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?
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.
