The definition I am using is $$N_G(K) = \{ g \in G : gkg^{-1} \in K \ \forall k\in K \} $$
I am trying to show that if the element $g\in N_G(K)$ then $g^{-1} \in N_G(K)$.
Proof Idea: If $k_1\in K$ our claim is that $g^{-1}k_1g \in K$. I can seem to get from the fact that $gkg^{-1} \in K$ for all $k\in K$ to $g^{-1}k_1g \in K$ thus I am stuck.
Is what I am trying to prove even true ?
I saw this proof that is similar to mine but it is wrong(I think so):
Let $g$ be in $N_G(K)$.
Note that for any $k$ in $K$, $g^{(-1)} k (g^{(-1)})^{(-1)} = (g k^{(-1)} g)^{(-1)}$. (Wrong identity thus proof fails)
Since $g$ is in $N_G(K)$, $g k^{(-1)} g$ is in $K$ (because $k^{(-1)}$ is in $K$), and thus $(g k^{(-1)} g)^{(-1)}$ is also in $K$.
Therefore, $g^{(-1)}$ is also in $N_G(K)$.
Please use this definition when answering the question and tell me why the proof above fails.
The reason you can’t prove it is that the “definition” you’re using is wrong.
What you are using is $N(K)=\{g\in G: \forall k\in K, gkg^{-1}\in K\}$, or in other words, $N(K)=\{g\in G: gKg^{-1}\subset K\}$. But the right definition is $$ N_G(K)=\{g\in G:gKg^{-1}=K\}\,. $$ Here’s a (fairly recondite) example that gives a case where the set defined in your formulation is not closed under inverse: for the underlying sets, take $\Bbb Z\times\Bbb Q$, but with the unusual law of combination depending on a fixed positive integer $p$ (need not be prime) $$ (m,a)\star(n,b)=(m+n,a+p^mb)\,. $$ If I have my group theory right, this is an example of a semidirect product. At any rate, you see that no matter how you parenthesize the three-fold product, you get $$ (m,a)\star(n,b)\star(r,c)=(m+n+r,a+p^mb+p^{m+n}c)\,. $$ So, this law of combination is associative, clearly noncommutative, and the $\star$-inverse of $(m,a)$ will be $(-m,-a/p^m)$. Thus we have a good group. Let’s look at the set $K$ of all $(0,k)$ for $k\in\Bbb Z$. You see that it’s a good subgroup, isomorphic to the additive group $\Bbb Z$. Now, according to your formulation, $g=(1,0)$ is in the pseudo-normalizer of $K$, since for every $k$, $g\star(0,k)\star g^{-1}=(1,0)\star(0,k)\star(-1,0)=(1,0)\star(-1,k)=(0,pk)\in K$. But you check that $g^{-1}\star(0,k)\star g=(0,k/p)\notin K$, certainly not for every $k$, anyway.