If $a$ conjugates with $b$, then $a^{-1}$ conjugates with $b^{-1}$ If $a$ conjugates with $b$, then $a^{-1}$ conjugates with $b^{-1}$ in a group $G.$
This seems like a pretty trivial statement but I'm trying to figure out how to prove it. So far I have that if $b=gag^{-1},$ * $b^{-1}=g^{-1}a^{-1}g,$ but we want **  $b^{-1}=ga^{-1}g^{-1}.$ Are * and ** equivalent?
 A: Be careful:

*

*"$a$ conjugates with $b$" means "there is a $g$ such that $b=gag^{-1}$".

*"$a^{-1}$ conjugates with $b^{-1}$" means "there is a $g$ such that $b^{-1}=ga^{-1}g^{-1}$".

However, these are separate definitions, so when both of these statements are in play at the same time, the $g$'s don't have to be the same. For the purposes of clarity, it is better to pick a second variable in the second statement. That is, you are assuming there is a $g$ such that $b=gag^{-1}$; you want to prove there is an $h$ such that $b^{-1} = ha^{-1}h^{-1}$.
If you've shown that $b^{-1} = g^{-1} a^{-1}g$, then you have done this! Let $h = g^{-1}$: now you have $b^{-1} = ha^{-1}h^{-1}$.
P.S. Having read user2661923's answer, I now believe that if you think you've shown $b^{-1} = g^{-1}a^{-1}g$ from $b = gag^{-1}$ you've made a mistake (which is a separate issue); you should actually be getting $b^{-1} = ga^{-1}g^{-1}$ with the same $g$. On the other hand, from $b = gag^{-1}$ you can get $a = g^{-1}bg$, which can help you prove that if $a$ conjugates with $b$, then $b$ conjugates with $a$.
A: In Group Theory, $(rs)^{-1} = s^{-1} \times r^{-1}.$
Therefore, if you start with the equation:
$$b = ga\left(g^{-1}\right), \tag1 $$
then you can simply take the inverse of both sides.
That is, if $r = s$, then $r^{-1} = s^{-1}.$
So, taking the inverse of both sides of (1) above yields
$$b^{-1} = g\left(a^{-1}\right)g^{-1}.$$
