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Let $H=\left\{ z\in\mathbb{C}\mid\Im\left(z\right)>0\right\}$ be the upper-half Poincare plane. Let $GL\left(2,\mathbb{R}\right)$ be the general linear group, $Z\left(GL\left(2,\mathbb{R}\right)\right)$ be the center of the general linear group and $O\left(2,\mathbb{R}\right)$ be the orthogonal subgroup of $GL\left(2,\mathbb{R}\right)$.

What does it mean to say $H=GL\left(2,\mathbb{R}\right)/\left(Z\left(GL\left(2,\mathbb{R}\right)\right)\cdot O\left(2,\mathbb{R}\right)\right)$? The left-hand side is a metric space and the right hand side is a set of cosets of $GL\left(2,\mathbb{R}\right)$. So I'm confused about what it means to write that they are equal or to say "the upper half plane is..." It seems like this would be the group of orientation preserving isometries of H, but I still find the terminology confusing.

I've been trying to figure out what this could possibly mean, but my searches on the internet have not been fruitful. I've also looked at 2 sources on standard modular groups but they make no mention of this fact. An explanation or reference would be greatly appreciated.

Motivation: I am reading a paper titled "On Modular Functions in characteristic p" by Wen-Ch'ing Winnie Li which can be found at http://www.jstor.org/stable/1997973. The claim appears on page 3 of the pdf (page 232 of the journal). It is also stated on the wikipedia page: http://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model

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My comments were getting too long, so I will post this as an answer: the isomorphism referred to is an isomorphism of $G$-sets. This works in great generality: let $H$ be a set, let $G$ act on $H$ transitively. Pick an arbitrary point $z\in H$ and let $K=\text{Stab}_G(z)$ be the point stabiliser in $G$. Note that $K$ is not normal in general, since the group $gKg^{-1}$ stabilises the point $g(z)$ (my action is on the left). So, $K$ is normal if and only if it acts trivially on $H$.

Nevertheless, the set of cosets $G/K$ is always a $G$-set, i.e. a set with an action of $G$: $g: hK\mapsto (gh)K$ for all $g\in G$ and $hK\in G/K$. This is the usual coset action. Now, check that the map $\phi:G/K\rightarrow H,\;gK\mapsto g(z)$ is a bijection of $G$-sets, i.e. a bijection of sets that respects the $G$-action.

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    Yes, although I am not a big fan of this notation. There are various different notational conventions to distinguish between isomorphisms and canonical isomorphisms.2010-11-30
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H is not a group, but rather a space with a group acting on it.

If K ≤ G is a subgroup, then G/K is not usually a group, but rather a set of cosets on which G acts. G/K is only a group if K is normal.

In your situation, the subgroup is not normal, so you only get a group action.

One keyword form the wikipedia article is "isotropy subgroup". If G acts transitively on a space H, then the set K of elements of G that fix some particular point of H forms a subgroup (the "isotropy subgroup"), and the action of G on H is isomorphic to the action of G on the cosets in G/K.

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    See my answer. It's more natural to define the map the other way round.2010-11-30