# Tag Info

Let $\Omega = \{1,2\}$, take $G$ to be any group and set $x \cdot g = 1$ for all $x \in \Omega$ and $g \in G$. Then of course $(x \cdot g) \cdot h = 1 \cdot h = 1 = x \cdot (gh)$, but $2 \cdot 1_G = 1 \neq 2$ .
Here is a very elementary proof, which uses Lagrange's Theorem, but not Sylow's Theorem. It is enough to prove that any two subgroups $P,Q$ of order $p$ are equal, because then we must have $gPg^{-1}=P$ for all $g \in G$, so $P$ is normal. So suppose that $P \ne Q$. Then $P \cap Q = \{1 \}$ by Lagrange. They are both cyclic so $P = \{x^i : 0 \le i < p ... 3 The point is that axiom (2) alone tells you that the map $$\varphi : g \mapsto (\omega \mapsto \omega \cdot g)$$ is a homomorphism of$G$into the semigroup$M(\Omega)$of maps on$\Omega$, i.e.$\varphi(g h) = \varphi(g) \circ \varphi(h)$. (Thanks to Derek Holt for correcting my previous mistake, I had written monoid instead of semigroup, of course a ... 3 Claim Let$g_1H, \ldots, g_mH$be all the cosets of$H$in$G$. Then any two of the following$g_1Hg_1^{-1}, \ldots, g_kHg_m^{-1}$intersect only in the identity. Proof. This uses the hypothesis that$H\cap gHg^{-1}=\{e\}$if$g\notin H$. Suppose$a\in g_1Hg_1^{-1}\cap g_2Hg_2^{-1}$. Then we have $$g_1^{-1}ag_1\in H\cap\ (g_1^{-1}g_2)H(g_1^{-1}g_2)^{-1}$$ ... 3 FACT 1: Group of order$15$is cyclic. [without Sylow theory] By Cauchy theorem,$G$contains a subgroup$H$of order$5$; if$H_1$is another subgroup of order$5$then$|HH_1|=|H|.|H_1|/|H\cap H_1|=5.5/1>|G|$, contradiction. So subgroup of order$5$is unique, hence normal. Let$K$be a subgroup of order$3$. For every$k\in K$, define ... 2 This isn't true at all. In fact, there need not exist any submodule ($G$-invariant or not)$N'\subset M$such that$M=N\oplus N'$. Take, for instance,$R=M=\mathbb{Z}$and$N=2\mathbb{Z}$. You can make this example equivariant for any group$G$by just letting$G$act trivially. (By the way, it also isn't true in general when$R$is a field. For ... 2 If$H$is such a subgroup, then a Theorem of Frobenius tells us that there is$K \lhd G$with$G = HK$and$H \cap K = 1$. Furthermore,$K= \{1_{G} \} \cup (G \backslash \cup_{g \in G} g^{-1}Hg)$. Hence$|G| = |K| + [G:H](|H|-1).$It follows that your seond inequality is violated precisely when$|K| \geq \frac{|G|}{2}.$But$K$is a subgroup of$G$, so the ... 2 For convenience let me write the action as$\phi^n : X \to X$for each$n \in \mathbb{Z}$, where$\phi : X \to X$is some homeomorphism. For (1), if the action is not free then there exists$n \ne 0$and$x \in X$such that$\phi^n(x)=x$. The finite set$\{x,\phi(x),...,\phi^{n-1}(x)\}$is therefore an invariant subset. Finite subsets being closed and ... 1 HINT: Just prove it directly. Suppose that$\sigma\in D$and$\tau\in S_A$, and consider which elements of$A$are moved by$\tau\sigma\tau^{-1}$. Specifically, what does$\tau\sigma\tau^{-1}$do to the set$\tau[F(\sigma)]$? 1 For a counterexample, how about$M = \mathbb{R}$,$G =$the group of affine isomorphisms$g \cdot y = ay+b$of$\mathbb{R}$and so$G \cdot y = \mathbb{R}$for all$y$,$x=0$,$g \in G_x$is given by$g \cdot y = \theta_g(y) = 2y$. For these values of$x$and$g, the linear map d_x\theta_g : \mathbb{R}=T_x(G \cdot x) \to T_x(G \cdot x)=\mathbb{R} ... 1 \newcommand{\Size}[1]{\lvert #1 \rvert}I think the following more general result is true. Let G = P N be a group, where P is a p-group for some prime p, N is a p'-group, and N is normal in G. Then for each q dividing the order of N, there is a q-Sylow subgroup of N normalized by P. Let \Delta be the set of q-Sylow subgroups ... 1 You have \begin{align*} p(h.x)&=\sum_gg^{-1}.L(g.(h.x))\\&=\sum_g(hh^{-1}g^{-1}).L((gh).x)\\ &=\sum_gh.[(gh)^{-1}.L((gh).x)]\\ &=h.\left(\sum_g(gh)^{-1}.L((gh).x)\right)\\ &=h.p(x) \end{align*} where the last equality follows from the fact that\sum_g(gh)^{-1}.L((gh).x=\sum_{gh^{-1}}g^{-1}.L(g.x)=\sum_{g}g^{-1}.L(g.x)=p(x).$$1 As noted by Geoff Robinson groups which have such a subgroup are called Frobenius groups, and in his answer he gave you the conditions when your seond inequality is true and when not. His cited results, namely that K (the Frobenius kernel) is a subgroup is a rather deep result. I add a more elementary proof. Suppose H \ne 1 (otherwise the inequality is ... 1 H \unlhd G and K \unlhd G are also sufficient conditions for there to be a single orbit. In both cases, the transporter is the whole of G. It might be hard to find necessary and sufficient conditions for there being a single orbit. Here is an example in which there are two orbits. Let G = S_4, H = \langle (1,3)(2,4) \rangle, and K = \langle ... 1 The right space is the second! For completeness, let \mathbb{V} be a complex vector space of dimension n; I recall that a flag of \mathbb{V} is a strictly increasing sequence of vector subspaces of \mathbb{V}$$ \{\underline{0}\}=\mathbb{V}_0<\mathbb{V}_1<\dots<\mathbb{V}_r\leq\mathbb{V}\,\text{where:}\,\forall ... 1 Lett$be any element of$G \setminus H$. Then, since you have already proved that$G = H \cup HgH$, we must have$t \in HgH$and hence$HgH=HtH$. Let$T = \langle t \rangle$. Then$1 \in T$and so$H = H1H \le HTH$, and also$HgH = HtH \le HTH$, so$G=HTH$. This applies in particular when$t$and$T$have order$2$. The fact that$G$is doubly transitive ... 1 I have answered your first and second questions in the comments. Let me answer the third question: We have$S_{xy}\times S_{x}=\bigcup_{u\in S_{xy}}\phi^{-1}\left( u\right) $, and the union is a disjoint union (i.e., the sets$\phi^{-1}\left( u\right) $for$u\in S_{xy}$are pairwise disjoint). Thus, (1)$\left\vert S_{xy}\times S_{x}\right\vert ...