Proving subgroups Question: Let $H$ be a subgroup of $G$ and let $K=\{x \in G: xax^{-1} \in H\ \iff\ a \in H\}$. 
Prove:
(a) K is a subgroup of G 
So for (a):
For closure, I need to show:
(i) if $a \in H$  then $xax^{-1}\in H$
(ii) if $xbx^{-1} \in H$ then $b \in H$

So, for (i) the problem I am having is, how do I know if x and x$^{-1}$ is $\in$ H. The only reason I can think of is that H is a subgroup of G, so that means it holds for closure and it has an inverse. Therefore, the inverse would be x$^{-1}$ and the original would be x. So if it has an inverse, then it has an identity but then:

if a $\in$ H then (xx$^{-1}$)a $\in$ H so how can I say they commute? Also, how would I obtain the inverse for (a)?
 A: I take your definition to mean 
$$
K=\{x \in G: \text{for each $a \in G$, $xax^{-1} \in H\ \iff\ a \in H$}\}.
$$
I claim that 
$$
K = N_{G}(H) = \{ x \in G : xHx^{-1}= H\}
= \{ x \in G : \text{for each $a \in H$, we have $x a x^{-1}, x^{-1} a x \in H$} \},
$$
 the latter subgroup being the normalizer of $H$ in $G$, which is known to be a subgroup of $G$.
In fact, $N_{G}(H) \le K$. Let $x \in N_{G}(H)$, and $a \in G$. If $a \in H$, then $x a x^{-1} \in H$. If $x a x^{-1} \in H$, since $x^{-1} \in N_{G}(H)$, we have $a = x^{-1} (x a x) x^{-1} \in H$.
Now let us prove $K \le N_{G}(H)$. Let $x \in K$, and $a \in G$. If $a \in H$, then $x a x^{-1} \in H$. Conversely, if $a \in H$, then $a = x (x^{-1} a x) x^{-1} \in H$, thus $x^{-1} a x \in H$, so that $x \in N_{G}(H)$.

If you want a direct proof, note that clearly $1 \in K$.
Then, if $x \in K$, we have
$$
a = x (x^{-1} a x) x^{-1} \in H \iff x^{-1} a x \in H
$$
that is, $x^{-1} \in K$.
And finally, if $x, y \in K$, then
$$
(xy) a (x y)^{-1} = x (y a y^{-1}) x^{-1} \in H 
\iff y a y^{-1} \in H \iff a \in H.
$$
A: To show that $K = \{x : x a x^{-1} \in H \text{iff} a\in H\}$, first we should understand what this group is. In this case, this is the group of elements that fix both $H$ and its complement under conjugation. So now all that is left to do is, let $x,y \in K$ show that $xy^-1 \in K$, and this is just a standard procedure:
$$(x y ^{-1}) a (x y ^{-1})^{-1} = x (y^{-1} a y) x^{-1}$$
Now since $y \in K$, and $y (y^{-1} a y ) y^{-1} = a \in H$, then $ y^{-1} a y \in H$. Finally since $x \in K$, as with $y$ it only fixes $H$ by conjugation.
$$x (y^{-1} a y) x^{-1} \in H$$
and so $xy^{-1}\in K$ for each $x,y\in K$, implying that $K$ is a subgroup of $G$
A: You need to show the elements $xax^{-1}$ form a group, which means they satisfy the group axioms.
The first one is closure:
The product of two numbers $xax^{-1}$ and $xbx^{-1}$ = $xax^{-1}$ * $xbx^{-1}$
= $xax^{-1}$ * $xbx^{-1}$
= $xabx^{-1}$
Which is of the form xcx^-1 and so an element of K..
Next is associativity, again the x and $x^{-1}$ terms cancel out so that
$xax^{-1}$ * $xbx^{-1}$ * $xcx^{-1}$ = $xabcx^{-1}$,  and as the original group is associative so are the terms $xax^{-1}$.
Now check each element has an inverse in K. You can verify that $xa^{-1} x^{-1}$ is the inverse of $xax^{-1}$ and is in K.
Finally the identity element of K is $xIx^{-1}$ which is in K and equal to I (I is the identity element for G).
As K meets the four requirements of a group it is a group. 
