Show that $\langle x,y,z \rangle$ has order $42$ for $x,y,z$ with given properties A similar question to what I asked here $|x|=7$,$|y|=3$ then $|\langle x,y \rangle |=21$
Let $G$ be a group of order $336$ and $x,y,z \in G$ with $|x|=7,|y|=3,|z|=2$ and the following relations: $yxy^{-1}=x^2$, $yz=zy$, $zxz^{-1}=x^6$. I also know for example $(zy)^2=y^2, (zy)^3=z.$ Now I have to show that the group $\langle x,y,z \rangle$ is of order $42$.
So in the previous question, it was shown that $\langle x,y \rangle$ is of order $21$. Following a similar style,$zyx=yzx=yx^6z=(yx)x^5z=x^2yx^5z=x^2(yx)x^4z=x^4yx^4z=x^4(yx)x^3z=x^6yx^3z=x^6(yx)x^2z=xyx^2z=x(yx)xz=x^3yxz=x^5yz$ so any element can be written below as
 $$\langle x,y,z \rangle=\{x^iy^jz^k: 0\leq i \leq 6, 0\leq j \leq 2, 0\leq k \leq 1\}.$$
Is this correct?
 A: Except for the common identity element between them the cyclic subgroups $\langle x\rangle,\langle y\rangle,\langle z\rangle$ are pairwise disjoint because their sizes are coprime. $\langle x\rangle$ is a normal subgroup of $G$ because $x,y,z\in\mathcal N(\langle x\rangle)$ ; i.e. $yxy^{-1},zxz^{-1}\in\langle x\rangle$. Therefore, it suffices to show that the quotient group $\frac{G}{\langle x\rangle}$ has size $6$. Naturally, $G\over\langle x\rangle$ is generated together by $yG,zG$ which commute it was given $yz=zy$ yielding the internal direct group product $G\over\langle x\rangle$ $=\langle yG\rangle\oplus\langle zG\rangle$ where $\text{ord}(y)=\text{ord}(yG)=3$ and $\text{ord}(z)=\text{ord}(zG)=2$ so that $|\langle yG\rangle\oplus\langle zG\rangle|=3\cdot 2=6$. 
A: By $\langle x, y, z\rangle$, I'm assuming you mean the cyclic subgroup $\langle (x, y, z)\rangle $ of $G\bigoplus G\bigoplus G$, in which case 
$$\begin{align}
|\langle (x, y, z)\rangle|& = |(x, y, z)| \\
&={\rm lcm}(|x|, |y|, |z|)\\
& ={\rm lcm} (7, 3, 2)\\
& = 42.
\end{align}$$
