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Let $GL_n(k)$ be the $n$ by $n$ general linear group over $k$, $B_n(k)$ be the subgroup of $GL_n(k)$ consisting of all upper triangular matrices, and $U_n(k)$ be the subgroup of $B_n(k)$ whose diagonal elements are all $1$.

To show $B_n(k)$ is solvable, I'm proving it now by following steps:

  1. $U_n(k)$ is a subgroup of $B_n(k)$. (done)
  2. $U_n(k)$ is normal in $B_n(k)$. (done)
  3. $U_n(k)$ is solvable. (question)
  4. $B_n(k) / U_n(k)$ is also solvable. (not yet)
  5. $B_n(k)$ is solvable. (by the below thm)

I'll use a theorem to verify $B_n(k)$ is solvable.

$G$ is solvable if and only if $H$ and $G/H$ are solvable for some normal subgroup $H$ of $G$.

So, I have to prove both step 3 and step 4. But I have no idea about them. How to prove them? Since my knowledge is not enough, I don't want to show them using Lie theory.

Thanks in advance.

Shaun
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JeongHobin
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2 Answers2

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You can prove $U_n$ solvable by induction on $n$. Consider $H_n$, the group of matrices of the form $$ \begin{pmatrix} 1 & & & a_1 \\ & \ddots & & \vdots \\ & & 1 & a_{n-1} \\ & & & 1 \end{pmatrix} $$ It is then easy to see that $H_n \triangleleft U_n$, that $H_n$ is abelian (in fact isomorphic to $k^{n-1}$), and that $U_n/H_n\cong U_{n-1}$.

hmakholm left over Monica
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  • Could you explain the reason/motivation behind such a choice for Hn? TIA. – Arkya Nov 16 '17 at 13:34
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    @ArkyaChatterjee: My reason for trying that is that the matrices of that shape constitute the difference between $U_n$ and those upper diagonal matrices whose last column is trivial (which are naturally isomorphic to $U_{n-1}$). From there it is just a matter of "... and it turned out to work when I thought it through". – hmakholm left over Monica Nov 16 '17 at 13:59
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If $A$ is an $n\times n$ matrix, I call the entries $a_{i,i+\ell}$ of $A$ the $\ell$-th diagonal of $A$. In particular, the $0$-th diagonal of $A$ is its actual diagonal.

You can prove that if $A,B \in U_n(k)$ and $C = [A,B]$ then the first diagonal of $C$ is $0$. More in general, if the first $\ell$ diagonals (except the $0$-th) of $A$ are $0$ and the first $m$ diagonals (except the $0$-th) of $B$ are $0$, then the first $\ell+m+1$ diagonals of $[A,B]$ are $0$.

In order to prove the statement about the quotient, you can observe that $B_n/U_n$ is indeed isomorphic to the group of diagonal matrices.

Shaun
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fulges
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  • I cannot understand the generalization but I prove that if A,B are in U_n, then the first diagonal entry of AB is a sum of the first diagonal entries of A and B. So, the commutator of A and B has the first diagonal entry 0. Inductively, we'll get n-th derived subgroup of U_n is trivial because the (n-1)-th entry of the commutator of two elements of (n-1)-th derived subgroup is 0! Thus, we have the solvability of U_n, right? – JeongHobin Apr 22 '16 at 01:49
  • I am not sure to understand. But if $A,B$ have the diagonals from $1$ to $\ell$ that are $0$, then the entries on the $\ell+1$-th diagonal of $AB$ are sums of the ones of $A$ and the ones of $B$. If this is what you mean, I agree. In particular the $k$-th derived subgroup of $U_n(k)$ is trivial (but indeed already the $k'$-th one is trivial, with $k' \approx \log(k)$). – fulges Apr 22 '16 at 04:28
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    Hi, do you have a source for this? – Georgi Kocharyan Aug 23 '21 at 09:11