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3

The action is non-trivial because for a fixed $p$, all Sylow-$p$ subgroups of a group $G$ are conjugate to one another. This is the second Sylow theorem. So the kernel of $G\to S^3$ is proper, since there are elements of $G$ that does a non-trivial permutation of the three Sylow-$2$ subgroups. The kernel is also non-trivial, since it's a homomorphism from a ...


3

$\zeta$ is a cycle if and only if there exists a unique $x \in E$ such that $$\text{Card}\Big(\overline{\zeta}.x\Big) > 1,$$ No, $x$ is not unique. What is unique is just the orbit $\overline \zeta\cdot x$, as a subset of $E$. If $y$ is another element of that orbit, we have $\overline \zeta\cdot x = \overline\zeta \cdot y$, yet $x\ne y$.


3

Let G be a group acting on A. The kernel of the action is the set $K =\{g \in G; g \cdot a = a , \forall a \in A\}$. Now the corresponding permutation representation is a group homomorphism $\psi : G \to S_A$ given by $\psi (g)(a) = g \cdot a $. $K \subseteq \ker \psi$ Let $k \in K$, then for all $a \in A$ we have that $\psi (k) (a) = k \cdot a = a$ ...


3

Your equation $\lvert N \rvert=\lvert Z\rvert-1$ is wrong. Not every element of $Z$ maps to a distinct cyclic subgroup. For every cyclic subgroup of order $p$, we have $e_G$ and then we have $p-1$ other elements of order $p$. Thus, for each cyclic subgroup, we add $p-1$ elements to $Z$. Therefore, $Z$ has $1$ element for $e_G$ and $p-1$ elements for each ...


3

The action of $SU(3)$ on $\mathbb C^3$ is far from being transitive. In any linear representation $0$ is a fixed point, but in addition, maps in $SU(3)$ are norm-preserving. Linear algebra shows that the orbits of $SU(3)$ in $\mathbb C^3$ are the spheres of radius $r$ for $r\geq 0$. The stabilizers of all vectors $x\neq 0$ are isomorphic: Any transformation ...


2

The following reformulation of properness is often very useful in practice. Proposition. (Lee, Introduction to smooth manifolds, Proposition 21.5) Let $M$ be a manifold and $G$ a Lie group acting continuously on $M$. The following are equivalent. The action is proper. If $(p_i)$ is a sequence in $M$ and $(g_i)$ is a sequence in $G$ such that ...


2

Let $G$ be the subgroup generated by $\sigma.$ Then $\{1,\cdots,n\}$ can be decomposed as the disjoint union of orbits of $G.$ Separating those orbits of size $1,$ i.e. those $i$ with $\sigma(i)=i,$ we have $$n=\sum\limits_{\text{orbits not fixed}}\mid\text{orbit }\mid+\mid\{\sigma(i)=i\}\mid.$$ If an orbit $\mathcal O$ is not fixed by $G,$ then $\mid\...


2

Let $C_1=G^0,...,C_n$ be the connected component of $G$ and $g_i\in C_i$, $G=\bigcup G^0g_i$. If $x\in X$, the orbit $G(x)$ is the union $G^0(x), G^0(g_1(x)),....,G^0(g_n(x))$ since $G^0(g_i(x))$ is the orbit of $g_i(x)$ by $G^0$, it is closed, so the orbit of $x$ is closed since it is a finite union of closed subsets.


2

I now redo the calculation (correctly, I hope), using the notation of the original question. The symmetry group of the hexagon is the dihedral group $D_6$, $$ D_6 = \langle r,s \mid r^6 = s^2 = 1, r^s = r^{-1} \rangle, $$ where $r$ is a rotation by $\frac{2\pi}{6}$, and $s$ a reflection in an axis connecting two opposite vertices. The conjugacy classes of $...


2

For future reference I would like to document how we can do this calculation using a cycle index. The key observation here is the following: the cycle structure of a rotation (but not a reflection) acting on the vertices and edges is the same for edges and vertices. So we may compute the cycle index by duplicating the cycle ...


2

The kernel of a group action is defined as the set of all group elements which act as the identity. The problem is asking you to show that this definition is related to the kernel of a homomorphism by showing that the kernel of the group action is isomorphic to the kernel of the homomorphism from $G$ onto the permutation group of the set. Please correct me ...


2

Consider the canonical action of $SO(n+1)$ on $R^{n+1}-\{0\}$. It defines an action of $SO(n+1)$ on $RP^n$ as follows: let $x$ be an element of $R^{n+1}-\{0\}$, denote by $[x]$ the class of $x$ in $RP^{n+1}$, for $g\in SO(n+1)$, write $g([x])=[g(x)]$. This action is transitive, since for $[x], [y]\in RP^n$, you can assume $\|x\|=\|y\|=1$, there exists $g\in ...


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The category of affine schemes is the opposite of the category of commutative rings, and taking opposites switches limits and colimits, so in general the categorical quotient $\left( \text{Spec } R \right) / G$ is the spectrum $\text{Spec } R^G$ of the fixed point subring. Hence you want to compute the subring of $R = k[x, y]$ ($k$ a field containing $\omega$...


