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If A is a matrix in SL(2,Z) is it directly obvious the stabilizer of A , G_A = {B in SL(2,Z) | A.B = A}, is the set containig only the Identity? or is this not true?

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The elements of SL(2,Z) are invertible, so you can multiply A.B = A on the left by A^-1, yielding B = I (and a resounding 'yes' to your first question).

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    Thanks, I did just that a $f$ew seconds ago and was about to remove the question. sorry $f$or asking the obvious, but thanks anywho.2011-04-03
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At the risk of further damage to this particular moribund equine:

The question has nothing to do with $\operatorname{SL}_2(\mathbb{Z})$. If $G$ is any group, then it has a natural left action on its itself: $g.x := gx$.

Consider now a general group action: a group $G$ acts on the left on a nonempty set $X$. The action is said to be transitive if for all $x,y \in X$ there exists $g \in G$ such that $g.x = y$. The action is said to be simply transitive if for all $x,y \in X$ there exists a unique $g \in G$ such that $g.x = y$. Equivalently, a simply transitive action is a transitive action in which for every $x \in X$, the point stabilizer

$G_x = \{g \in G \ | \ g.x = x\}$

is the trivial group $\{e\}$.

The left (or right!) action of any group $G$ on itself is simply transitive: this is an immediate calculation done in the same way as in yatima2975's answer. But perhaps more importantly, there is a converse: if $G$ acts simply transitively on a set $X$, then $X$ is isomorphic as a $G$-set to the left action of $G$ on itself. To get an isomorphism, fix any element $x_0 \in X$, and define a map $X \rightarrow G$ by sending $x$ to the unique element $g \in G$ such that $x = g.x_0$.