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6

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 ...


3

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.


3

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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