, 2019

Algebra Difficulty 5.8 AIME, harder Prove it Romania

Let $n$ be a positive integer, and let $G$ be a finite group of order $n$ . A function $f: G \to G$ is a pseudoendomorphism if $f(xyz) = f(x)f(y)f(z)$ , for all $x, y, z$ in $G$ .

a) If $n$ is odd, show that every pseudoendomorphism of $G$ is an endomorphism.

b) If $n$ is even, is every pseudoendomorphism of $G$ an endomorphism?

Solution

a) Let $e$ denote the unit of $G$ . Let $x = y = z = e$ to write $f(e)^3 = f(e)$ , so $f(e)^2 = e$ . Since $n$ is odd, it follows that $f(e) = e$ .

If $x$ and $y$ are members of $G$ , write $f(xy) = f(xye) = f(x)f(y)f(e) = f(x)f(y)$ , to conclude that $f$ is indeed an endomorphism of $G$ .

b) The answer is negative. Let $a$ be an order $2$ element of $G$ , and let $f: G \to G$ , $f(x) = a$ . If $x, y, z$ are elements of $G$ , then $f(xyz) = a = a^3 = f(x)f(y)f(z)$ , so $f$ is a pseudoendomorphism. However, $f$ is not an endomorphism, since $f(e) = a \neq e$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.