Prove normalizer is a subgroup(the inverse part) 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.
 A: 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.
