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Just a quick question. I'm trying to understand the answer to one of my previous questions. The precise problem I want to show is as follows.

Let $G$ be a group acting faithfully on a manifold $X$. If $G$ is such that $\dim G$ makes sense (for example, $G$ is also a manifold), then $\dim G \leq \dim X$.

Suppose not, i.e., $\dim G > \dim X$. Then we wish to show that the kernel of the homomorphism $G \to \operatorname{Aut}(X)$ is nontrivial. This should follow if $\dim \operatorname{Aut}(X) = \dim X$, but in general $\operatorname{Aut}(X)$ can be much larger than $X$. (Also, does $\dim \operatorname{Aut}(X)$ make sense?)

Any suggestions?

EDIT: It turns out that this proposition is false as the examples below show, and the answer I linked to has been retracted (and is being rewritten?). Thanks to everyone for the help.

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This is not true. For example, $SO(3)$ acts faithfully on the $2$-sphere $S^2$, but $SO(3)$ has dimension $3$, and $S^2$ only has dimension $2$.

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This is not true. $G=GL_2(\mathbb{R})$ acts faithfully on $\mathbb{R}^2$, no?

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