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.

  • $\begingroup$ 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. $\endgroup$ – YCor May 25 at 22:14
  • $\begingroup$ 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)? $\endgroup$ – user10354138 May 26 at 0:38
  • $\begingroup$ Sorry, that's a typo in the title, it's supposed to be as in the body. Fixed now. $\endgroup$ – pizzaroll May 26 at 9:09
  • $\begingroup$ 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. $\endgroup$ – 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:

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


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

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.


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