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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?

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$b,c$ are correct –  Bunuelian Trick Jan 19 '13 at 16:35
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$$\left( \begin{array}{cc} b_{11} &b_{12}\\ 0 & b_{22} \end{array} \right)^{-1}\left( \begin{array}{cc} a_{11} &a_{12}\\ 0 & a_{22} \end{array} \right)\left( \begin{array}{cc} b_{11} &b_{12}\\ 0 & b_{22} \end{array} \right)=\left( \begin{array}{cc} a_{11} &\frac{b_{22} a_{12}+a_{11} b_{12}-a_{22} b_{12}}{b_{11}} \\ 0 & a_{22} \end{array} \right)$$

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.

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