Let $(T,*)$ be the subgroup of $\text{GL}_2(\mathbb{R})$ consisting of nonsingular upper-triangular $2\times 2$ matrices with entries in $\mathbb{R}$; that is the matrices of the form $$ \begin{pmatrix} a & b\\ 0 & c \end{pmatrix} $$ with $a,b,c$ elements of $\mathbb{R}$ and $ac$ not equal to $0$. Let $U$ consist of matrices of the form \begin{pmatrix} 1 &x\\ 0 &1 \end{pmatrix} with $x \in \mathbb{R}$. Show that $U$ is a normal subgroup of $T$.

To show that $U$ is a normal subgroup of $T$, I multiplied the matrices, $UTU^{-1}$ and showed that $UTU^{-1}$ is a subset of $T$. I believe my calculations are correct, but from my calculations it seems that $UTU^{-1}$ is not a subset of $T$.

I know there are other ways to show that $U$ is a normal subgroup of $T$. Any suggestions and corrections would be much appreciated!

  • $\begingroup$ (a b 0 c) and (1 x 0 1) are supposed to be 2x2 upper triangular matrices, but I am not sure how to write them in the correct format here. I apologize! $\endgroup$ – Monica Oct 17 '16 at 4:24
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    $\begingroup$ can you define a homomorphism with $U$ as the kernel? $\endgroup$ – AnotherPerson Oct 17 '16 at 5:28
  • $\begingroup$ The title mentions a normal subgroup of GL2, but the question is about a normal subgroup of T... You should fix at least one of the two. $\endgroup$ – Mariano Suárez-Álvarez Oct 17 '16 at 5:35
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    $\begingroup$ You should check in your notes or textbook what the definition of a normal subgroup is, because you seem to not have gotten that right... $\endgroup$ – Mariano Suárez-Álvarez Oct 17 '16 at 5:37
  • $\begingroup$ @Monica I edited your post to add LaTeX formatting. You can learn more about typesetting math here. $\endgroup$ – Viktor Vaughn Oct 17 '16 at 6:19

A subgroup $H < G$ is normal if it's invariant by conjugation, so if for any $g \in G$ , $gHg^{-1} = H$.

Checking this is pretty easy:

Pick an arbitrary matrix in $T$, $\begin{pmatrix} a & b\\ 0 & c \end{pmatrix}$, compute its inverse $\begin{pmatrix} 1/a & -b/(ac)\\ 0 & 1/c \end{pmatrix}$ (note that, since the matrix is an element of the general linear group, it is invertible, so neither $a$ nor $c$ can be zero ) and verify that $$\begin{pmatrix} a & b\\ 0 & c \end{pmatrix} \begin{pmatrix} 1 & x\\ 0 & 1\end{pmatrix}\begin{pmatrix} 1/a & -b/(ac)\\ 0 & 1/c \end{pmatrix} = \begin{pmatrix} 1 &*\\ 0 & 1 \end{pmatrix}$$ (I'll leave the actual multiplication to you).


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