I would like to find the kernel of the following homomorphisms and show that those kernels have trivial intersection.
$H:=\mathbb{Z}*\mathbb{Z}/n\mathbb{Z}=\langle p,q| q^n=1\rangle$
$BS(1,-1)=\langle c,d| dcd^{-1}=c^{-1}\rangle$
$G:=BS(n,-n)=\langle a,b| ba^nb^{-1}=a^{-n}\rangle$
$\phi_1' :BS(n,-n)\rightarrow H$ defined by $\phi_1'(a)=q$ and $\phi_1'(b)=p.$
$\phi_2':BS(n,-n)\rightarrow BS(1,-1)=\langle c,d| dcd^{-1}=c^{-1}\rangle$ defined by $\phi_2'(a)=c$ and $\phi_2'(b)=d$
We have that $N≤\ker(\phi_1')$ where $N$ is the normal closure of $\langle a^n\rangle$ since $a^n\in\ker(\phi_1')$, but I think $N\neq\langle a^n\rangle$ since $a^n$ and $b$ don't commute. So what is the kernel of $\phi_1'?$
For $\phi_2',$ we have $N′≤ker(\phi_2')$ where $N′$ is the normal closure of $bab^{−1}a$ since $bab^{−1}a\in\ker(\phi_2').$ But is $N′$ equal to $\langle bab^{−1}a\rangle$ and is $N'$ the kernel of $\phi_2'?$
By advance thank you.