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6

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 $2 \cdot 1_G = 1 \neq 2$ .


5

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

Let $t$ 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 ...



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