# What are the semi-direct products of $\mathbb{Z}$ with itself? (Check my work please)

I am just starting out with semi-direct products. I would like to list and describe the semi-direct products of $$\mathbb{Z}$$ with itself.

I first need to find the automorphisms $$\varphi$$ from $$\mathbb{Z}$$ to $$\mathbb{Z}$$. These automorphisms are determined by $$\varphi(1)$$ by the additive property of such morphisms : $$\varphi(n+m)=\varphi(n)+\varphi(m).$$ Therefore, I only need to determine the possible values of $$\varphi(1)$$. But I know that an automorphism must send generators on generators, and the generators of $$\mathbb{Z}$$ are $$\pm1$$, so there are at most two automorphisms : $$Id:x \mapsto x$$ and $$-Id:x \mapsto -x$$ (and they are automorphisms, so these are the only ones).

Therefore, a semi-direct product $$\mathbb{Z} \rtimes_{\psi}\mathbb{Z}$$ is given by a morphism $$\psi:\mathbb{Z}\rightarrow Aut({\mathbb{Z}}) \cong \mathbb{Z}/2\mathbb{Z}$$. The only possibilites, if I didn't make any mistakes, are $$\psi:n \rightarrow Id$$ (constant morphism) and $$\psi:n\rightarrow (-1)^nId$$ ($$\psi$$ is determined by $$\psi(1)$$ which is either $$Id$$ yielding the constant morphism or $$-Id$$ which yields the second one.)

So there are only two semi-direct products possible, one of which is the direct product (I guess?).

$$\mathbb{Z} \rtimes_{n \rightarrow id}\mathbb{Z}\cong\mathbb{Z}^2$$

and

$$\mathbb{Z} \rtimes_{n \rightarrow (-1)^nId} \mathbb{Z}$$

Did I miss any of them, and is there anything to say about this last semi-direct product? Is $$\mathbb{Z} \rtimes_{n \rightarrow (-1)^nId} \mathbb{Z}$$ isomorphic to any known groups? I'm not sure what to say now, I should give it a brief description but I don't know what else there is to say than $$\mathbb{Z} \rtimes_{n \rightarrow (-1)^nId} \mathbb{Z}$$. Its multiplicative law is $$(x,y)*(z,t)=(x+\psi(y)(z),z+t)=(x+(-1)^yz,z+t)$$, so it kind of "oscillates" and it doesn't sound familiar to me... I don't recall any groups with such a weird multiplication law. It's not even commutative I guess, since $$(1,1)*(2,2)=(1-2,4)=(-1,4)$$ but $$(2,2)*(1,1)=(2+1,2)=(3,2) \neq(-1,4)...$$

I don't really know a common name for the second semidirect product. Another way to think of it is via a presentation: it is generated by two elements $$a$$ and $$b$$ with the relation $$bab^{-1}=a^{-1}$$ (here $$a=(1,0)$$ and $$b=(0,1)$$). This group arises naturally in topology: it is isomorphic to the fundamental group of the Klein bottle.
• The first part was just meant to determine $Aut(\mathbb{Z})$. The homomorphism from $\mathbb{Z}$ to $Aut(\mathbb{Z})$ was meant to be $\psi$. I agree this was confusing. Thank you very much for your insight, especially the last part! – Evariste Jan 30 at 21:41