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