Show that $K:=\bigcap\{H\leq G:\forall_{x,y\in G}:xyx^{-1}y^{-1}\in H\}$ is a normal subgroup of $G$ $G$ is a group,I used $\leq$ to say that $H$ is a subgroup of $G$.
Can somebody please help me to understand the Question? I know I have to show $\forall_{g\in G}:gK=Kg$, but what is $K$, what Special properties do the elements have? A hint how I could start to understand what $K$ is, why it is a subgroup in the first place and what Special properties the Elements of $K$ have would be much appreciated.
Thank you so much for helping me. 
 A: Oh dear. Your comments got quite off track. First of all, $\newcommand\inv{^{-1}}(xy)\inv =y\inv x\inv \ne x\inv y\inv$ (where by $\ne$ I don't mean never equals, but just that this equality doesn't hold generally).
Second $K$ is the subgroup generated by all elements of the form $xyx\inv y\inv$, in general there are elements that are not of this form themselves, although I don't have an example of this off of the top of my head.
The trick is the following. Suppose we have a set $S\subseteq G$ with $gSg^{-1}\subseteq S$ for all $g\in G$. Then the subgroup generated by $S$,
$$\langle S\rangle := \bigcap_{\substack{H\le G\\ S\subseteq H}} H$$
is normal.
Proof:
First note that since $gSg\inv \subseteq S$ for all $g$ in $G$, we also have $g\inv S g \subseteq S$ for all $g$ in $G$, or $S\subseteq gSg\inv$ for all $g\in G$. Hence we in fact have $gSg\inv = S$ for all $g\in G$.
Then we have
$$g\langle S\rangle g\inv = g\newcommand\of[1]{\left({#1}\right)}\of{\bigcap_{\substack{H\le G\\S\subseteq H}} H }g\inv=\bigcap_{\substack{H\le G\\ S\subseteq H}} gHg\inv.$$
Reindexing this intersection with $H'=gHg\inv$, we have
$$g\langle S\rangle g\inv 
= \bigcap_{\substack{H'\le G \\ S\subseteq g\inv H' g}} H'
=  \bigcap_{\substack{H'\le G \\ gSg\inv \subseteq H'}} H'
=  \bigcap_{\substack{H'\le G \\ S \subseteq H'}} H'
= \langle S \rangle.
$$
Thus $\langle S \rangle$ is normal as desired. $\blacksquare$
Now relating it back to your problem:
Now can you show that $g(xyx\inv y\inv)g\inv$ can be written in the form $aba\inv b\inv$ for some $a$ and $b$ in $G$? Do you see how this together with the lemma above proves what you want?
