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i think it is unlikely that this is true. Take a dumbbell type shape with rotational symmetry. Make the bar bit pinch in. The smallest circle is a totally geodesic submanifold and the invariant under rotations. Now deform the rest of the barbell whilst leaving a small neighbourhood of the smallest circle unchanged. If you make it lumpy enough there won't be ...

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This simply cannot be true. A generic Riemannian manifold does not have any isometries, but still the image of any geodesic is a totally geodesic submanifold.

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There are counting-proof methods for this, but here's the group action proof using Burnside's Lemma: Suppose we have a $4\times4$ checkerboard with 16 squares. Let $S$ be the set of all colorings with 8 black and 8 white colors. How many different colorings are there? It is obvious that there are $\binom{16}{8}=12870$ different ways to color the board. ...

3

This argument is due to Golomb. Let $p$ be a prime, and consider the set of all sequences of length $p$ with elements taken from $\{1, 2, \dots, n\}$. We can view $Z_p$ as acting on these sequences by cyclic rotation (e.g. rotating $12523$ by $2$ yields $23125$). This action partitions the $n^p$ sequences of length $p$ into orbits, including $n$ orbits ...

3

It's the action $Sym(S) \times S \to S$ given by $(f,x) \mapsto f(x)$. Every action $G \times S \to S$ defines a homomorphism $G \to Sym(S)$ and vice-versa. The natural action defined above corresponds the identity homomorphism. This makes it "natural". An "unnatural" group action $Sym(S) \times S \to S$ could for instance correspond to an nontrivial ...

2

For comparison's sake, we can try to build an "unnatural" group action. Given a group $G$ and a set $S$, we know that the identity $e \in G$ must act as the identity map and that the action of two elements is the same as the action of their composition. I propose as our "unnatural action" we take $S_n$ as our group and $S$ to be a set of $n$ labeled ...

2

By orbit-stabilizer, the dimension of the orbit should be the dimension of the group minus the dimension of the stabilizer, so we should compute the dimension of the stabilizer. Suppose $$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & ... 2 Take any group H acting faithfully on a set X and any noninjective group homomorphism G\to H. Then G acts on X as well, but not faithfully. This may sound contrived, but actually any non-faithful action is of this kind (we can simply let H be G modulo the kernel of the action). 2 Theorem. Considering {\rm GL}(V)'s induced action on {\Bbb P}(V), for any A\in{\rm GL}(V) we have$${\Bbb P}(V)^A=\bigsqcup_{\lambda\ne0}\Bbb P(V_\lambda). \tag{$\circ$}$$Here V is a finite-dimensional complex vector space and V_\lambda is the \lambda-eigenspace for A (the space of all eigenvectors of A with eigenvalue \lambda). Note ... 2 First, let facet be a lists of lists representing facets: gap> facets := [[1, 3, 5, 7], [2, 4, 6, 8], [1, 2, 5, 6], > [3, 4, 7, 8], [1, 2, 3, 4], [5, 6, 7, 8]]; [ [ 1, 3, 5, 7 ], [ 2, 4, 6, 8 ], [ 1, 2, 5, 6 ], [ 3, 4, 7, 8 ], [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ] ] Now we will form a direct product of as many copies of the group of order 2 as the ... 2 A group action on a finite set of k elements is nothing but a homomorphism from the group to S_{k}. Now, since all the 7 sylow subgroups are conjugate, then G acts by conjugation on them. This gives rise to a homomorphism from G to S_{8}. Let K be the kernel, then by the first isomorphism theorem [G:K] divides 8!. Since 49 does not divide 8!, then 49 ... 2 The scope of potential answers to this question is truly vast. I suggest this MSE Meta Link as an entry point for additional exploration. By way of enrichment I would like to present another proof that choosing q objects from m distinct objects where the order matters is$$\frac{m!}{(m-q)!}$$using the Polya Enumeration Theorem (PET). ... 2 You should start off by answering question #1 yourself; this is a great collection of questions, and you're definitely capable of it. Every group G acts on itself by conjugation; x \mapsto x^g = g^{-1}xg\  (prove this). Under this action, you should figure out what the more common names for "orbit" and "stabilizer" of a group element are. This will ... 2 When a group acts on a graph (in the usual sense), there is more structure than just a permutation of the vertices. Specifically, it not only maps vertices to vertices, but preserves the property that if there is an edge (a,b) in the graph, then there will be an edge (g(a), g(b)) as well. That is, it permutes both the vertices and edges in a compatible ... 1 The proof is based on a basic fact about finite groups. Theorem Let H \subseteq G be a subgroup of index n. Then G/core_G(H) is isomorphic to a subgroup of S_n. Proof See I.M. Isaacs, Finite Group Theory, Theorem 1.1. Note, core_G(H):=\bigcap_{g \in G}H^g, which is a normal subgroup contained in H. Now let us have a look at the question. Let P ... 1 H acts on X by conjugation. This follows from hxh^{-1} \in X: it's easy to check that this implies h^ix^jh^{-i} \in X for all i, j, i.e. h'x'h'^{-1} \in X for all h' \in H, x' \in X. Checking that this action satisfies the other definitions of an axiom is also easy to check (it's the same as checking that any conjugation action is a valid ... 1 \newcommand{\Span}{\langle #1 \rangle}You've done an excellent job. Perhaps it's worth noting that since y = (3 5) (2 6) inverts by conjugation x = (123456) (that is, y^{-1} x y = x^{-1}), the group has order 12, and its elements can be written uniquely in the form x^{j} y^{i}, for 0 \le i < 2 and 0 \le j < 6. (This is used in (sub) ... 1 unless I am overlooking something, I think you could use projection in place of saturation showing the answer is no. Let G be the real line and X be the plane, with the action defined as (g,(x,y))\to(x,g+ y). Take any A\subset X that is Borel, but whose projection to the x-axis is not Borel (it would only be analytic). The projection of A to ... 1 A group G acting on a set X means each element of G leads to a permutation of the set. The effect of a permutation corresponding to g\in G on an element x\in X is written as g.x or simply as gx. We can compose permutations and also apply group law on two elements of G. The permutations associated to these elements have to be such that an ... 1 Note that another way to understand group actions is by "currying" the action into a function G\to(S\to S); in this language a group action is just a homomorphism from G to the permutation group on S. Thus in many ways the theory of group actions does not add anything more than you will get from understanding permutation groups and homomorphisms. ... 1 The group SL_2(\mathbb{R}) has a natural but nonfaithful action on the projective space \mathbb{R}P^1. Letting M \in SL_2(\mathbb{R}) and letting \ell_{\vec v} \in \mathbb{R}P^1 be the line through the origin with a direction vector \vec v, the action is given by$$M \cdot \ell_{\vec v} = \ell_{M \vec v}  The kernel of this action is all ...

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