Trouble with a group presentation

Question

I'm trying to prove that the elements of the group $\langle h_{0},h_{1} | h_{1}h_{0}h_{1}^{-1}=h_{0}^{2}\rangle$ can be expressed uniquely as $h_{0}^{n}h_{1}^{m}$ for some $n$, $m\in \mathbb{Z}$. In order to prove that, I only need to prove that $h_{1}h_{0}$, $h_{1}^{-1}h_{0}$, $h_{1}h_{0}^{-1}$ and $h_{1}^{-1}h_{0}^{-1}$ can be written in that way. I know that $h_{1}h_{0}=h_{0}^2h_{1}$ and $h_{1}h_{0}^{-1}=h_{0}^{-2}h_{1}$, but I'm not able to prove the other two cases. Can anyone help me, please?

Answer 1

The answer is that it's not possible. Let's use $a$ and $b$ for our group generators, so our group is $$ G = \langle a,b \mid bab^{-1} = a^2\rangle. $$ Let $$ A = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix},\qquad B = \begin{bmatrix}2 & 0 \\ 0 & 1\end{bmatrix}. $$ It is easy to check that $BAB^{-1}=A^2$, so there exists a homomorphism $\varphi\colon G\to \mathrm{GL}(2,\mathbb{R})$ satisfying $\varphi(a)=A$ and $\varphi(b)=B$.

Now, if $m$ and $n$ are integers, then $$ \varphi(a^mb^n) = A^mB^n = \begin{bmatrix}1 & m \\ 0 & 1\end{bmatrix}\begin{bmatrix}2^n & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}2^n & m \\ 0 & 1\end{bmatrix}. $$ However, $$ \varphi(b^{-1}a) = B^{-1}A = \begin{bmatrix}1/2 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}1/2 & 1/2 \\ 0 & 1\end{bmatrix}. $$ Since $m=1/2$ is not an integer, it follows that $b^{-1}a$ cannot be written as $a^mb^n$ for any $m,n\in\mathbb{Z}$.

Answer 2

Using $a$ and $b$ instead of $h_0$ and $h_1$ for obvious reasons:

We have the group $\langle a,b \mid bab^{-1} = a^2 \rangle$.

So, $ba=a^2b$.

We show that:

$$\begin{array}{lcl} 1 &=& a^0b^0 \\ a^m b^n a &=& a^{m+2^n}b^n \\ a^m b^n b &=& a^mb^{n+1} \\ a^m b^n a^{-1} &=& a^{m-2^n}b^n \\ a^m b^n b^{-1} &=& a^mb^{n-1} \end{array}$$

That means, the identity is expressible in $a^xb^y$ and for any expressible word, so is its right product with $a$, with $b$, with $a^{-1}$, and with $b^{-1}$.