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11

Since you're working on $S_6$ for class, I won't answer that one for you. But I have another easy example of an outer automorphism. The Quaternion group $Q_8$ can be represented as $\{1, -1, i, -i, j, -j, k, -k\}$, where $ij=k$, $jk=i$, $ki=j$, and $ji=-k$, $kj=-i$, $ik=-j$. The $-1$ element acts pretty obviously on the rest, e.g. $(-1)j=-j$. I think ...


8

Here is a context of group action where ergodicity arises naturally. As pointed out by Martin, the definition given in the grey box applies to more general situations where the transformations are not required to be measure-preserving, and the group is not countable discrete. Take a probability measure space $(X,\mathcal{A},\mu)$ and let a countable ...


8

We have $\lvert G\rvert = 28 = 2^2\cdot 7$. Let $a_7$ be the number of $7$-Sylow groups in $G$. By Sylow $$a_7\equiv 1\mod 7\qquad\text{and}\qquad a_7 \mid 4\text{.}$$ This implies $a_7 = 1$, so there is a unique subgroup $H$ of $G$ of order $7$. For all $g\in G$, $gHg^{-1}$ is again a subgroup of order $7$, which forces $gHg^{-1} = H$ for all $g\in G$. So ...


8

Well, the category of schemes is a completely different category than the category of locally ringed spaces. It doesn't have colimits (including quotients by group actions), limits, whereas the category of locally ringed spaces has these. Don't forget the forgetful functor here. If you have a group acting on a scheme, then the quotient of the underlying ...


8

More naturally, the action of an algebraic group $G\times X\rightarrow X$ on an algerbaic variety is called algebraic, if $(g,x)\mapsto g.x$ is a morphism satisfying the two usual axioms $e.x=x$ and $(gh).x=g.(h.x)$. Every finite group is algebraic, so this makes sense.


7

For a finite group acting on a scheme, if each orbit is contained in an open affine, then one can form the quotient scheme, by reducing to the affine case (where one just takes invariants). For example, since any finite set of points in a projective space are contained in the complement of a(n appropriately chosen) hyperplane, we see that for finite groups ...


7

That's a feature, not a bug. Strong continuity is actually supposed to be a weaker condition than continuity (see also strong operator topology, which is weaker than the norm topology). For example, the action of $\mathbb{R}$ on $L^2(\mathbb{R})$ by translation is strongly continuous if $L^2(\mathbb{R})$ is given the norm topology but not continuous.


7

This is tricky - the two cases look the same, but they're not. The first one is a right action $v \cdot (p_1 \cdot p_2) = (v \cdot p_1) \cdot p_2$, while the second one is a left action $p_1 \cdot (p_2 \cdot f) = (p_1 \cdot p_2) \cdot f$. To see why, consider these 2 permutations: $$p_1(1) = 1, p_1(2) = 3, p_1(3) = 2 $$ and $$p_2(1) = 3, p_2(2) = 2, ...


7

Let $G=\{1,a,b,ab\}$ be a Klein $4$-group, and consider the following two actions on $\{1,2,3,4,5,6\}$. In the first action $a \mapsto (1,2)(3,4)$, $b \mapsto (1,3)(2,4)$, $ab \mapsto (1,4)(2,3)$, and in the second action $a \mapsto (1,2)(3,4)$, $b \mapsto (3,4)(5,6)$, $ab \mapsto (1,2)(5,6)$. In both actions, all three involutions have two orbits of ...


6

Differentiate $AU_t=U_tA$ and evaluate at $t=0$. (Relevant: matrix calculus.)


6

What normality of the stabiliser says is exactly that every group element $g\in G$ that fixes $s$ also fixes the entire orbit $Gs$ pointwise. Conversely any $g\in G$ that fixes any element of the orbit $Gs$ will also fix $s$. These two parts are equivalent, although the first sentence says that every conjugate of $G_s$ contains $G_s$, while the second ...


6

Another different proof: since as you said you know that $G$ has an element of order $7$, by Cauchy, you also know that there's at least one subgroup of order $7$, let's call it $H$. Suppose that $K \leq G$ is another subgroup of order $7$, then we can consider the subset $HK$ that has order $|HK|=|H||K|/|H \cap K|$. If $H$ and $K$ were distinct then $H ...


6

There are many sets of more than $n$ elements on which $\mathfrak S_n$ acts transitively. The largest possible example is that of the $n!$ total orderings of the set of $n$ elements (the one used to define $\mathfrak S_n$). One can deduce numerous smaller examples from this one.


5

This question is equivalent to asking what the conjugacy classes of subgroups of $S_{n}$ are, since each transitive permutation representation of any finite group is permutation equivalent to the action on the cosets of one of its subgroups, and conjugate subgroups lead to equivalent permutation representations. Since every finite group embeds in some ...


