# Tag Info

6

The category of left $G$-sets is equivalent to the category of left $H$-sets if and only if $G$ is isomorphic to $H$. Indeed, consider $G$ as a left $G$-set. It has the property that $\mathrm{Hom}_G (G, -) : G \textbf{-Set} \to \mathbf{Set}$ preserves all colimits. Moreover, up to isomorphism, $G$ is the unique such left $G$-set, i.e. if $X$ is a left ...

5

Suppose the normal subgroup $H$ fixes $x$, let $y$ be in the set and let $g$ be such that $y=gx$. Then $gHg^{-1}=H$ fixes $y$, so $H$ fixes every point.

4

Observe that: $F(S^m,2)/\mathbb{Z}_2$ is the space of all unordered pairs of points on $S^m$. $\mathbb{R}P^m$ is the space of all unordered pairs of antipodal points on $S^m$. The first space deformation retracts onto the second. Specifically, given an unordered pair $\{a,b\}$ of points on $S^m$ that are not antipodal, let $C$ be the great circle ...

4

Apply Burnside's Lemma: Consider the action of $G$ over $[n]$. The number of orbits is $$|[n]/G|=\sum_{g\in G}\frac{[n]^g}{|G|}$$ where $[n]^g=\{k\in[n]:g(k)=k\}$. The hypothesis says that $|[n]^g|=1$ for all $g\neq e$. Moreover, $[n]^e=[n]$. Then $$|[n]/G|=\frac n{|G|}+\sum_{g\in G,g\neq e}\frac{1}{|G|}=\frac{|G|+n-1}{|G|}>1$$ This means two things: ...

3

Define $i:S^m\to F(S^m,2)$ by $i(a)=(a,-a)$. This map is equivariant with respect to the $\mathbb{Z}_2$-actions on both sides, and I claim it actually realizes $S^m$ as a $\mathbb{Z}_2$-equivariant deformation retract of $F(S^m,2)$. Indeed, given a point $(a,b)\in F(S^m,2)$, we can continuously move $a$ and $b$ in opposite directions from each other along ...

2

$Z_n\left(M\right)$ is the $n$-th homogeneous component of the $0$-th cyclic homology of the tensor algebra $T\left(M\right)$. There is a nice expository note about this by Clas Löfwall, including the relation to the logarithm: Clas Löfwall, Cyklisk homologi, 28 Aug 2012. Despite the title, it is in English.

2

I think the question is probably too broad to have a very satisfying general answer. But (assuming you're looking for a smooth Lie group action) here are a couple of necessary, but certainly not sufficient, conditions: Each level set of $f$ must be a smooth submanifold, because every orbit of a smooth Lie group action is an (immersed) submanifold. Each ...

2

A permutation group is automatically a subset of $S_A$. Equivalently, it is an embedding $G\to S_A$, which means this group homomorphism must be injective (aka one-to-one). In general, a group action needn't be specified by an injective map $G\to S_A$. If the map is not $1$-to-$1$, then different elements of $G$ are sent to the same permutation, so they act ...

2

Let $D_{4n}=\langle a,b\rangle$ with $a^{2n}=1$ and $b^2=1$ and let $D_{2n}=\langle c,d\rangle$ with $c^{n}=1$ and $b^2=1$. Define $\phi : D_{4n} \to D_{2n}$ by $\phi(a)=c^2$ and $\phi(b)=d$. This is a homomorphism with kernel $\{a^ib^j \in D_{4n} : \phi(a^ib^j)=1\}$, and $$a^ib^j \in \ker\phi \iff 1=\phi(a^ib^j)=\phi(a)^i\phi(b)^j=c^{2i}d^j$$ If $j$ is ...

2

Since the category of finite sets with bijections is equivalent to the permutation groupoid $\coprod_{n \geq 0} S_n$, two combinatorial species are isomorphic if and only if the corresponding $S_n$-sets are isomorphic for all $n$. If you do not believe this, here is a direct argument: For each $n$, let $\alpha_n : F([n]) \to G([n])$ be an isomorphism of ...

1

Your claim is not true. For example, consider the subgroup $G < S_5$ generated by $(123)$ and $(12)(45)$, which is isomorphic to $S_3$. Here every element of $G$ fixes a point but there is no common fixed point for $G$.

1

Yes, the orbital mapping is your $\phi_x$ Faithful does not mean that all $\phi_x$ are injective. For example $G$ acts naturally on itself by left multiplication and we can extand this action to an action on $G\cup\{\infty\}$ (where $\infty$ is an arbitrary object $\notin G$) by declaring $\alpha \infty=\infty$ for all $\alpha \in G$. Then this action is ...

1

Statement (c) does not make sense (because $G$ and $G_a$ are not subsets of $A$ and so cannot be blocks). The other statements, however, are true and would remain true even if "Let $G$ be a transitive permutation group" were replaced by "Let $G$ be a group acting transitively". So the fact that you didn't need to assume $G$ has a trivial kernel does not ...

1

About point c): $p$ being a submersion means exactly $p_{*,g}:T_gG\to T_{g\cdot x_o}E$ being surjective for every $g\in G$; but $p_{*,g}= p_{*,e}\circ L_{*,g^{-1}}$ from the action property, where $L_{g^{-1}}:G\to G$ is left translation, hence the claim. About claim 1) The compostion of differentiable maps is differentiable, right? And projection of a ...

1

If $S^3/G$ was not orientable, then the orientation double cover would be non-trivial. From the correspondence between covers of $S^3/G$ and subgroups of $\pi_1(S^3/G)\cong G$, the cover would correspond to an index two subgroup of $G$ (call this subgroup $H$). But now $G/H \cong \Bbb Z/2\Bbb Z$ is a non-trivial abelian quotient of $G$, so $G$ cannot have ...

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