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Thread: prove subgroup of GL2(R)

  1. #1
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    prove subgroup of GL2(R)

    Hi, I think I have this correct but I need someone to confirm my working.

    Let G be the set of matrices of the form:
    [a b]
    [0 1]
    for $\displaystyle a,b \in \mathbb{R}$
    Prove that G is a subgroup of GL2(R).
    1. if $\displaystyle x,y \in G$, then $\displaystyle xy \in G$
    Let x=[a b : 0 1] and y =[c d : 0 1]
    Then
    xy=[ac ad+b : 0 1]
    $\displaystyle \implies xy \in G$

    2. If a=1, b=0, then I = [1 0 : 0 1]
    $\displaystyle \implies I \in G$

    3. if $\displaystyle x \in G$, then $\displaystyle x^{-1} \in G$.
    x=[a b : 0 1], then x^{-1} = [1 -b : 0 a]
    thus, $\displaystyle x^{-1} \in G$.

    Thus, G is a subgroup of GL2(R).

    Thanks
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: prove subgroup of GL2(R)

    Quote Originally Posted by shelford View Post
    Let G be the set of matrices of the form:
    [a b]
    [0 1]
    for $\displaystyle a,b \in \mathbb{R}$
    Prove that G is a subgroup of GL2(R).
    It should be ... with $\displaystyle a\neq 0$

    1. if $\displaystyle x,y \in G$, then $\displaystyle xy \in G$
    Let x=[a b : 0 1] and y =[c d : 0 1]
    Then
    xy=[ac ad+b : 0 1]
    $\displaystyle \implies xy \in G$
    ... because $\displaystyle ac\neq 0$

    2. If a=1, b=0, then I = [1 0 : 0 1]
    $\displaystyle \implies I \in G$
    ... i.e. the identity element of $\displaystyle \textrm{GL}_2(\mathbb{R})$ belongs to $\displaystyle G$ .

    3. if $\displaystyle x \in G$, then $\displaystyle x^{-1} \in G$.
    x=[a b : 0 1], then x^{-1} = [1 -b : 0 a]
    thus, $\displaystyle x^{-1} \in G$.
    $\displaystyle x^{-1}=\dfrac{1}{a}\begin{bmatrix}1&-b\\0&a \end{bmatrix}$ ... i.e. $\displaystyle x^{-1}\in G$

    P.S. If you use the characterization theorem of subgroups you only need to prove $\displaystyle G\neq \emptyset$ and $\displaystyle xy^{-1}$ for all $\displaystyle x,y\in G$ .
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  3. #3
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    Re: prove subgroup of GL2(R)

    Thanks FernandoRevilla, I understand it now.
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