Let $H=\langle\sigma\rangle$ be the infinite cyclic group. Consider $G = (\mathbb{Z}\times \mathbb{Z})\rtimes H$. Determine $G'$. 
Let $H=\langle\sigma\rangle$ be the infinite cyclic group. Consider $G = (\mathbb{Z}\times \mathbb{Z})\rtimes H$ with action given by $$ \sigma:\mathbb{Z}\times\mathbb{Z}\to \mathbb{Z}\times\mathbb{Z}:(x,y)\mapsto (x+y,y).$$
a) Compute the derived subgroup $G'$.
b) Is $G$ nilpotent? Is $G$ solvable?

My attempt:
In general, if $G = N\rtimes H$, then $(h,n)\cdot (k,m) = (hk,n^km)$ for $h,k\in H$ and $n,m\in N$, and where $n^k$ is the image of $n$ under the action of the element $k\in H$.
Let $g = (\sigma^i, (a,b))$ and $h = (\sigma^j, (c,d))$, then $[g,h] = g^{-1}g^h = (1,(jb-id,0)) $. I'm not sure what these elements generate. We have $G' \cong \langle jb-id \mid b,d,i,j\in \mathbb{Z}\rangle \overset{?}\cong \mathbb{Z} $.
Any help is wanted. Thanks.
 A: The commutator subgroup of a group $G$ is the smallest normal subgroup of $G$ for which the quotient is abelian.
Since $G/(\mathbb{Z}\times\mathbb{Z})\cong H$ is abelian, the commutator subgroup is necessarily contained in $\mathbb{Z}\times\mathbb{Z}$. On the other hand, the group is not abelian (multiply $(0,(0,1))$ by $(1,(0,0))$ to get $(1,(1,1))$, but multiplied in the othe order we get $(1,(0,1))$), so the commutator subgroup is not trivial.
It is not hard to verify that the subgroup $\mathbb{Z}\times\{0\}$ of the normal subgroup is normal in $G$ (since it is invariant under the action of $H$) and the modding out $G$ by this subgroup yields an abelian group. So again we know the commutator subgroup is contained in this subgroup of $G$. (Your calculations also show as much, by the way, but I've arrived at this without having to compute an arbitrary commutator).
So, the commutator must be a nontrivial subgroup of $\mathbb{Z}\times\{0\}$. All of them are normal ($N$ is abelian, and the action of $H$ on this subgroup is trivial). Which one is it? Well, remember that $[x,y] = x^{-1}y^{-1}xy$ is the unique element such that $xy=yx[x,y]$. If we take the two elements I used above, $x=(0,(0,1))$ and $y=(1,(0,0))$, we have $xy=(1,(1,1))$ and $yx=(1,(0,1))$. The unique element such that $xy=yx[x,y]$ is $(0,(1,0))$, sincee
$$yx(0,(1,0)) = (1,(0,1))(0,(1,0)) = (1,(1,1)) = xy.$$
Thus, the commutator subgroup contains $(0,(1,0))$, and hence we conclude that the commutator subgroup is precisely the subgroup $\mathbb{Z}\times\{0\}$ of $N$. That is,
$$G' = \{ (0,(a,0))\mid a\in\mathbb{Z}\}\leq (\mathbb{Z}\times\mathbb{Z})\rtimes \mathbb{Z}.$$
(If you happen to use the commutator convention that defines $[x,y]=xyx^{-1}y^{-1}$, then the computations above need to be adjusted, but the conclusion will be the same; this commutator is the unique element such that $xy=[x,y]yx$.)
You don't ask for help with the other questions, but they should be straightforward; the first can be answered without even bothering to compute exactly what $[G,G]$ is. The second can now be answered by noting that $[x,y]=x^{-1}x^y$, and verifying that $[G,G]$ happens to be central in $G$.
