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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 step3 and step4. But I have no idea about them. How to prove? Since my knowledge is not enough, I don't wanna show using Lie theory. Thanks in advance.

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    $\begingroup$ It is not hard to see that $U_n$ is solvable! $\endgroup$ – Mhenni Benghorbal Apr 22 '16 at 1:41
  • $\begingroup$ Its commutator subgroup is a $p$-group - upper triangular matrices with diagonal entries equal to 1. $\endgroup$ – Chris Godsil Apr 22 '16 at 1:52
  • $\begingroup$ @Chris: do you have in mind the case that $k$ is a finite field here? I think this result is just true for all $k$. $\endgroup$ – Qiaochu Yuan Apr 22 '16 at 2:43
  • $\begingroup$ @Qiachu Yuan: you are right twice, I was thinking $k$ was finite, and the result is true in general. In the latter case the commutator is nilpotent, which might help. $\endgroup$ – Chris Godsil Apr 22 '16 at 12:05
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If $A$ is a $k\times k$ 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.

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  • $\begingroup$ 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? $\endgroup$ – JeongHobin Apr 22 '16 at 1:49
  • $\begingroup$ 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 $k' \approx \log(k)$). $\endgroup$ – fulges Apr 22 '16 at 4:28
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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 $U_n/H_n\cong U_{n-1}$.

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  • $\begingroup$ Could you explain the reason/motivation behind such a choice for Hn? TIA. $\endgroup$ – Arkya Chatterjee Nov 16 '17 at 13:34
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    $\begingroup$ @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". $\endgroup$ – hmakholm left over Monica Nov 16 '17 at 13:59

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