# Tag Info

9

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

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

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 ...

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

Two examples of purely group-theoretic theorems proved using representation theory are: Burnside's theorem, which states that a finite group having no more than 2 distinct prime divisors must be solvable. (Much later, a proof without representation theory was found in the 1970's.) Theorem about Frobenius groups: Assume a finite group $G$ contains a ...

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 $$p_2(1) = 3, p_2(2) = 2, ... 6 That's a feature, not a bug. Strong continuity is actually supposed to be a weaker condition than continuity (see also strong operator topology, which is weaker than the norm topology). For example, the action of \mathbb{R} on L^2(\mathbb{R}) by translation is strongly continuous if L^2(\mathbb{R}) is given the norm topology but not continuous. 6 Another different proof: since as you said you know that G has an element of order 7, by Cauchy, you also know that there's at least one subgroup of order 7, let's call it H. Suppose that K \leq G is another subgroup of order 7, then we can consider the subset HK that has order |HK|=|H||K|/|H \cap K|. If H and K were distinct then H ... 5 If you didn't invert and you tried to turn x \mapsto xg into a left-action then you'd get into trouble, because (if say g \star x = xg) then you'd have$$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

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

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

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 ...

5

To address the stated subtext of your post: many people, myself included, take the position that groups are important because they act on things. A representation is just a group action on a vector space (by linear operators). And whenever you have a group action, even if it isn't on a vector space, there is often a closely related representation lurking ...

5

Just a few unsorted comments to add to what's been said above... The classification of finite simple groups relied heavily on modular representation theory, so there is that. Many of the sporadics are only described reasonably through their representations. Symmetric groups have nicely describable representations that correspond to a combinatorics ...

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

$\def\R{\mathbb{R}} \def\SL{\text{SL}} \def\SO{\text{SO}}$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

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

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

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

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 ...

4

In the semidirect product $N \rtimes \operatorname{Inn} N$, the normal subgroup $N \times 1$ is centralized by the subgroup $\{ (h^{-1},h) \mid h \in N \}$, which is isomorphic to $N$. So this subgroup is equal to the second factor $1 \times N$ in the direct product $1 \times N$. The isomorphism $N \times N \to N \rtimes \operatorname{Inn} N$ is given by ...

4

The fundamental lemma for a group action is the following. Let $x$ be in $X$, a set on which $G$ acts. There is a map $f: G \rightarrow X$ defined by $g \mapsto g.x$. By definition, the image of this map is the orbit of $x$, denoted by $G.x$. Morever, let $H$ be the subset of $G$ of elements $h$ satisfying $h.x = x$; $H$ is called the stabilizer of $x$ and ...

4

The general theorem that you need here, and which Steve aludes, is: Theorem: Let $\,P\,$ be a finite $\,p-\,$ group , $\,X\,$ a finite set, and suppose $\,P\,$ acts on $\,X\,$ . Define $\,X_P:=\{x\in X\;:\;ax=x\,\,\forall\,a\in P\}\,$ . Then, $\,|X_P|=|X|\pmod p\,$ Proof: We know that $\,|X|=\sum|\mathcal Orb(x)|\,$ , where the sum runs over different (and ...

4

I think, it is quite important to carefully write down what one is supposed to do in a situation like this one, as it is really easy to get confused. You are given a subbasis element $S(C,U)$ with $C\subset X$ compact, and $U\subset X$ open. To prove continuity of $\gamma_2$ it suffices to show that for any $g\in G$ satisfying $g(C)\subset U$ there exists a ...

4

Yes if you assume a group has a normal subgroup $Z$ of finite index which is infinite cyclic then it's much easier. If $Z$ is central, then a classical theorem shows that the derived subgroup is finite. So modding out by the derived subgroup, you get an abelian group, and eventually get that the group has a homomorphism onto $\mathbf{Z}$ (with finite kernel, ...

4

You aren't computing $\sigma(\tau(v_1 \otimes ... \otimes v_4))$ correctly. We have $$\tau(v_1 \otimes ... \otimes v_4) = v_3 \otimes v_2 \otimes v_4 \otimes v_1.$$ Now rename $w_1 = v_3, w_2 = v_2, w_3 = v_4, w_4 = v_1$. Then $$\sigma(w_1 \otimes ... \otimes w_4) = w_2 \otimes w_3 \otimes w_1 \otimes w_4$$ which is equal to v_2 \otimes v_4 \otimes ...

4

One important thing that comes out of this argument is that $SO(4)$ almost factors as a direct product, namely its universal (double) cover Spin$(4)$ factors as the direct product of Spin$(3)$ (which is $S^3$, or equivalently $SU(2)$) with itself. So regarding question (3), in some sense your are asking whether there is a way to see that $SO(4)$ should ...

4

The proof is correct, but one can generalize and shorten it as follows: Let $G$ be a finite group and assume that the intersection of all non-trivial subgroups of $G$ is non-trivial, i.e. contains some $g \neq 1$. If $G$ acts on a set $A$ with $|A|<|G|$, then for every $a \in A$ we have $|G:G_a|=|Ga| \leq |A| < |G|$. Therefore, $G_a$ is non-trivial, ...

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