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 $(2,2)*(1,1)=(2+1,2)=(3,2) \neq(-1,4)…$
Best Answer
Your work is correct. The first semidirect product is indeed just the direct product: a semidirect product is just the direct product if the homomorphism used is trivial (since then the group operation on the semidirect product reduces to just the usual coordinatewise operation).
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.