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## Hot answers tagged group-actions

4

Let $U$ be an open neighborhood of the $G$-invariant subset $A$. Now define $W=\bigcap_{g\in G}gU$. Can you show that this set is a $G$-invariant neighborhood of $A$?

3

It's not clear by your armgument, why there must exists a $\sigma$ for each $m$ such that $\sigma(m)=k$. On the other hand, there's no need for a counting argument. Given, $1\le m<k\le n$, let $\sigma$ be the transposition $(mk)$. That is, the bijection which switches $m$ and $k$ and leaves every other number fixed.

3

The transitive action of $A_4$ of degree $6$ on the cosets of the subgroup $\langle (1,2)(3,4) \rangle$ is a counterexample. Here we have $|S_\alpha|=2$, $|S|=4$, and $S$ has three orbits of zise $2$. You could take $$G = \langle (1, 2)(5, 6), (1, 3, 5)(2, 4, 6) \rangle$$ with $$S = \{ 1,\,(1,2)(3,4),\,(1,2)(5,6),\,(3,4)(5,6) \}.$$ Another ...

3

This action of $Z^2$ on $R$ is not proper. Since $\alpha$ is irrational, $n+m\alpha, n,m\in Z$ is dense, thus the orbit of $0$ is dense, thus $R/Z^2$ is not separated. Let $x\in R$ and $[x]$ the class of $x$ in $R/Z^2$ every neighborhood of $[x]$ is equal to $p(U)$ where $U$ is a neighborhood of an element $y\in p^{-1}([x])$, there exists $g\in Z^2$ such ...

2

Of course if $x=y$, then $\psi_g(x)=\psi_g(y)$! This is true for any function. What you need to show for injectivity is the converse of this, that is, if $\psi_g(x)=\psi_g(y)$, then $x=y$. The second part about comparing cardinalities is correct (for finite sets $X$), but needs to be justified.

2

Hint in both cases: directly check the four defining properties of a group operation until you hit one that is violated.

2

First prove the following exercise: $$gHg^{-1}=g\mathrm{Stab}_G(a)g^{-1}=\mathrm{Stab}_G(g.a)$$ Since the action of $G$ is transitive, every $b\in A$ is equal to $g.a$ for some $g\in G$, so we have $$\cap_{g\in G}gHg^{-1}=\cap_{a\in A}\mathrm{Stab}(a)=:K$$ Now, $K$ is the set of elements $g\in G$ such that $g.a=a$ for all $a\in A$. But $G\leq S_A$, and the ...

2

For arbitrary subgroup $H$ of $G$, the homomorphism $\phi\colon G\rightarrow S_X=S_{G/H}$ is not necessarily injective or surjective. Consider $G$ with $|G|=9$ and $|H|=3$. We will get a homomorphism $\phi\colon G \rightarrow S_{G/H}\cong S_3$. Comparing orders, you will see that it is neither injective nor surjective.

2

The image of $\phi$ is a transitive subgroup of $S_{|G/H|}$, but that's all you can say about it; any transitive subgroup can appear (exercise). The kernel of $\phi$ is the intersection $\bigcap_{g \in G} gHg^{-1}$ of all of the conjugates of $H$ (exercise).

2

Orbits have lengths dividing $p^n$, so either $1$ or a multiple of $p$. The sum of all orbit lengths must be $pm$. $F$ collects precisely those elements of orbit length $1$.

1

if the group is $\{1,a\}$ with $a^2=1$ then over $F_2$ $$(1+a)^2= 1 + 2a+a^2 \equiv_2 0$$ since the element $1+a$ is nilpotent the group algebra is not semisimple

1

In fact the conclusion of Maschke's theorem never holds in the modular case, that is, suppose that ${\rm char}\, k =p\mid G$. Consider the element $\eta=\sum_{g\in G}g\in kG$. Then $g\eta=\eta$ for every $g\in G$ and then the ideal generated by $\eta$ is the same as the $k$-subspace generated by $\eta$, i.e. $\langle \eta\rangle =(\eta)$. Moreover ...

1

I showed how to prove a) in the comments. Here is an outline proof of b) and c). Grün's Theorem says that $G' \cap S = S_0 := \langle N_G(S)' \cap S, T' \cap S \mid T \in {\rm Syl}_p(G) \rangle$. We shall show that $S_0 \le G_\alpha$. We proved $N_G(S)' \le G_\alpha$ in a). Since $S$ acts semiregularly on $\Omega \setminus \{ \overline{\alpha} \}$, the ...

1

The description of all three homogeneous spaces can be derived from the standard linear action of $U(3)$ on $\mathbb C^3$. Since unitary maps preserve the lenght of vectors, this restricts to a transitive action on the Unit sphere in $\mathbb C^3$ which is $S^5$. The stabilizer of a point in $S^5$ is easily seen to be isomorphic to $U(2)$, whence ...

