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The fixed-point set can be extremely wild. For example, every closed subset of $\mathbb R^n$ is the fixed point set of some smooth $\mathbb R$-action. To see this, let $S\subseteq\mathbb R^n$ be closed, and let $f\colon \mathbb R^n\to\mathbb R$ be a smooth nonnegative function whose zero set is exactly $S$. (Such a function always exists; see, for example, ...

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Let $x \in M$ be a fixed point of a $G$-action, $G$ acting properly on $M$ (meaning the map $G \times M \to M \times M$, $(g,m) \mapsto (m,gm)$, is proper). We'll use the following tubular neighborhood theorem: If $M$ is a smooth proper $G$-manifold without boundary, and $N$ is a closed $G$-invariant submanifold, then there is an open neighborhood $U ... 2 We study the action of$H$on$X = G/P$. Since$(|X|,p) = 1$, it follows that the number of classes in some orbit$\mathcal{O}$in$X$under this$H$-action is relatively prime to$p$. If$H_x \subseteq H$denotes the stabiliser of any element$x \in \mathcal{O}$, then, check that$H_x$has the desired form, so that$H_x$is a$p$-group and$([H: H_x], p) = ...

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Just knowing about the action in a neighborhood of one point of the manifold tells you nothing about whether the action is transitive. For example, consider the standard action of $\operatorname{GL}(n,\mathbb R)$ on $\mathbb R^n$. This action is not transitive, because it fixes the origin. On the other hand, we can restrict the action to $\mathbb ... 1 You have the action of$S_n$on itself by conjugation. The equivalence classes, or orbits, under an action by conjugation are called conjugacy classes. What you need to show is that two elements in$S_n$are conjugate if and only if they share the same cycle type. You can find a proof of that on page 126 of Dummit and Foote under Proposition 11, for one. ... 1 This is not a definite answer, but it's too long for a comment. Let$S$be the set of roots of$f$in$M$. Then$G$acts transitively on$S$, as is well-known. You may consider the$H$-orbits of$S$; call them$S_i$. Then you can form the polynomial$f_i = \prod_{\lambda\in S_i} (X-\lambda)\in L[X]$. It's easy to see that$f=\prod f_i$is a decomposition ... 1 The basic problem is that the adjoint maps need not be injective. That is, the extra group relations may collapse all our part of your group into something much smaller than the monoid. The right adjoint takes monoid elements to units. (This behavior is analogous to constructing the quotient of a ring with one of its maximal ideals.) For the ... 1 Hint: show that the subset of permutations whose signature is 1 and the subset whose signature is$-1$are preserved by the action. 1 O(n) acts on n-dimensional space by rotations. Rotations preserve length. Now remark that given two vectors of the same length they can be rotated onto one another. This Is easy to see by taking the plane through the two vectors. So the problem is now to see that two vectors in the plane can be rotated onto another and this is trivial, just rotate the plane ... 1 Hint If one writes$M$in terms of its column vectors, i.e., as $$\newcommand{\mbx}{{\mathbf x}} M = \pmatrix{\mbx_1 & \cdots & \mbx_n},$$ and substitutes this expression in the defining equation$M^T M = I$of$O(n)$, we see that$M$is orthogonal iff$(\mbx_1, \ldots, \mbx_n)$is an orthonormal basis of$\Bbb R^n\$, which is exactly the output of ...

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