If $G \subset GL_n(\mathbb R)$ is a group consisting of upper-triangular matrices under matrix multiplication and $H$ is subgroup of $G$, but with $1$s on diagonal, prove that $H$ is normal in $G$.

The definition $ghg^{-1} \in H$, for every $g$ and $h$ seems a bit tedious, especially because the matrices are $n \times n$ , for some natural number n.

I tried finding a homomorphism such that $H$ is its kernel, which would imply it's normal, but I failed to do so.

Any help?

  • $\begingroup$ When you multiply two (or three) upper triangular matrices together, what happens to the diagonal entries? $\endgroup$ – Max Nov 21 '17 at 21:14
  • $\begingroup$ @Max they're the product of respective diagonal entries? So using definition $ABA^{-1}$, I'd have to prove that $A^{-1}$ has $\frac{1}{a_i}$ on diagonal if $A$ has $a_i$ on diagonal... Doesn't seem obvious to me $\endgroup$ – windircurse Nov 21 '17 at 21:34
  • $\begingroup$ @windircurse See my edit. $\endgroup$ – Alex Clark Nov 21 '17 at 21:41
  • 1
    $\begingroup$ So the homomorphism you are looking for just maps the upper triangular matrix to the diagonal matrix having the same entries on the diagonal. $\endgroup$ – Derek Holt Nov 21 '17 at 22:40

Hint: You have that $H\subset G\subset U(n)$ where $U(n)$ is the group of $n\times n$ invertible matrices. $H\subset N(n)$, the group of unipotent $n\times n$ matrices.

Consider the commutator subgroup $[U(n),U(n)]$, what do you see?

(Expanded hint, $[G,G]\subset [U(n),U(n)]\implies ghg^{-1}\in [U(n),U(n)]h$ for $g,h\in G$)

In regard to your comment. Sometimes it is much easier to look only at an entry of the matrix.

You know that $A=(a_{ij})_{ij}$ and $B=(b_{ij})_{ij}$, and $AB=(\sum_{k=1}^n a_{ik}b_{kj})_{ij}$. If they are upper triangular, then $a_{ij}=0=b_{ij}$ for $i>j$, in which case $\sum_{k=1}^n a_{ik}b_{ki}$ only has nonvanishing terms where $i\leq k$ and $k\leq i$. You should be able to compute $ghg^{-1}$ and such now.


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