Let $H$ be a subgroup of $G$, and let $K=\{x∈G:xax^{−1}∈H\text{ for every }a∈H\}$, prove that $K$ is a subgroup of $G$. Let $H$ be a subgroup of $G$, and let $K=\{x\in G:xax^{-1}\in H \text{ for every a}\in H\}$, prove that $K$ is a subgroup of $G$.

I don't know how to prove the inverse part of (a), that $x\in K \implies x^{-1}\in K$.
This is an exercise from Pinter's book, and I found an anwser from here, but I don't get the iff part of the proof, i.e., the following argument:

Let $p\in K$. Choose an element $a\in G$, and let $b=pap^{-1}$. Then, $b\in H$ iff $a$ is.



*

*If $a\in H$, then $b\in H$: ok, this is obvious.

*If $b\in H$, then $a\in H$: why?


Note that the problem on the 1st edition of the book is the 1st paragraph of this post, whereas in the answer page, the exercise 7 is stated differently (i.e., added iff part) as shown in the 2nd edition:


  
*Let H be a subgroup of G and let K = {x∈G: xax-1∈H iff a∈H}.
  

So it seems there were some upates...
 A: Note: as pointed out in the comment below this answer, this reasoning only really applies to the case where $G$ is a finite group, which I believe is what the author had in mind.  

Here's one argument that works. Note that the map $f_p:H \to H$ given by
$$
f_p(a) = pap^{-1}
$$
is invertible (in the infinite case, injective with left-inverse) with $(f_p)^{-1} = f_{p^{-1}}$. Thus, $f_{p^{-1}}$ defines a map from $H$ to $H$ which is to say that if $pap^{-1} \in H$, we have
$$
f_{p^{-1}}(pap^{-1}) = a \in H
$$
as desired.  
In other words, the iff in the second definition is redundant.  Given that $xax^{-1} \in H$ for every $a \in H$, it must also hold that $xax^{-1} \in H$ IFF $a \in H$.

That being said: with the above argument, we have $p \in K \iff f_p$ maps $H$ to $H$.  From this perspective, the fact that $p \mapsto f_p$ defines a homomorphism makes it clear that $K$ will be a subgroup.
A: Hint:
For your first question,use the defining property of $K$ with $a^{-1}$, which belongs to $H$ since $H$ is a subgroup.
Second question:  express $a$ in function of $b$.
Incidentally, $K$ is called the normaliser of $H$, as it is the greatest subgroup of $G$ containing $H$ in which $H$ is normal.
