# Kernel of homomorphisms of the Baumslag-Solitar group BS(n,n)

I would like to find the kernel of the following homomorphisms and show that thoses kernels have trivial intersection.

$H:=\mathbb{Z}*\mathbb{Z}/n\mathbb{Z}=\langle p,q| q^n=1\rangle$

$K=\mathbb{Z}^2=\langle c,d| cd=dc\rangle$

$G:=BS(n,n)=\langle a,b| ba^nb^{-1}=a^n\rangle$

$\phi_1 :G\rightarrow H$ defined by $\phi_1(a)=q$ and $\phi_1(b)=p.$

$\phi_2 :G\rightarrow K$ defined by $\phi_1(a)=c$ and $\phi_1(b)=d.$

It is easy to see that ker($\phi_1)\leq\langle b^ka^nb^{-k} \mid k \in {\mathbb Z} \rangle.$ I wonder if this is an equality.

If $g=a^{k_1}b^{l_1}\dots a^{k_r}b^{l_r}\in G$ with $k_1,\dots,k_r,l_1,\dots, l_r\in\mathbb{Z}$, then I proved that $g\in$ ker($\phi_2$) if $k_1+\cdots k_r=0$ and $n$ divides $l_1+\cdots+l_r.$ Is there a better description of ker($\phi_2$)? By advance thank you.

• But $b^ka^nb^{-k} = a^n$. In fact $\langle a^n \rangle = Z(G)$ and this is the kernel of $\phi_1$. What you have written about $\ker(\phi_2)$ does not look right because $\phi_2(b^n)=d^n \ne 1$. In fact $\ker(\phi_2) = [G,G]$ and it is true that $\ker(\phi_1) \cap \ker(\phi_2)=1$. – Derek Holt Jul 23 '15 at 16:17
• But how do I prove ker($\phi_1)\leq\langle a^n\rangle$ and ker($\phi_2)\leq[G,G]$? – Edgar Ndie Jul 24 '15 at 12:13
• See my answer below. – Derek Holt Jul 24 '15 at 12:45
• I would like to do the same thing for $BS(n,-n)$ i.e find the kernel of $\phi_1' :BS(n,-n)\rightarrow H$ defined by $\phi_1'(a)=q$ and $\phi_1'(b)=p$ but there is a problem. Ok We still have that $N \le \ker(\phi_1')$ where $N$ is still the normal closure of $\langle a^n\rangle$ since $a^n\in\ker(\phi_1'),$ but I think in the case of $BS(n,-n),$ $N\neq\langle a^n\rangle$ since $a^n$ and $b$ don't commute in this case. So in this case wat is the kernel of $\phi_1'?$ – Edgar Ndie Aug 3 '15 at 7:43
• I would also like to find the kernel of $\phi_2':BS(n,-n)\rightarrow BS(1,-1)=\langle c,d| dcd^{-1}=c^{-1}\rangle$ defined by $\phi_2'(a)=c$ and $\phi_2'(b)=d.$ Again $N'\leq\ker(\phi_2')$ wher $N'$ is the normal closure of $bab^{-1}a$ since $bab^{-1}a\in\ker(\phi_2').$ But is $N'$ equal to $\langle bab^{-1}a\rangle?$ Because if if it's the case, then in using your method below, we'll have $\ker(\phi_2')=\langle bab^{-1}a\rangle.$ By advance thank you. – Edgar Ndie Aug 3 '15 at 7:56

Clearly $a^n \in \ker(\phi_1)$. So $N \le \ker(\phi_1)$, where $N$ is the normal closure of $\langle a^n \rangle$ in $G$. But in fact, since $a^n$ is centralized by both $a$ and by $b$, we have $a^n \le Z(G)$, so $N = \langle a^n \rangle$.
So $\phi_1$ induces a homomorphism $\bar{\phi}_1: G/N \to H$. But $$G/N = \langle a,b \mid b^{-1}a^nb=a^n, a^n=1 \rangle = \langle a,b \mid a^n=1 \rangle$$ and hence $\bar{\phi}_1$ is an isomorphism, so $N = \ker(\phi_1)$.
The proof that $\ker(\phi_2) = [G,G]$ is similar, because $$G/[G,G] = \langle a,b \mid b^{-1}a^nb=a^n, ab=ba \rangle = \langle a,b \mid ab=ba \rangle.$$