

(2) If $a, b, c$ are elements of $A$ such that $a * c=b * c$, then $a=b$;

(3) There exists an element $e$ of $A$ such that $a * e=a$ for all $a$ in $A$; and

(4) If $a$ and $b$ are distinct elements of $A \backslash\{e\}$, then $a^{3} * b=b^{3} * a^{2}$, where $x^{k}=x * x^{k-1}$ for all integers $k \geq 2$ and all $x$ in $A$.

Determine the largest cardinality $A$ may have.

## BoJan Bašić, Serbia


Solution. The required maximum is 3 , in which case $A$ is necessarily a copy of the additive group of residue classes modulo 3 ; that is, $A=\left\{a, a^{2}, a^{3}=e\right\}$, which clearly satisfies the conditions.

We show that if $|A| \geq 3$, then $A$ has the above form. For convenience, write $x y=x * y$ for all $x, y$ in $A$. Let $a$ and $b$ be distinct elements of $A \backslash\{e\}$.

We first prove that either $a^{2} \neq a$ or $b^{2} \neq b$. If $a^{2}=a$, then $a^{3}=a$, and $b a=b a^{2}$ implies $b=b a$ by (2). Similarly, if $b^{2}=b$, then $b^{3}=b$, and $a=a b$. Hence, if $a^{2}=a$ and $b^{2}=b$, then $a=a b=a^{3} b=b^{3} a^{2}=b a=b$ by (4) in the middle, a contradiction.

Assume henceforth $a^{2} \neq a$.

Next, we show that $a^{2} \neq e$. Suppose $a^{2}=e$ and use (3) and (4) to write $a b=(a e) b=$ $\left(a a^{2}\right) b=a^{3} b=b^{3} a^{2}=b^{3} e=b^{3}$, so $a=b^{2}$ by (2). Then $b^{5}=b^{3} b^{2}=b^{3} a=a^{3} b^{2}=\left(b^{2}\right)^{3} b^{2}=b^{8}$ by (4) in the middle, so $b=b^{4}$ by (2), whence $b=\left(b^{2}\right)^{2}=a^{2}=e$, a contradiction.

We now prove that $a^{3}=e$. Suppose $a^{3} \neq e$ and notice that $a^{2}$ and $a^{3}$ are distinct elements of $A \backslash\{e\}$ (otherwise, (2) would imply $a=a^{2}$ ), to use (4) and write $a^{11}=\left(a^{3}\right)^{3} a^{2}=\left(a^{2}\right)^{3}\left(a^{3}\right)^{2}=a^{12}$, so $a=a^{2}$, a contradiction.

Finally, we show that $b=a^{2}$, so $A$ consists of $e, a$ and $a^{2}$ alone. Use (4) to get $e b=a^{3} b=b^{3} a^{2}$ and $e b^{2}=a^{3} b^{2}=b^{3} a$. This yields $e b^{2}=b^{3} a=b^{2}(e b) a=b^{2} b^{3} a^{2} a=b^{3} b^{2}$, so $e=b^{3}$ by (2). Then $e b=b^{3} a^{2}=e a^{2}$, so $b=b^{4}=e b=e a^{2}=a^{3} a^{2}=a^{2} e=a^{2}$. This completes the proof.

## COMBINATORICS
