For the elements of the group, we introduce the notation $h^{g}=g^{-1}h g$ and $u \leftrightarrow v$ will mean that the elements $u, v$ commute: $uv = vu$. Suppose that in some group $G$ for fixed elements $a, b\in G$, the following relations hold: $a^b\leftrightarrow b^a$ and $a^b\leftrightarrow b^{a^2}$. Prove that the relations $a^{b^m}\leftrightarrow b^{a^n}$ hold for any natural numbers $m$ and $n$.
I have been trying to solve this problem for a long time, I want to present my solution, and also see an "alternative" solution based on general considerations.
- Let us introduce the notation $x_1=a^{-1}$ and $x_0=ba^{-1}$. It follows from them that $a=x_1^{-1}$, $b=x_0x_1^{-1}$. The relation $a^b\leftrightarrow b^a$ can be written as $a^{ba^{-1}}\leftrightarrow b$, that is, $(x_1^{-1})^{x_0}$ commutes with $x_0x_1^{-1}$. Passing to the converse, we see that $x_1^{x_0}$ commutes with $x_1x_0^{-1}$.
Note that if $u$ commutes with $vw^{-1}$, then $u^{vw^{-1}}=u$, that is, $u^v=u^w$. Thus, we have come to the relationship $x_1^{x_0x_1}=x_1^{x_0^2}$.
By definition, we put $x_2=x_1^{x_0}$, $x_3=x_2^{x_0}$, $x_4=x_3^{x_0}$, and so on, that is, $x_n=x_1^{x_0^{n-1}}$ for all $n\ge2$. In these notations, the relation from the previous point will take the form $x_2^{x_1}=x_3$. Let us now analyze what the second of the conditional relations gives us, that is, $a^b\leftrightarrow b^{a^2}$. Here $a^{ba^{-2}}\leftrightarrow b$, that is, $(x_1^{-1})^{x_0x_1}$ commutes with $x_0x_1^{-1}$. Passing to the converse, we find that $x_1^{x_0x_1}=x_2^{x_1}=x_3$ commutes with $x_1x_0^{-1}$, and then $x_3^{x_1}=x_3^{x_0}=x_4$.
Thus, we have two relations of the form $x_n^{x_1}=x_{n+1}$ for $n=2$ and $n=3$. We deduce from them the relation for any $n\ge2$, by conducting an induction on $n$. From here it will follow that $x_j^{x_i}=x_{j+1}$ for all $j > i\ge0$. Indeed, the case $i=0$ follows from the definition, and for $i\ge1$ it will be necessary to take the relation $x_n^{x_1}=x_{n+1}$, where $n=j-i+1$, and conjugate both of its parts by the element $x_0^{i-1}$. With this conjugation, all subscripts will increase by $i-1$, and one will turn into another.
Proving the statement $x_n^{x_1}=x_{n+1}$ for $n\ge4$, we can now assume by induction that $x_j^{x_i}=x_{j+1}$ for the case of $0 < j-i < n-1$ by virtue of the conjugation considered above, a suitable degree is $x_0$. In particular, $x_n=x_{n-1}^{x_{n-2}}$. Conjugating both sides by $x_1$ and using the induction assumption (taking into account the fact that $n-2 > 1$), we have $x_n^{x_1}=x_n^{x_{n-1}}=x_{n+1}$.
- Relations of the form $x_j^{x_i}=x_{j+1}$ for $j > i\ge0$, define the famous Richard Thompson group $F$. It is also given by two generating $x_0$, $x_1$ and two defining relations $x_2^{x_1}=x_3$, $x_3^{x_1}=x_4$, where $x_2$, $x_3$, $x_4$ are given by the equalities from point 2). Now it remains for us to check that all relations of the form $a^{b^m}\leftrightarrow b^{a^n}$ follow from what was deduced above.
So, $a^b=(x_1^{-1})^{x_0x_1^{-1}}=(x_2^{-1})^{x_1^{-1}}=x_1x_2^{-1}x_1^{-1}$, and then $a^{b^2}=(a^b)^{x_0x_1^{-1}}=(x_2x_3^{-1}x_2^{-1})^{x_1^{-1}}=x_1x_2x_3^{-1}x_2^{-1}x_1^{-1}$, from where by induction we have the equality $a^{b^m}=x_1x_2\ldots x_mx_{m+1}^{-1}x_m^{-1}\ldots x_2^{-1}x_1^{-1}$. This element must commute with $b^{a^n}=x_1^n(x_0x_1^{-1})x_1^{-n}$. Conjugating both sides by the element $x_1^{-1}$, we come to the condition $x_2\ldots x_mx_{m+1}^{-1}x_m^{-1}\ldots x_2^{-1}\leftrightarrow x_1^{n-1}x_0x_1^{-n}$. Now we conjugate both sides by the element $x_1^{n-1}$, which on the left side increases all subscripts by $n-1$, and we get the equivalent condition $x_{n+1}\ldots x_{m+n-1}x_{m+n}^{-1}x_{m+n-1}^{-1}\ldots x_{n+1}^{-1}\leftrightarrow x_0x_1^{-1}$. But this is already obvious due to the remark at the end of point 1): it is clear that the conjugation of the left part by both the element $x_0$ and the element $x_1$ leads to an increase in all subscripts by one, that is, to the same result.