# $Aut(\mathbb{Z}G)=?$ for $G=\mathbb{Z}^2\rtimes_n\mathbb{Z}$

I am interested in the automorphism group of the group ring $\mathbb{Z}G$ for some noncommutative group $G$ of the form $\mathbb{Z}^2\rtimes_n\mathbb{Z}$, say $$\mathbb{Z}^2\rtimes_n\mathbb{Z}=\langle x,y,z: xz=zx, zy=yz, xy=yz^nx\rangle$$ for some natural number $n>0$.

For example, $G$ could be the discrete Heisenberg group, i.e., $$G=\langle x,y,z: xz=zx, zy=yz, xy=yzx\rangle$$

Then it is clear that we can extend linearly an automorphism $\phi$ of $G$ to an automorphism of $\mathbb{Z}G$, we can also define an automorphism by sending $x$ to $\pm x$, $y$ to $\pm y$, and fixing $z$, but are there any other nontrivial automorphisms?

How to find $\operatorname{Aut}(\mathbb{Z}G)$ for $G=\mathbb{Z}^2\rtimes_n\mathbb{Z}$?

• It would be useful to describe the automorphism group of $G_n$ at the first place. This group lies in an extension $1\to\mathbf{Z}^2\to\mathrm{Aut}(G_n)\to\mathrm{GL}_2(\mathbf{Z})\to 1$. The right map is the action on the quotient by the central and characteristic subgroup $\langle z\rangle$. The kernel $\mathbf{Z}^2$ is an overgroup of index $n$ of the group of inner automorphisms: these are the automorphisms $(x,y,z)\mapsto (xz^a,yz^b,z)$ for $(a,b)\in\mathbf{Z}$. – YCor Nov 13 '13 at 14:43

Let $G_n$ the group generated by $x$, $y$, $z$ with $z$ central and $yx=xyz^n$; this is your group up to renaming; I like this one better :-)
The only finite conjugacy classes of the group $G_n$ are those of the powers of $z$. It follows that the center of the group algebra $\def\ZZ{\mathbb Z}\ZZ G$ is $\ZZ Z$ with $Z=(z)\subseteq G_n$ the center. An automorphism $\alpha$ of $\ZZ G_n$ must preserve the center of the ring, so it induces an automorphism of $\ZZ Z$, and the restriction of $\alpha$ to $\ZZ Z$ must in turn restrict to the group of units of $\ZZ Z$, which is $\pm Z$. The group structure of $\pm Z$ is very simple, and allows us to see that $\alpha$ must map $z$ to $z$ or to $-z$.
Let me assume now that $n$ is even. In this case, the ring $\ZZ G_n$ has an automorphism which maps $x$, $y$ and $z$ to $-x$, $-y$ and $-z$. Composing $\alpha$ with it, we can suppose that in fact $\alpha(z)=z$.
In that case, $\alpha$ preserves the ideal $I\subset\ZZ G_n$ generated by $z-1$, so passes on to the quotient $\ZZ G_n/I$, which is isomorphic to the group algebra of the abelian free group generated by the images of $x$ and $y$. The automorphisms of this quotient are easily described, I guess, so one can lift this information to $\ZZ G_n$. I'll try to check/fill in the details later.
• there seems some tricky points concerning lifting automorphism from the quotient algebra to the original one, since $x\to x+y, y\to y$ gives us an automorphism of the quotient algebra $\mathbb{Z}G_2/I$, but we can check that $\alpha(x)=x+y, \alpha(y)=y, \alpha(z)=z$ is not an automorphism of $\mathbb{Z}G_2$ since it does not preserve the skew relation $yx=xyz^2$. So, in general, we have to determine two noncommutative polynomial $p, q$, s.t., $\alpha(x)=x+y+(z-1)p, \alpha(y)=y+(z-1)q, \alpha(z)=z$ and it preserves the above skew relation. – Jiang Nov 13 '13 at 14:01
• @Jiang, In the quotient algebra, $x$ is a unit but $x+y$ is not, so there is no such authomorphism. – Mariano Suárez-Álvarez Nov 13 '13 at 14:21
• But then, it should be clear the only units of this quotient ring consist of $\pm 1, \pm x^{\pm 1}, \pm y^{\pm 1}$ and their products. – Jiang Nov 13 '13 at 14:43