Let $H$ be the subgroup of $GL(n,\mathbb{R})$ such that all the elements of $H$ are upper triangular matrices

Question

Let $H$ be the subgroup of $GL(n,\mathbb{R})$ such that all the elements of $H$ are upper triangular matrices with all the diagonal entries equal to $1$. Is $H$ a normal subgroup of $GL(n,\mathbb{R})$?

I tried by using definition of normal subgroup but the calculation is some cubersome.

Answer 1

$H$ is not normal in $\text{GL}(n, \Bbb R)$. Recall that $H<G$ is normal iff $x\,H\,x^{-1}=H$ for every $x\in G$, that is $xyx^{-1}\in H$ for every $y\in H$.

Let $\begin{pmatrix} 1&1\\ 0&1\end{pmatrix}\in H$. Pick a matrix $A=\begin{pmatrix} a&b\\ c&d\end{pmatrix}\in \text{GL}(n, \Bbb R)$. Assume $c\neq 0$, otherwise $A\in H$. We may further assume that $\det(A)=1$ for making the computation easier. Then

$$\begin{pmatrix} a&b\\ c&d\end{pmatrix}\begin{pmatrix} 1&1\\ 0&1\end{pmatrix}\begin{pmatrix} d&-b\\ -c&a\end{pmatrix}=\begin{pmatrix} 1-ac&a^2\\ -c^2&1+ac\end{pmatrix}\notin H$$

This provides you a counterexample in dimension $2$. It is easy to extend this counterexample to a generic $n$.

Answer 2

Hint:

You are probably aware that we can "triangularize" matrices via gaussian elimination. That is, we can write a non upper triangular matrix $A$ as

$$ A = C^{-1} U C$

where $U$ is upper triangular.

Can you use this fact to determine if $H$ is or isn't normal?

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I hope this helps ^_^