1

A different way, $S_3$ is a subgroup of $S_4$ and $A_3=S_3'\leq S_4'$. Thus, $|S_4'|$ is divisible by $3$. As it is contained by $A_4$ ,$|S_4'|\in \{3,6,12\}$. it is not $3$ as the subgroups of order $3$ is not normal and $A_4$ does not have a subgroup of order $6$. Hence, $S_4'=A_4$.

1

@mesel's answer is correct, but here's a (maybe?) more easy proof. You already proved $S_{4}' \leq A_{4}$, so let's try to prove $A_{4} \leq S_{4}'$. As mentionned in the comment, $A_{4}$ is generated by the $3$- cycles of $S_{4}$, thus you only have to show that each $3$-cycle is a commutator of $S_{4}$. But $(abc) = (acb)(acb) = (ac)(bc)(ac)(bc) = ... 1 For question 1. Take$f$an element of$G$acting trivially. Then for all$a$in$A$,$f.a=a$. Now$f.a:=f(a)$so that$f(a)=a$. This is true for all$a$so$f$is the identity function on$A$. So$f$is indeed the neutral element of$S_A$. For the second question demonstrate then use the following result : When$G$acts on a set$X$then for all$g$in$G$... 1 Here is an example: Let$X$denote the set of integers,$G$the group of integer translations,$D_1=\{1\}$,$D_2=\{0, 1\}$,$A= [2, \infty)\cap X, B=[2, \infty)\cap X$. Then$D_1\cup A$and$D_2\cup B$are equidecomposable (with$n=1$,$g_1$is the unit translation), but$D_1$cannot be equidecomposable to$D_2$since they have different cardinalities. And ... 1 Suppose that the vectors are in$V=\mathbb{R}^3$. 1) The dot product is an operation that take two vectors and gives a scalar as a result so$V$cannot be a group with respect such operation because the result is not in$V$. 2) the cross product gives a vector as a result, but it's not associative (see Jacobi identity) as required by the axioms of a ... 1 If$\psi_g(x)=\psi_g(y)$then$gx=gy$. Then, applying$\psi _{g^{-1}}$and using the associativity of the action, we have $$\psi _{g^{-1}}(gx)=\psi _{g^{-1}}(gy)\Rightarrow g^{-1}(gx)=g^{-1}(gy)\Rightarrow (g^{-1}g)x=(g^{-1}g)y\Rightarrow x=y$$ so$\psi_g$is injective. If$x\in X$, then, again, using associativity, we have$\psi _g(g^{-1}x)=x$which ... 1 A linear complex structure on$\mathbb{R}^{2n}$is the structure of a complex vector space on it compatible with its real vector space structure.$J$is multiplication by$i$. Since there is only one$n$-dimensional complex vector space up to isomorphism, any two such complex structures give rise to two complex vector spaces$V, V'$such that there must be ... 1 Answer that lacks the application of group actions (so it might be useless for you). For a proof of: $$\binom{2p}{p}=\sum_{k=0}^p\binom{p}{k}^2=2+\sum_{k=1}^{p-1}\binom{p}{k}^2$$ have a look here. Evidendly prime$p$divides$\binom{p}{k}$for each$k\in\{1,\dots,p-1\}$. I don't know, but maybe some action can be found on a set of cardinality ... 1 We may consider every subset$S$of$\{1,\ldots,2p\}$with$p$elements as a couple$\left(S\cap \{1,p\},S\cap\{p+1,2p\}\right)$, then consider the action (by shifting) of$C_p\times C_p$on such couples. Iff both$S\cap\{1,p\}$and$S\cap\{p+1,2p\}$are non-empty, the equivalence class of a couple has exactly$p^2$elements, hence we get:$$p^2 \mid ... 1 Hint: Consider the rotations of a regular polygon with$2p$vertices,$p$of which are to be colored black and the others white. Of the$2p\choose p$colorings, there are$2$that alternate black and white. Can you show that the rest come in groups of$p$? 1 By definition,$G_i:= \{s \in S \mid s \cdot i = i\}$, that is,$G_i$consists of the permutations in$S_n$that fix$i$. You can view such permutations as just permutations of the other$n-1$elements of$A$. 1 When you showed that$C_G(A)\cap G_{\alpha}=1$, noticed that you showed it for any$\alpha \in \Omega$. Thus, orbit of$\alpha$has exactly$|C_G(A)|$elements under the action of$C_G(A)$on$\Omega$. It directly showed that$|C_G(A)|$divides$|\Omega|\$.

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