Unipotent matrices are conjugate to their inverses

Question

An element $a$ in a ring $R$ is called unipotent if $(1-a)^k=0$ for some positive integer $k$. Let $F$ be a field and $n$ a positive integer. It is not difficult to see that each unipotent element in the ring of matrices over $F$ is of forms an upper triangular matrix whose main diagonal entries are $1$. According to 4', Page 302, each unipotent matrix is a product of at most two elements of order $2$. Recall that an element $a$ of order $2$ means $a^2=1$. Observe that a product of at most two elements of order $2$ is conjugate to its inverse, and thus any unipotent matrix is conjugate to its inverse.

I would like to ask whether it is true in general, that is, any unipotent element in an arbitrary ring is conjugate to its inverse. Observe that if $x$ is an unipotent element in a ring $R$, then $$x^{-1}=(1-(1-x))^{-1}=1+(1-x)+(1-x)^2+\cdots+(1-x)^{k-1}.$$ Therefore, we need find an element $a$ in $R$ such that $x=ax^{-1}a^{-1}$, that is, $xa=ax^{-1}$. However, I do not know the existence of such an element. Any counterexample or reference or technique is very much appreciated. Thank you in advance.

Answer 1

I would like to ask whether it is true in general, that is, any unipotent element in an arbitrary ring is conjugate to its inverse.

For a counterexample we only need to find an unipotent element of a commutative ring that is not self-inverse. We can consider for simplicity the rings $ℤ/nℤ$. For these rings, the smallest counterexample are the elements $4$ and $7$ in $ℤ/9ℤ$.

Addendum

I would like to ask whether each unipotent element in a non-commutative ring can be expressed as a product of finite elements $a_1, a_2, \dotsc, a_m$ in which each $a_i$ is conjugate to $a_i^{-1}$.

Take some noncommutative ring $S$ and consider $R ≔ ℤ/9ℤ × S$. The ring $R$ is also noncommutative, and the elements $(4, 1)$ and $(7, 1)$ of $R$ are unipotent.

If $(4, 1)$ had the desired form, then by projecting on the first coordinate, $4$ would also have the desired form. That is, there would exist elements $a_1, \dotsc, a_n$ of $ℤ/9ℤ$ such that $4 = a_1 \dotsm a_n$, each $a_i$ is invertible, and each $a_i$ is conjugated to its inverse. As $ℤ/9ℤ$ is commutative, this would mean that each $a_i$ is self-inverse.

The invertible elements of $ℤ/9ℤ$ are $1, 2, 4, 5, 7, 8$, of which only $1$ and $8$ are self-inverse. But the only elements that can be represented as elements of the form $b_1 \dotsm b_m$ with $b_i ∈ \{1, 8\}$ are $1$ and $8$, precisely because these two elements commute and are self-inverse. Therefore, $4$ is not of the desired form. (We could also have used $7$ instead of $4$.)