Is this group a semidirect product? $G=\langle x,y,z:xy=yx,zxz^{-1}=x^{-1},zyz^{-1}=y^{-1}\rangle$, could you help me to understand if this group is a semidirect product of the type $\langle x,y\rangle\rtimes_\varphi\langle z\rangle$.
I was trying to prove that $\langle x,y\rangle\triangleleft G$ and $\langle x,y\rangle\cap\langle z\rangle=\{1\}$, but I'm having trouble with the second, and actually I even don't know if this is true, it's possible that this group is not a semidirect product. Could you help me?
 A: Showing that the intersection of two subgroups is trivial in a group described by generators and relations is a little tricky.
Clearly, it is enough to show that if $i,j,k$ are integers and $x^i y^j z^k = 1$, then $i=j=k=0$.  This is of course equivalent to showing that if $i,j,k$ are not all zero, then $x^i y^j z^k \ne 1$ in $G$.
In a group described by generators and relations, in order to show that some word $w \ne 1$, you need to show that there is some group $H$ such that: (1) $H$ contains elements that satisfy the relations; (2) $w \ne 1$ in $H$.  (This proves that the relations together with the group axioms do not force $w=1$; hence $w \ne 1$ in $G$.)
So we have to show, for each $(i,j,k)$, there is some group $H$ such that (1) $H$ contains three elements $x,y,z$ satisfying the given relations; (2) in $H$, we have $x^i y^j z^k \ne 1$.  
You need such an $H$ for each nonzero triple $(i,j,k)$, so you'll need to find lots of groups containing 3 elements satisfying the given relations.  A good source of such groups are the dihedral groups: $x,y$ can be any 2 rotations, and $z$ any reflection.  It is easy to check that $x,y,z$ satisfy the given relations.
The dihedral groups should give you enough $H$'s to rule out $x^i y^j z^k = 1$ for any $(i,j,k) \ne (0,0,0)$.
