If group $G = \langle a,b\mid bab^{-1}=a^{-1}\rangle$, then how to verify every element can be written uniquely as form $a^ib^j$?

Question

If group $G = \langle a,b\mid bab^{-1}=a^{-1}\rangle$, then how to verify every element can be written as form $a^ib^j$?

My attempt is that I can use the relation $ba = a^{-1}b$ to change the order of $a$ and $b$. But I think it fails to something like $ba^{-1}$, moreover I don't know how to verify the uniqueness. So how can every elements be written as $a^ib^j$? Any help and hints will be appreciated!

Best regards!

Answer 1

Let's take a look at our relation and observe that $$\begin{align} bab^{-1} = a^{-1} &\iff b = a^{-1}ba^{-1} \\ &\iff b = aba. \end{align}$$ The same properties work for $\ b^{-1}$: $$\begin{align} bab^{-1} = a^{-1} &\iff b^{-1} = a^{-1}b^{-1}a^{-1} \\ &\iff b^{-1} = ab^{-1}a. \end{align}$$

Now, for every word $\ x_1^{w_1}x_2^{w_2}\dots x_n^{w_n}$ where $x_i \in \{a, b\}$ and $w_i = \pm 1,$ find the last $i$ such that $x_i = b,\, x_{i+1} = a$ and replace $x_i^{w_i}$ with one of the above to destroy $a$ after it. Do this until you will have no such $i$, and now you have it in the form $a^ib^j$.

Some examples: $$\begin{align} ba^{-1} &= (aba) a^{-1}\\ & = ab. \end{align} $$ $$\begin{align} b^{-1}b^{-1}aa &= b^{-1}(a^{-1}b^{-1}a^{-1})aa \\ &= b^{-1}a^{-1}b^{-1}a \\ &= b^{-1}a^{-1}a^{-1}b^{-1}\\ &= ab^{-1}a^{-1}b^{-1}\\ &= aab^{-1}b^{-1}. \end{align}$$ Using this operation, we can always replace $a$ and $b$ whatever powers they have; this means that we can anways reach the final form of $a^ib^j$.

Now suppose that we have two different elements that are equal: $a^ib^j=a^ub^v$. This means that for some $k, w$ where $k$ and $w$ are not zero at the same time, $a^kb^w = 1$, but this is impossible because $G$ is a free group with the only relation $bab^{-1} = a^{-1}$ (these relations are not equivalent).

Answer 2

Without uniqueness it's trivial because $ba=a^{-1}b$ and $a$ and $b$ are written in different orders on different sides of the equation. This enables you to "swap", whenever $a$ and $b$ appear next to each other.

Otoh, that all or some $a^ib^j$ are different is not as trivial, and without some more information amounts to the word problem, which says that there is no way to tell, in general, if a word is equal to $e$, the trivial word. The word problem is solvable for one relator groups though.

In this case, it is easy to see that the group is infinite. For instance, $a$ has infinite order. In general, if the deficiency of a group is $\ge1$, it is infinite. See this. In this case we have two generators and one relator.

Answer 3

This follows directly from the uniqueness of the normal form of HNN extensions, where your group is the extension of $G = \mathbb{Z}$ and the subgroups $A, B$ are both $G$ and the morphisms are $\phi_a(1) = 1$ and $\phi_b(1) = -1$