For all $x,y\in G$ we have: $f(xf(y))=f(x)y$. Prove that $f$ is an isomorphism?

Question

Let $G$ be a group and $f : G \to G$ a function such that for all $x,y\in G$: $$f(x f(y)) = f(x) y.$$

Prove that $f$ is an isomorphism.

There are two problems here: we don't know that $f$ is a group homomorphism, and we don't know that it is bijective, the two conditions required for $f$ to be an isomorphism.

It is difficult to solve this problem because we don't know the formula for $f$, only an equation it satisfies. Directly proving that the kernel is trivial, or that every element is in the image will be difficult.

Answer 1

Note first that $f(e) = e$.

This is because $$f(f(e)) = f(e f(e)) = f(e) e = f(e),$$ and then also $$f(f(e)) = f(f(f(e))) = f(e f(f(e))) = f(e) f(e) = f(e)^{2}.$$

Then show that $f \circ f = 1$, the identity map.

This is because $f(f(y)) = f( e f(y)) = f(e) y = e y = y$.

Thus $f$ is bijective, and a homomorphism, as $$ f(x y) = f(x f(f(y))) = f(x) f(y). $$

Answer 2

We can start by proving that $f$ is a bijection. Let $y\in G$ be arbitrary. Then, $$f(e f(f(e)^{-1}y))=f(e)f(e)^{-1}y = y$$ holds, so $f$ is surjective. (Where $e$ is the identity element.)

To prove that $f$ is also injective, suppose $f(x) = f(y)$. Then we have $$f(ef(x))=f(ef(y))$$ which means that $$f(e)x=f(e)y.$$ It follows that $x=y$, so $f$ is injective.

Next, observe that $$f(f(e))=f(ef(e))=f(e)e=f(e).$$ Since $f$ is injective, we may conclude that $f(e)=e$. This further implies that $$f(f(x))=f(ef(x))=f(e)x=x.$$ Finally, $$f(xy)=f(xf(f(y)))=f(x)f(y),$$ so $f$ is a homomorphism and we are done.