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Let $G$ be a group. Prove that $N=\langle x^{-1}y^{-1}xy \vert x,y \in G \rangle$ is a normal subgroup of $G$.

My attempt:

Assume $N$ is normal. Then for some $n \in N$ and some $g \in G$, we have $$ gx^{-1}y^{-1}xyg^{-1}=n $$ now let $g=x$ since $x$ is an arbitrary element of $G$. We have $$ xx^{-1}y^{-1}xyx^{-1}=y^{-1}xyx^{-1}=n $$ and this gives us $$ (y^{-1}xyx^{-1})^{-1}=xy^{-1}x^{-1}y=n^{-1}\in N $$ so $gNg^{-1} \subseteq N$

To show the reverse inclusion we have by hypothesis for some $n \in N$ and any $g \in G$ that $$ x^{-1}y^{-1}xy=gng^{-1} $$ which gives us $$ x(x^{-1}y^{-1}xy)x^{-1}=x(gng^{-1})x^{-1} $$ and finally we have $$ (y^{-1}xyx^{-1})^{-1}= (xgng^{-1}x^{-1})^{-1}=xgn^{-1}g^{-1}x^{-1}=xgn^{-1}(xg)^{-1} \in gNg^{-1} $$ so $N \subseteq gNg^{-1}$

Have I actually proved this or is this just circular reasoning? My goal was to show that if $N$ is normal then $N=gNg^{-1}$. If I haven't proved this, am I at least on the right track?

Edit: (Second Attempt)

First note that $$g[x,y]g^{-1}=g(x^{-1}y^{-1}xy)g^{-1}=gx^{-1}g^{-1}gy^{-1}g^{-1}gxg^{-1}gyg^{-1}=(gxg^{-1})^{-1}(gyg^{-1})^{-1}(gxg^{-1})(gyg^{-1})=[gxg^{-1},gyg^{-1}] $$ and since $gxg^{-1} \in G$ for all $g \in G$ then $[gxg^{-1},gyg^{-1}]\in N$ for all $g \in G$ and we have $g\langle [x,y] \vert x,y \in G \rangle g^{-1} =\langle [gxg^{-1},gyg^{-1}]\vert x,y \in G\rangle \subseteq N $ for all $g \in G$. Thus $N$ is normal since $gNg^{-1} \subseteq N$ for all $g \in G$.

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    How come you need to prove *something* and you begin your proof by **assuming** that something?! – 2017-01-02
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    @AnalysisStudent If you are proving a statement of the form $P\implies Q$, there is no such method of proof that start by assuming $Q$. – 2017-01-02
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    Ok I knew something wasn't right about that one..thank you – 2017-01-02

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Hint: For all $\;g\in G\;,\;\;[x,y]:=x^{-1}y^{-1}xy\;$ :

$$[x,y]^g:=g^{-1}[x,y]g=[g^{-1}xg,\,g^{-1}yg]$$

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    And this also works with arbitrary endomorphisms in place of conjugation. – 2017-01-02
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    @DonAntonio Thank you for your answer I woke up this morning and immediately saw the path from what you wrote. I guess there is something to be said about the significance of the subconscious in solving problems. Anyways, I edited my post with a second attempt, could you verify if it is correct or at least closer to the correct proof? – 2017-01-03
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    This is one of the best examples of a result which is proved more easily by first proving stronger result i.e $[G,G]$ is a characteristic subgroup - or even stronger, a fully invariant subgroup. – 2017-01-03