5

You can indeed work by analogy with the real case. To do so, you must think of $\mathbb{CP}^1$ as the quotient of the complex sphere by the aciton of the unit length complex numbers: $$\mathbb S^3\times \mathbb S^1\to\mathbb S^3, ((z,z'),\lambda)\mapsto (z\lambda,z'\lambda)$$ There is an obvious identification $$\mathbb S^3/\mathbb S^1\simeq\mathbb{CP}^1$$ ...


5

$\def\R{\mathbb{R}} \def\SL{\text{SL}} \def\SO{\text{SO}}$Often you use the group action to study $G$ and not just to study $X$. Here is an example: what does $\SL_2(\R)$ look like as a manifold? You can solve this by thinking of the group directly, but an easier way is to note that it acts transitively on the upper half plane by Mobius transformations. ...


5

I believe it's more the other way around: a group action of $G$ on a space $X$ allows to construct a new space, the quotient $G\backslash X$ (of course the quotient space is "nice" only under some technical conditions). Recognizing that a space $Y$ is actually realized as the orbit space $G\backslash X$ of some simpler space $X$ can help understand better ...


5

The non-trivial version of the statement is that $N$ can be taken to be a subgroup of $H$. In other words, the collection of finite index normal subgroups is cofinal in the collection of all finite index subgroups.


5

Yes, the structure group is not unique. For example, a vector bundle $E$ has, by definition, a structure group $GL(n)$. Some additional structures on $E$ are equivalent to reductions of the structure group to a subgroup of $GL(n)$ and (even if they exist) you may not want to specify these additional structures and thus may not care about getting a smaller ...


5

I'd say originally the action of a group on a set is what motivates many groups. Thus the symmetric group $S_n$ is in fact defined by its action on the set $\{1,\ldots ,n\}$. Or the symmetry group of an object (a regular dodecahedron, say) is inherently given by the way how this (abstract) group acts on the (more concrete) object. Often one gets insights out ...


5

The group action is faithful if the action of an element $x$ is trivial iff $x=1_G$. That is, if $$(xgH=gH\text{ for each }g\in G)\iff x=e $$ Of course, the condition $xgH=gH$ can be rewritten as $$g^{-1}xgH=H\iff g^{-1}xg\in H\iff x\in gHg^{-1}.$$ Is the path forward clear now?


5

Gromov in his original 1987 book (Section 3.1) wrote a classification for arbitrary isometric group actions on hyperbolic spaces (with no further assumption) into 5 main classes. It goes at follows (the terminology is borrowed from here) 1: bounded: orbits are bounded 2: horocyclic: orbits are unbounded, $G$ acts with no hyperbolic isometry (hence there's ...


5

There is no direct relation between the uses of "transitive" in "transitive relation" and in "transitive action". The relation of being in the same orbit is transitive, but that is true for any group action. However both senses are derived from the latin verb "transire", meaning something like "going across" or "going through". A transitive action allows ...


5

Let $S_1$ and $S_2$ be two of the Sylow $3$-subgroups. If $a \in \ker \phi$, then in particular $$aS_1a^{-1} = S_1 \Rightarrow a \in N_G(S_1).$$ The same holds for $S_2$, so $$\ker \phi \subset N_G(S_1) \cap N_G(S_2).$$ Now, since the Sylow $3$-subgroups aren't normal, what is $N_G(S_i)$?


5

$x \mapsto x^{-1}$ is an isomorphism of $G$-sets, not an isomorphism of groups. (A $G$-set is a set equipped with an action of $G$.)


5

Hint: $G$ acts on the right cosets of $U$ by right multiplication. This provides you a homomorphism from $G$ to $S_{index[G:U]}$, the permutation group on $index[G:U]$ elements. The kernel of the action is exactly $core_G(U)$, whence $|G/core_G(U)|$ divides $index[G:U]!$, in particular is finite.


5

This is not true. For example, $SO(3)$ acts faithfully on the $2$-sphere $S^2$, but $SO(3)$ has dimension $3$, and $S^2$ only has dimension $2$.


5

Let me clarify the confusion. Elements $g \in G$ act on the functions on $V$. Thus, if we have $$g_{1} \cdot f(g_{2}^{-1} x)$$ we need to identify what the the function is. Is the function here $f$? No, because the function is meant to take as input elements of $V$, but inside of $f$ we have a group element $g_{2}^{-1}$. The actual function here is $$h = ...


5

If you didn't invert and you tried to turn $x \mapsto xg$ into a left-action then you'd get into trouble, because (if say $g \star x = xg$) then you'd have $$g \star (h \star x) = (xh)g = x(hg) = (hg) \star x \ne (gh) \star x$$ however if you had $g \star x = xg^{-1}$ then you'd have $$g \star (h \star x) = (xh^{-1})g^{-1} = x(h^{-1}g^{-1}) = x(gh)^{-1} = ...


5

If you have a subgroup $\overline{M}$ of a quotient group $G/N$, the lift of $\overline{M}$ is a subgroup $M$ of $G$ such that the $\overline{M}$ is the image of $M$ under the projection homomorphism $G\rightarrow G/N$. (This is guaranteed to exist by the correspondence theorem.) So speaking of the lifted action of some group action implies that you are ...



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