Showing f is a Bijection on a Group

[Gale]

1. The problem statement, all variables and given/known data
Let (G,*) be a group, and denote the inverse of an element x
by x'. Show that f: G to G defined by f(x) = x' is a bijection,
by explicitly writing down an inverse. Given x, y in G, what is
f(x *y)?

Homework Equations

The Attempt at a Solution

Okay, I think I'm just over thinking this... because it seems obvious that f is a bijection, since any function that has an inverse is bijective, and f obviously has an inverse where f^-1(x')=x. And since x' and x are both in G, and G is a group then f^-1 is well defined. But I keep having trouble with how to write proofs like this. Could someone help me out? Thanks!

 

[jgens]

Try writing something like: "Notice that $f \circ f = \mathrm{id}_G$ and this implies that $f$ is a bijection."

 

[HallsofIvy]

Yes, since the inverse of the inverse of a group element is the group element itself so this function is its own inverse.