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Let $G=\langle x,y,z:xy=yx,zx=x^2z,zy=yz\rangle $, I think that this group is a semidirect product, my first attempt was to prove that this was isomorphic to $\mathbb{Z}^2\rtimes_\varphi\mathbb{Z}$, where $\varphi(1)((1,0))=(2,0)$ and $\varphi(1)((0,1))=(0,1)$. If $x=((1,0),0),y=((0,1),0),z=((0,0),1)$, then $x,y,z$ satisfies those relations. But I noticed that $\varphi(1)\not\in\mathrm{Aut}(\mathbb{Z}^2)$. How can I fix this? (actually I'm not sure that $G$ is a semidirect product, so if you can prove that it is not, that's fine).

EDIT: the subgroup $\langle x,y\rangle $ is normal in G, and $\langle x,y\rangle \cap\langle z\rangle=1$ because if we add the relations $x=y=1$ we obtain the cyclic group generated by $z$. So $G=\langle x,y\rangle\rtimes_\varphi\langle z\rangle $ with $\varphi(z)(x)=zxz^{-1}=x^2$ and $\varphi(z)(y)=y$, but I want to be more explicit and find out to what semidirect product $G$ is isomorphic

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The group is a direct product, $\langle x,z\rangle\times\langle y\rangle$. –  Arturo Magidin Jan 29 '12 at 4:42
    
yeah but I want to understand if $G$ is a semidirect product of the type $<x,y>\rtimes_\varphi<z>$, I edited the question –  John Jan 29 '12 at 4:53
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That is: it is a common mistake to think that if $G=\langle X\rangle$, then in order to show $N\triangleleft G$, it suffices to check that $xNx^{-1}\subseteq N$ for all $x\in X$. You either need to show $xNx^{-1}=N$ for all $x\in X$, or you need to show $xNx^{-1}\subseteq N$ and $x^{-1}Nx\subseteq N$ for all $x\in X$. –  Arturo Magidin Jan 29 '12 at 5:37
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Killing $y$ gives the group $\langle x,z\ |\ zx=x^2z\rangle$. By the Freiheitssatz, $x$ has infinite order. –  user641 Jan 29 '12 at 18:57
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@John: What did you study? What course is this for? You are asking big questions, and then asking people to walk you through every single step one question at a time. Perhaps if you gave context, people wouldn't be talking past you and wasting everybody's time. –  Arturo Magidin Jan 29 '12 at 22:26

1 Answer 1

up vote 2 down vote accepted

Let's put all the information together...

The fact is that the subgroup $\langle x,y\rangle$ is not normal in $G$, contrary to the assertion in the edit. Although $z\langle x,y\rangle z^{-1}\subseteq \langle x,y\rangle$, we do not have equality, and this implies that the subgroup is not invariant under conjugation by $z^{-1}$.

To see this, we first show that each of $x$, $y$, and $z$ are of infinite order.

It is easy to see that $z$ and $y$ have infinite order: add the relations $x=y=1$ to get an infinite cyclic group generated by $z$. Add the relation $x=z=1$ to get an infinite cyclic group generated by $y$.

Showing $x$ has infinite order seems a bit harder to me to do from first principles. There may very well be something easier, but here is what I came up with:

For each $n\gt 0$, consider the cyclic group $C_{2^n-1}$ of order $2^n-1$, written multiplicatively and generated by an element $a$. The homomorphism given by $a\mapsto a^2$ is an automorphism, since $\gcd(2,2^n-1) = 1$, so $a^2$ is also a generator of the group. Thus, we can form the semidirect product $C_{2^{n}-1} \rtimes C_n$, with $b$ being the generator of the cyclic group of order $n$, where $bab^{-1} = a^2$. Since $a$ and $b$ in this group satisfy $ba = a^2b$, the group $G$ has $C_{2^n-1}\rtimes C_n$ as a quotient, mapping $x\mapsto a$, $z\mapsto b$, $y\mapsto 1$. Thus, $x$ has order at least $2^n-1$ for each $n\gt 0$, and hence $x$ must be of infinite order as well.

Now note that $\langle x\rangle\cap\langle y\rangle = \{1\}$ (again, set $x=z=1$ to see this), it follows that $\langle x,y\rangle \cong \mathbb{Z}\times\mathbb{Z}$; the action of $z$ on this group maps $(1,0)\mapsto (2,0)$, and $(0,1)\mapsto (0,1)$. This is a non-surjective endomorphism, so $\langle x,y\rangle$ is not normal in $G$.

The group is a direct product, $\langle y\rangle \times \langle x,z\rangle$.

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