# Thread: prove subgroup of GL2(R)

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

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

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

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

Thanks

2. ## Re: prove subgroup of GL2(R)

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

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

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

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

P.S. If you use the characterization theorem of subgroups you only need to prove $G\neq \emptyset$ and $xy^{-1}$ for all $x,y\in G$ .

3. ## Re: prove subgroup of GL2(R)

Thanks FernandoRevilla, I understand it now.