Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be the group of all invertible $2×2$ upper triangular matrices (under matrix multiplication). Pick out the normal subgroups of $G$ from the following:
(a) $H=\{A\in G:a_{12}=0\}$;
(b) $H=\{A\in G:a_{11}=1\}$;
(c) $H=\{A\in G:a_{11}=a_{22}\}$.

After calculation I get only (b) is right. Am I correct?

share|cite|improve this question
$b,c$ are correct – Un Chien Andalou Jan 19 '13 at 16:35

$$\left( \begin{array}{cc} b_{11} &b_{12}\\ 0 & b_{22} \end{array} \right)^{-1}\left( \begin{array}{cc} a_{11} &a_{12}\\ 0 & a_{22} \end{array} \right)\left( \begin{array}{cc} b_{11} &b_{12}\\ 0 & b_{22} \end{array} \right)=\left( \begin{array}{cc} a_{11} &\frac{b_{22} a_{12}+a_{11} b_{12}-a_{22} b_{12}}{b_{11}} \\ 0 & a_{22} \end{array} \right)$$

You can see that $a_{11}=a_{22}$ works, as well as $a_{11}=1$. However you can still have $a_{12}=0$ and a nonzero $_{12}$ entry on the RHS.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.