Given the subgroup $H\leq GL_n(\mathbb{R})$ the set of upper triangular matrices, I want to show that $H$ is not nilpotent.

The first part of the problem is showing $H$ is solvable. Define $G_1 = H$, $G_2=[G_1, G_1]$, and $G_{i+1} = [G_i, G_i]$, $H$ is solvable if at some point we have $G_N = \{id\}$. We see that $G_2 = SL_n(\mathbb{R})$, in $G_3$ the matrices have $1$'s on the diagonal, for each step after, one further upper diagonal will become zero and we have $G_{n+2} = \{id\}$.

Using the similar definition for nilpotent, define $G_1 = H$, $G_2=[G_1, G_1]$, and $G_{i+1} = [G_1, G_i]$ we need $G_N = \{id\}$. But in general, I don't quite see what $[G_1, G_i]$ looks like except $i=1$.

I also have another definition for nilpotent, that is, going from below from $G_1 = {id}, G_2 = Z(H)$, and $G_{i+1}$ is chosen such that $G_{i+1}/G_i = Z(H/G_i)$, $H$ is nilpotent if at some point, we have $G_N = H$.

Note: $[X,Y]$ is the commutator group generated by $xyx^{-1}y^{-1}$ for all $x,\in X, y\in Y$. $Z(G)$ is the center of a group $G$.

I posted my answer below, could someone give me some feedback, I am not sure if it is correct.

  • $\begingroup$ Why the first part is to show $\;H\;$ is solvable? What do we care about that? We indeed have that any nilpotent group is solvable, so from showing the latter the former doesn't follow... unless there's some key idea used in showing solvability that can help us. $\endgroup$ – DonAntonio Oct 12 '16 at 20:23
  • $\begingroup$ @DonAntonio showing it is solvable was first part of the problem, I thought it might give me some insight about how to argue for the second part. $\endgroup$ – Xiao Oct 12 '16 at 20:26
  • $\begingroup$ @Xi Ok, I see. You may be right $\endgroup$ – DonAntonio Oct 12 '16 at 20:30

I think I got it.

First Note that $Z(H)$ are only the scaler matrices.

We use the second definition for nilpotent, given $G_1 = \{id\}, G_2=Z(H)$ we show that $G_3 = Z(H)$ then by induction, all $G_n = Z(H)$ so the sequence $G_n$ will never reach $H$.

From $G_2/Z(H) = Z(H/Z(H))$, suppose given $bZ(H)\in G_2/Z(H)$, then we have $$bZ(H)hZ(H) =hZ(H)bZ(H) \quad \text{ for all $h \in H$} $$ the above implies $$bhZ(H) = hbZ(H) \quad \text{ for all $h \in H$} $$ then $b$ has to be a scaler matrix.


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