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Recall that $GL_2(\mathbb{R})$ is the group of all invertible 2x2 matrices with real entries. Let:

$G = (\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \in GL_2(\mathbb{R}) : ac \neq 0$)

and

H = ($\begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix}: x\in\mathbb{R}$)

H is a subgroup of G.

Show that every left coset of H in G is equal to the right coset of H in G.

First, I thought of assuming that G is abelian. But clearly that failed for me because G isn't abelian. So the other thing I thought of trying was to show $ghg^{-1}\in H$ where $g\in G$ and $h\in H$. Now with this idea, I'm stuck in showing that $ghg^{-1}\in H$. Can anyone help me out?

Johnny
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    Can you not calculate $g^{-1}hg$, when $g$ is the typical matrix in $G$, and $h$ the typical matrix in $H$? – Gerry Myerson Oct 13 '13 at 06:18
  • I thought of that. But how can I really show that $ghg^{-1}$ is in H? It's true that $g^{-1}$ is in G since G is a subgroup. But show do I know that $g^{-1}$ is in H? :P – Johnny Oct 13 '13 at 06:22
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    Let me ask again: can you not calculate $g^{-1}hg$, where $g$ is the typical matrix in G, and $h$ is the typical matrix in H? You are told what the matrices in $G$ look like, and you are told what the matrices in $H$ look like; do the calculation! – Gerry Myerson Oct 13 '13 at 06:26
  • I read it wrong the first time woops, that makes sense now. Thank you! – Johnny Oct 13 '13 at 06:31

3 Answers3

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Evaluate the multiplication $\displaystyle\begin{pmatrix}a & b \\ 0 & c\end{pmatrix}\begin{pmatrix} 1 & x \\ 0 & 1\end{pmatrix}$ explicitly. What happens as $x$ varies over $\Bbb R$?

And then what if the matrices are reversed - what do the products look like then as $x$ varies?

anon
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If $g = \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \in G$, then $$ g^{-1} = \begin{pmatrix} 1/a & -b/ca \\ 0 & 1/c \end{pmatrix} $$ so if $h = \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} \in H$, then $$ g^{-1}hg = \begin{pmatrix} 1 & -cx/a \\ 0 & 1 \end{pmatrix} \in H $$ Thus, if $g\in G, h \in H$, there is an $h' \in H$ such that $$ hg = gh' $$ Thus, $$ Hg \subset gH $$ and the other containment holds similarly, so $gH = Hg$.

Prahlad Vaidyanathan
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The map $\varphi\colon G\to\mathbb{R}^*\times\mathbb{R}^*$ (where $\mathbb{R}^*$ denotes the multiplicative group of nonzero real numbers) defined by $$ \varphi\colon\begin{pmatrix}a & b \\ 0 & c\end{pmatrix} \mapsto (a,c) $$ is a homomorphism because $$ \begin{pmatrix}a_1 & b_1 \\ 0 & c_1\end{pmatrix} \begin{pmatrix}a_2 & b_2 \\ 0 & c_2\end{pmatrix} = \begin{pmatrix}a_1a_2 & a_1b_2+b_1c_2 \\ 0 & c_1c_2\end{pmatrix} $$ Since $H=\ker\varphi$, $H$ is normal in $G$.

egreg
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