$B_n$ is the subgroup of upper triangular matrices in $GL_n(\mathbb{R})$. $T_n$ is the group of diagonal matrices in $GL_n(\mathbb{R})$, $U_n \subset B_n$ is the subgroup of matrices whose diagonal entries are 1.

I was trying to prove that $B_n$ is a semidirect product of $U_n$ and $T_n$. There are similar questions on the site, but none of them address my question.

My approach is to prove:

  1. $U_n \cap T_n = \{e\}$ which is easy to see
  2. $U_n \triangleleft B_n$
  3. $B_n = U_nT_n$

I have problems with 2 and 3. For 2, I used the definition of the normal subgroup (conjugation) and tested the $2 \times 2$ case (Normal Subgroup of T (upper triangular matrices under multiplication)) but I was unable to prove it in general, and I don't know how to do 3.

Any help will be appreciated.


1 Answer 1


Hints: 2. Observe that the diagonal elements of the product of two upper triangular matrices are simply the products of the diagonal elements.
3. For a given upper triangular matrix with diagonal entries $d_i$, multiply it by $\mathrm{diag}(1/d_1,\dots, 1/d_n)$ to obtain a matrix in $U_n$.

  • $\begingroup$ that helps thank you! $\endgroup$ Commented Sep 14, 2019 at 21:40

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