# Finding a surjective homomorphism $G \to \mathbb{Z}[\frac12, \frac13] \rtimes \mathbb{Z}$ which is not iso

Let $$G = \langle a,t \mid t a^2 t^{-1} = a^3 \rangle$$. I am asked to find a surjective group homomorphism $$G \to \mathbb{Z}[\frac12, \frac13] \rtimes \mathbb{Z}$$, where $$\mathbb{Z}[\frac12, \frac13] := (\{ a 2^b 3^c \mid a,b,c \in \mathbb{Z}\}, +)$$ and we can choose the action in the semidirect product (and I need to prove it is a surjective homomorphism).

My first thought is that $$G$$ is the HNN extension $$\mathbb{Z} \ast_{2\mathbb{Z} = 3\mathbb{Z}}$$.

I think a good action would be $$\phi(n)(a 2^b 3^c) = a 2^{b-n} 3^{b+n}$$. Then we can define a map $$\psi$$ by $$t \mapsto (0,1)$$ and $$a \mapsto (1,0)$$. The relation is satisfied, since $$\psi(t a^2 t^{-1}) = (0,1)(2,0)(0,-1) = (0,1)(2,-1) = (3,0) = \psi(a^3)$$. I am not sure how to show this is surjective and maybe it isn't, I can't really see how to get all of the elements $$(a 2^b 3^c, 0)$$. It is possible I have the map wrong.

I am also asked to prove this is not an isomorphism. I can only guess these groups are not isomorphic, I am not sure my action is correct, so I am a bit stuck here. Neither group is abelian, both have elements that don't commute, I don't know what I can use.

Help finishing this off would be greatly appreciated.

• You just need to find a generating pair in the target group, satisfying the given relator. (BTW, $Z[1/2,1/3]$ can more briefly written as $Z[1/6]$). It seems you chose the good action and found one. – YCor May 25 at 22:14
• What is the range of the homomorphism suppose to be, $\mathbb{Z}[1/2,1/3]$ (as in title) or $\mathbb{Z}[1/2,1/3]\rtimes\mathbb{Z}$ (as in body)? – user10354138 May 26 at 0:38
• Sorry, that's a typo in the title, it's supposed to be as in the body. Fixed now. – pizzaroll May 26 at 9:09
• To prove that it's not an isomorphism, you need to find an element in the kernel of the map and then show that element is nontrivial in $G$. For that you need to know something about HNN extensions, such as Britten's lemma. – Derek Holt May 26 at 9:19

Using the map $$\psi$$ in the question body given by $$\psi(a) = (1,0)$$ and $$\psi(t) = (0,1)$$ and the action for the semidirect product where $$1 \in \mathbb{Z}$$ acts by multiplication by $$\frac32$$, I came up with the following proof. For surjectivity of $$\psi$$:

We need generators for $$\mathbb{Z}[1/6] \rtimes \mathbb{Z}$$, which are $$(1/6^n, 0)$$ for all $$n \in \mathbb{Z}$$ and $$(0, n)$$ for all $$n \in \mathbb{Z}$$.

We can get $$(1/2^n, 0)$$ by noting we already have it for $$n=0$$, $$\psi(a) = (1,0)$$. Then inductively we get the rest by doing the following for $$n=1,2,\ldots$$. Define $$x_n = \frac{3^n - 1}{2}$$, which is in $$\mathbb{Z}$$ since $$3^n - 1$$ is even. Then:

$$\left(\frac32\right)^n - \frac{x_n}{2^{n-1}} = \frac1{2^n}$$

So:

$$(1/2^n, 0) = (0,n)(1,0)\left(-\frac{x}{2^{n-1}}, 0\right)(0,-n)$$

Then also:

$$(1/6^n, 0) = (0,-n)(1/2^{2n}, 0)(0,n)$$

So the map hits all generators, is surjective.

It's not isomorphic since we have a nontrivial element in the kernel. Write $$\mathbb{Z} = \langle a \rangle$$, $$H = \langle a^2 \rangle$$, $$\theta(a^2) = a^3$$. So $$\theta$$ is a monomorphism defining the HNN extension: $$G = \mathbb{Z} \ast_H$$.

Britton's lemma means the element $$g = tat^{-1}a^{-1}tat^{-1}a^{-2}$$ is nontrivial. Then since:

$$tat^{-1}a^{-1} \mapsto (0,1)(1,0)(0,-1)(-1,0) = (3/2, 0)(-1, 0) = (1/2, 0)$$

Adding that to itself and adding $$\psi(a^{-1}) = (-1,0)$$ gives $$(0,0)$$ as desired.

Thanks to the comments by YCor and Derek Holt for helping me to come up with this solution.