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Let $G$ be a group, $k$ a field and $T(n,k)\subset Gl(n,k)$, the group of invertible upper triangular $n\times n$ matrices. I know that if $\rho : G\rightarrow T(n,k)$ is faithful (i.e. into) then $G$ is resoluble, we have also theorems for the converse of the form (Lie-Kochin)

$G\mathrm{\ is\ resoluble\ +\ conditions}(G,k)$ $\Longrightarrow$
$\mathrm{it\ exists}$ $n$ $\mathrm{and\ a\ faithful\ representation}$ $\rho : G\rightarrow T(n,k)$

What is known if "finite" is one of the conditions ? (i.e. for finite resoluble groups)

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1 Answer 1

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A finite group of upper triangular matrices has a nilpotent derived group, so a solvable group whose derived group is not nilpotent can't have a faithful representation by upper triangular matrices. One such group is ${\rm GL}(2,3)$ whose derived group is ${\rm SL}(2,3)$, which is certainly not nilpotent (actually $S_{4}$ also works and is a more or less equivalent (but simpler) example).

Later edit: When $k$ has finite characteristic $p >0$, we in fact need $[G,G]$ to be a $p$-group to have any chance of a faithful representation by upper triangular matrices. It is the case that when $G$ is finite and $[G,G]$ is a $p$-group, then $G$ does have a faithful representation by upper triangular matrices over some finite field $k$ of characteristic $p$. For $G/O_{p}(G)$ is then a finite Abelian $p^{\prime}$-group, so all absolutely irreducible representations of $G$ in characteristic $p$ are $1$-dimensional. It follows that there is a finite field $k$ of characteristic $p$ such that the regular representation of $G$ over $k$ (which is certainly faithful) is equivalent as a matrix representation to a representation of $G$ by upper triangular matrices over $k$.

Even later edit: I suppose I should remark for completeness that when $k$ has characteristic $0$, the only finite solvable groups which have a faithful representation by upper triangular matrices with entries in $k$ are Abelian. For if $G$ is such a group, then $[G,G]$ consists of unipotent matrices ( in the chosen faithful representation), and the identity matrix is the only unipotent matrix of finite order in ${\rm GL}(n,k)$ when $k$ has characteristic $0$.

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  • $\begingroup$ Ok, then not a general finite, thanks. How about $G$ s.t. $D(G)$ is nilpotent ? $\endgroup$ Jun 17, 2015 at 20:15
  • $\begingroup$ Is it clear in my text that I am asking for more than a non-go statement ? If somebody knows (pieces of) the state-of-the-art for $\mathrm{conditions(G,k)}$ when ``finite'' is one of the conditions, I would be grateful. $\endgroup$ Jun 17, 2015 at 20:34
  • $\begingroup$ I found it a bit unclear exactly what you were asking for. To be more precise than my answer, you certainly need the derived group $[G,G]$ to be a $p$-group when $k$ has characteristic $p >0.$ That may also be sufficient for $k$ large enough, I might expand my answer. $\endgroup$ Jun 17, 2015 at 20:41
  • $\begingroup$ The answer is now complete and satisfying. Thanks ! $\endgroup$ Jun 17, 2015 at 21:02
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    $\begingroup$ As a corollary a finite group has a triangular faithful representation over some finite product of fields (or equivalently, over some reduced commutative ring) iff its derived subgroup is nilpotent. $\endgroup$
    – YCor
    Jun 17, 2015 at 21:58

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