Having trouble proving that $N \trianglelefteq G$ 
I am having a bit trouble showing $N$ is normal in $G$.
The group $G$, of order 18, and generated by $x,y$ and $z$, of order 2, 3 and 3, respectively, satisfy the following conditions:
$$yz=zy, \quad x^{-1}yx=y^{-1}, \quad x^{-1}zx=z^{-1}$$
Let $N:=\langle y,z \rangle$. 
I know that for $N$ to be normal in $G$, I have to show that $gNg^{-1} \subseteq N$ for all $g\in G$, i.e. showing $xyx^{-1}, yyy^{-1}, zyz^{-1}, xzx^{-1}, yzy^{-1}$ and $zzz^{-1} \subseteq N$. I have already noticed that $yyy^{-1}, zyz^{-1},yzy^{-1}$ and $zzz^{-1} \subseteq N$, because $N$ is generated by $y,z$, and by the given conditions $y$ and $z$ commute. Now I only have to show that $xyx^{-1}, xzx^{-1} \subseteq N$, but I am kinda stuck. 
I have noticed by the given conditions, that $yz=zy \Longleftrightarrow y=zyz^{-1}$, and I have tried to substitute $y$ in $xyx^{-1}$, (and same for $z$), but that didn't get me far. I know that I am missing something simple, but I can't figure out what. Maybe somebody can help with a hint?
 A: You can do this almost hands-down. If $G=\langle x,y,z \rangle$, then a subgroup $N \leq G$ is normal if and only if $x,y$ and $z$ normalize $N$. Since in this case $N=\langle y,z \rangle$, you only need to verify this for $x$. But this follows from the second and third presentation equations.
A: Since $G=\langle x,y,z\rangle$ we must have $G=\langle x^{-1},y,z\rangle$ and thus $\langle T_{x^{-1}},T_{y},T_{z}\rangle$ is precisely the $G$ inner automorphism group $\text{Inn}(G)$ where  $$T_{x^{-1}}(g)=x^{-1}gx\;\;\;\;T_y=ygy^{-1}\;\;\;\;T_z(g)=zgz^{-1}$$
Therefore, to see $N\unlhd G$, it suffices to show $$T_{x^{-1}}[N]=T_y[N]=T_z[N]=N$$ but $y,z\in N=\langle y,z\rangle$ and thus $T_y[N]=T_z[N]=N$. Now, since $T_{x^{-1}}(y)=y^{-1}$ and $T_{x^{-1}}(z)=z^{-1}$, we must have $$\therefore\;T_{x^{-1}}[\langle y,z\rangle]=\langle T_{x^{-1}}(y),T_{x^{-1}}(z)\rangle=\langle y^{-1},z^{-1}\rangle=\langle y,z\rangle$$ $$\therefore\;x^{-1}\;\text{lies in the normalizer $\mathcal N(N)$ of $N$}$$
$$\therefore\;x\;\text{lies in the normalizer $\mathcal N(N)$ of $N$}$$
$$\therefore\;T_x[N]=N$$
