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I'm trying to go through Harvard's Abstract Algebra lectures on my own, and would like a little help with one of the homeworks. The problem asks:

Let $G$ be the group of invertible real upper $2 x 2$ matrices. Determine whether or not the following conditions describe normal subgroups H of G. If they do, use the First Isomorphism Theorem to identify the quotient group $G/H$. (a) $a_{11}$ = 1 (b) $a_{12}$ = 0 (c) $a_{11}$ = $a_{22}$ (d) $a_{11}$ = $a_{22}$ = 1

To do this, we'll need to go one by one and determine whether the subgroup described, H, is normal or not. If it's normal then it must be kernel of a surjective homomorphism. It can easily be shown that

$det: G \rightarrow R^{*}$

is a surjective homomorphism. Then, G/H must be isomorphic to $R^{*}$ by the First Isomorphism Theorem. So, we'll go one by one and see if they're normal.

NOTE: I have put *'s in places where computation would be too long simply to indicate the presence of some value determined through multiplying through.

(a) $H = \begin{bmatrix} 1 & b \\ 0 & d \end{bmatrix}$

$aha^{-1} = \begin{bmatrix} a & b\\ 0 & d \end{bmatrix} \begin{bmatrix} 1 & b'\\ 0 & d' \end{bmatrix} \begin{bmatrix} \frac{1}{a} & \frac{-b}{ad} \\ 0 & \frac{1}{d} \end{bmatrix}$ = $\begin{bmatrix} 1 & * \\ 0 & d' \end{bmatrix}$

This is a normal subgroup and therefore $G/H \simeq R^{*}$

(b) $H = \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix}$

$aha^{-1} = \begin{bmatrix} a & b\\ 0 & d \end{bmatrix} \begin{bmatrix} a' & 0\\ 0 & d' \end{bmatrix} \begin{bmatrix} \frac{1}{a} & \frac{-b}{ad} \\ 0 & \frac{1}{d} \end{bmatrix}$ = $\begin{bmatrix} a' & * \\ 0 & d' \end{bmatrix}$

This is NOT a normal subgroup.

(c) $H = \begin{bmatrix} a & b \\ 0 & a \end{bmatrix}$

$aha^{-1} = \begin{bmatrix} a & b\\ 0 & d \end{bmatrix} \begin{bmatrix} a' & b'\\ 0 & a' \end{bmatrix} \begin{bmatrix} \frac{1}{a} & \frac{-b}{ad} \\ 0 & \frac{1}{d} \end{bmatrix}$ = $\begin{bmatrix} a' & * \\ 0 & a' \end{bmatrix}$

This is a normal subgroup and therefore $G/H \simeq R^{*}$

(d) This is an instance of (c) and therefore it follows trivially that it is a normal subgroup.

Is this correct? Any and all help is greatly appreciated. Thanks.

  • 1
    why are you assuming that $a_{21}=0$? Is it given? – 2017-02-05
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    Forgot to include the words "invertible **real upper** 2x2 matrices" – 2017-02-05
  • 0
    $\begin{bmatrix}a&b\\0&d\end{bmatrix}^{-1}=\frac{1}{ad}\begin{bmatrix}d&-b\\0&a\end{bmatrix}$. The inverse of your matrices are incorrect. – 2017-02-05
  • 0
    @LivingRobot: Please edit your question with the correct version of the inverse as pointed out by Anuraf A. – 2017-02-06
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    Thanks for pointing out that stupid error. I've fixed it accordingly. – 2017-02-07
  • 0
    in the case (d) your conclusion for normality is correct but as subgroup of case (c) it has one dimension less so you should reconsider the quotient – 2017-02-07

0 Answers 0