Help with semidirect product I need help with this problem, i am trying to understand the semidirect product, so if anyine could help or give me some ideas
Let $G$ be the group generated by $<a,b>$ and the relations $aba^{-1}b=1$.


*

*Prove that G has a normal subgroup $H\simeq\mathbb{Z}$ such that $G/H\simeq\mathbb{Z}$

*Show that exist a short exact sequence of the form
$$0 \rightarrow \mathbb{Z}\oplus\mathbb{Z} \rightarrow G \rightarrow \mathbb{Z}_2 \rightarrow 0$$

*Show that $G$ is isomorphic to only one of the next groups:
$$\mathbb{Z} \times \mathbb{Z} ; \mathbb{Z} \times \mathbb{Z}_2;  \mathbb{Z}\ltimes\mathbb{Z} ;  \mathbb{Z}\ltimes\mathbb{Z}_2$$
 A: Notice that $aba^{-1}=b^{-1}$. This means that the cyclic subgroup $<a>$ action (by conjugation) on the cyclic subgroup $<b>$ is an automorphism. Now think about the definition of semidirect product. Which group is acting on which group, and in what way, in a semidirect product? Specifically, if $G=H\ltimes_{\Phi} N$, and if you take $h\in H$ and $n\in N$, what would you expect to be the result of $hnh^{-1}$, written in term of $\Phi$?
A: Cool group $G$ you have there, its $\mathbf{Z}\ltimes\mathbf{Z}$, the fundamental group of the Klein bottle.  
To see that $G$ is actually $\mathbf{Z}\ltimes\mathbf{Z}$ (and this is the hardest part) we check the conditions of the splitting lemma, we have to show that there is short exact sequence 
$$1\rightarrow\mathbf{Z}\rightarrow G\rightarrow\mathbf{Z}\rightarrow 1,$$
the first map will be denoted $\phi$ and the second by $\psi$. We have to show that there is a map $\eta:\mathbf{Z}\rightarrow G$ such that $\psi\circ\eta$ is the identity on $\mathbf{Z}$.
Define $\phi:\mathbf{Z}\rightarrow G$ to be the composite $\mathbf{Z}\rightarrow F(\{a,b\})\rightarrow G$, where $\mathbf{Z}\rightarrow F(\{a,b\})$ is defined by $1\mapsto b$ (this extends uniquely since $\mathbf{Z}=F(\{1\})$, and the second map is the quotient map.
Define  $\psi:G\rightarrow\mathbf{Z}$ as follows: send $a$ to $1$, $b$ to $0$; the induced map $F(\{a,b\})\rightarrow\mathbf{Z}$ is surjective and descends to a surjection $\psi:G\rightarrow\mathbf{Z}$ since the element $aba^{-1}b$ lies in his kernel. 
Define $\eta:\mathbf{Z}\rightarrow G$ to be the composite $\mathbf{Z}\rightarrow F(\{a,b\})\rightarrow G$, where the first map is by $1\mapsto a$, extending uniquely, and the second one is the quotient map. We have $\psi\circ\eta=\mathrm{id}_\mathbf{Z}$ since this holds already on generators, namely on $1\in\mathbf{Z}$.
Your 1) is trivial now: let $H$ be the image of $\phi$, we have $\mathbf{Z}\cong H$ since $\phi$ is injective, and $G/H\cong\mathbf{Z}$ by the exactness of the sequence. As for 3)
notice that the groups are all nonisomorphic (by comparing their group properties, e.g. the first one is free abelian which the second one is not, our group is not abelian but the first two are, etc; by the way the presentation of the first one $\left\langle a,b \ | \ aba^{-1}b^{-1}=1\right\rangle $). For 2) define the maps as I did above, e.g. the surjective homomorphism $F(\{a,b\})\rightarrow\mathbf{Z}/2\mathbf{Z}$ by $a\mapsto 1$ and $b\mapsto 0$ descends to a surjective map $G\rightarrow\mathbf{Z}/2\mathbf{Z}$, etc. etc.
