Let kQ be the path algebra of a Dynkin quiver over a finite field with $q=p^n$ elements.
Let $f(Q,p^n)$ denote the exponent of the unit group of kQ. Is there an explicit formula for $f(Q,q)$?



Exponent of unit group of Dynkin quiver algebras
If $Q$ is an acyclic quiver, then the exponent of the unit group of $kQ$ is $(q-1)p^m$ where $m$ is chosen smallest so that $p^m$ is greater than the length of the longest path in $Q$.


Here is a proof. A unit of $kQ$ is of the form 
$$u=\sum\_{v\in Q\_0}a\_v\epsilon\_v+r$$
 where $Q\_0$ is the vertex set, $\epsilon\_v$ is the empty path at $v$, $a\_v\neq 0$ for all $v$ and $r$ is a linear combination of paths of length $1$ or more. 


Then $u^{q-1}= 1+r'$ where $r'$ is a linear combination of paths of length $1$ or more. Thus $(1+r')^{p^m}=1+(r')^{p^m}=1$ as any product of $p^m$ paths of length $1$ or more is $0$ by choice of $m$. Thus $u^{(q-1)p^m}=1$.


Now let $a$ be a primitive element of $k$, that is, have order $q-1$. Consider $u=a(1+\sum\_{e\in Q\_1}e)$ where $Q\_1$ is the set of edges of $Q$. Then $u$ is a unit and we claim it has exponent $(q-1)p^m$. Indeed, let $u'=1+\sum\_{e\in Q\_1}e$. Then $u'$ has order a power of $p$ (by the argument in the previous paragraph any element of the form $1+r'$ with $r'$ a linear combination of paths of length 1 or more has order dividing $p^m$) and $a$ has order $q-1$, which is relatively prime to $p$. Since $a$ and $u'$ commute, the order of $u$ is the $(q-1)p^s$ where $p^s$ is the order of $u'$.


Let $\alpha$ be a path of length $p^s$. Then $\alpha$ appears with coefficient $1$ in $(u')^{p^s}$ by construction. Thus if $p^s$ is not larger than the longest path in $Q$, then $(u')^{p^s}\neq 1$. Thus $u'$ has order $p^m$ and so $u=au'$ has order $(q-1)p^m$.


