Let $f_a(x) = axa^{-1}$ be an inner automorphism of G, where $a,x$ $ \in$ G. What is $f_a^{-1}$?
Method 1.
$f_a^{-1}(x) = (axa^{-1})^{-1} = ax^{-1}a^{-1}.$
Method 2.
$f_a(x) = axa^{-1} = y$
$\implies xa^{-1} = a^{-1}y$
$\implies x = a^{-1}ya$
$\implies f_a^{-1}(x) = a^{-1}xa.$
Are they both equal? What is that I'm missing here?
On
Method 1 is wrong. The correct equation is
${f_{a}(x)}^{-1} = (axa^{-1})^{-1} = ax^{-1}a^{-1} $
But in general you have that $f_{a}^{-1}(x) \neq {f_{a}(x)}^{-1}$
(On the left-hand-side there is the image of x by the inverse function of the inner automorphism function while on the right there is the multiplicative inverse of the image of x by the inner automporhism function)
Method 2 instead is right.
In the first method, you are doing $(f_a(x))^{-1}$; that is, you are inverting the output of a homomorphism. By homomorphism properties, this is also the same as $f_a(x^{-1})$.
In the second method, you are actually solving for the inverse of the homomorphism $f_a$ itself, and not taking the inverse of a single element in its output. So, the second method is correct.