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$G=\langle x,y,z:xy=yx,zxz^{-1}=x^{-1},zyz^{-1}=y^{-1}\rangle$, 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?

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I changed $< \bullet >$ to $\langle \bullet \rangle$. The code is \langle \rangle. – Michael Hardy Jan 27 '12 at 6:39
up vote 5 down vote accepted

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 $(i,j,k) \ne (0,0,0)$.

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Showing that $\langle x,y \rangle \cap {\mathbb Z} = 1$ is actually much easier than that. Just add the extra relations $x=y=1$ and observe that the presentation of the quotient group defined by doing that reduces to the infinite cyclic group generated by $z$. So this cyclic group does not intersect the normal closure of $\langle x,y \rangle$. But of course, Ted's argument shows more than this. It shows that $x$ and $y$ generate a free abelian group. – Derek Holt Jan 27 '12 at 8:14

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