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If your generators are algebraically independent over $k$. Then it is enough, because construction of polynomial algebra gives rise to a functor: $$F:\mathbb{Set}\rightarrow \mathbb{Alg}_k$$ So if you have a group $G$ acting on a set by bijections $\{\phi_g\}_{g\in G}$, then you have action of $G$ via elements $\{F(\phi_g)\}_{g\in G}$ on $F(X)=k[X]$. On the ...


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If a function from the $2$-sphere to itself has no fixed points, then it is homotopic to the antipodal map. In particular it has degree $-1$.


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The obvious way to do this is to let the entries in $F$ be a set of $n^2$ unknowns to be solved for. It is necessary and sufficient for the equations to be satisfied for all $g$ in a generating set $S$ of $G$. So you get a system of $|S|n^2$ linear equations in $n^2$ unknown to solve. That should work up to dimension $22$. It depends a bit on the field. If ...


1

The second Sylow theorem says that all of the Sylow $2$-subgroups are conjugate to one another. If they are $H_1,H_2$, and $H_3$, there are $g_2,g_3\in G$ such that $H_2=H_1^{g_2}$ and $H_3=H_1^{g_3}$, and of course then $H_3=H_2^{g_2^{-1}g_3}$. Thus, the action is clearly not trivial. The kernel is proper, since the action doesn’t collapse the group, and ...


1

The action mentioned in the hint gives a homomorphism $\phi: \langle x \rangle \to S_3$. We must have $\phi(x)=1$, because the order of $\phi(x)$ must divide both $7$ and $3!=6$, which are coprime. In other words, $x$ acts on the cosets of $H$ like the identity, and so $xH=H$. This can only happen if $x \in H$.


1

Let $t$ denote the number of orbits of $G$ acting on $S$. By the Cauchy-Frobenius Lemma (the lemma that is not Burnside's), $t = \frac{1}{|G|} \sum_{g \in G} |fix(g)|$, where $|fix(g)|$ is the number of points in $S$ that are fixed by $g$. In other words, the number of orbits is equal to the average value of the number of fixed points of permutations in $G$....


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Let $f\colon G\rightarrow X$ be the map $f(g)=gx$, and $h\colon G\rightarrow Y$ the map $h(g)=gy$. Let $Z=X\coprod_{x=y} Y$ be the push-out or colimit of the diagram made out of $f$ and $h$ in the category of $G$-sets. Let $i\colon X\rightarrow Z$ and $j\colon Y\rightarrow Z$ be the maps that come along with $Z$. Then $i(x)=j(y)$, and the stabilizer of this ...


1

First off: what is the affine cone here? $v_d(\mathbb P^n)$ is the image of all monomials of degree $d$ in $n+1$ variables. So the affine cone is $Y = \mathrm{Spec}k[x^I]$, where $x^I$ ranges through all monomials of degree $d$. According to Wikipedia, we have to check two things. First, that $Y$ is invariant under $G$, and secondly that any invariant ...


1

By Burnside's lemma, $$ 1=|X/G|=\frac{1}{|G|}\sum_{g\in G}|X^g|, $$ the first equality holding since the action is transitive, i.e., there is one $G$-orbit. Now try to prove $|X^g|=0$ for some $g\in G$.


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The action has only one orbit, and since (this is Burnside's lemma) $$|X/G||G|= \sum_{g\in G} |X^g|$$ where $X^g$ is the set of fixed points of $g$, it is then clear that if every $|X^g|\geqslant 1$ we obtain a contradiction. Namely, that $|G| > |G|$.


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Hint: prove any such group action either does nothing or has the nontrivial group element switch two things (and leave the third fixed). With more algebra: a group action will be a homomorphism $\mathbb{Z}_2\to S_3$, which will be determined by where it sends the nontrivial element, which must be taken to either the identity or an element of order two. You ...


1

Hint: if $g\in\mathrm{SU}(3)$ stabilizes the $1$-dim complex subspace $\mathbb{C}\times\{(0,0)\}$ then it is a block matrix with blocks of sizes $1$ and $2$.


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If you write $\sigma$ as a product of cycles with disjoint supports, since the order of $\sigma$ is the prime number $p$, all the cycles have order $p$. Say there are $k$ of them. The set of fixed points is the complement of the union of the supports. So its cardinality is $n - kp$.


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What “the same” means is that there exists a bijection between the set of left actions of $G$ on $A$ and the set of right actions of $G$ on $A$. Thus we want to find a map $\lambda\mapsto\lambda^r$ and a map $\rho\mapsto\rho^l$ such that, for a left action $\lambda$, $\lambda^{rl}=\lambda$ and, for a right action, $\rho^{lr}=\rho$. Given a left action $\...


1

The answer depends on the meaning of "representation" here. In case you mean "unitary representations" by it, a possible reference is The Structure and Unitary Representations of $SU(2,1)$, where also the finite-dimensional representations of the Lie algebra $\mathfrak{sl}_3(\mathbb{C})$ are classified, and all of your bonus question are answered. A special ...



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