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3

Let $G$ be a finite group of order $n$. $G$ acts on itself by left multiplication. $g \star x=gx$. Now, this action is transitive:let $x$ and $y$ be in $G$.$y=(yx^{-1})x$. Now, if $x \in G$, $Stab(x)=\lbrace g \in G, gx=x\rbrace =\lbrace 1 \rbrace.$ This gives rise to homomorphism $ \rho:G \to Sym(G)$, and $ker \rho$ is trivial (will leave it for you to ...


3

Given any set $S$, any group $G$ can act on $S$ via the trivial action ($gs=s$ for all $g\in G$ and $s\in S$). Thus, any group can act on $\{1,\ldots,n\}$, regardless of whether it is an actual, literal subset of $S_n$ (the group of permutations of the set $\{1,\ldots,n\}$). On the other hand, every group does embed in (i.e., have an injective homomorphism ...


3

Yes, one easy example is $G$ acting by left multiplication on the disjoint union $$\large X=\bigsqcup_{\substack{\text{subgroups}\\ H\subseteq G}} G/H$$ Clearly $H$ is the stabilizer of the coset $eH\in G/H$, and since $X$ includes a copy of every $G/H$, every subgroup of $G$ occurs as a stabilizer at least once. This also doesn't require $G$ to be finite.


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Yes. Let $G$ act on the power set of its underlying set by elementwise left multiplication. (This does not require $G$ to be finite)


2

Might be helpful to consider the quotient topology. Let $\pi: X \to X/G$ be the projection mapping. Then $U \subseteq X/G$ is open if and only if $\pi^{-1}(U) \subseteq X$ is open. Using this idea we then have only a few subsets to check. Clearly $\pi^{-1}(\emptyset) = \emptyset$ and $\pi^{-1}(X/G) = X$. We then have that $\pi^{-1}(\{\{0\}\}) = \{0\}$ ...


2

Since we know that every group of order $n$ is isomorphic to a subgroup of $S_n$. So suppose $\phi:G\to S_{2m}$ is that homomorphism. Now since $|G|=2m~(even)$ hence $\exists$ an element $a$ whoes order is $2$. Define $a_l:G\to G$ as $$a_l (x)=ax$$then clearly $a_l$ is a bijection so $a_l\in sym(G)=S_{2m}$ Notice that $a_l$ gives a transposition with ...


2

Alternatively: As you noticed, $\{e\}$ is an orbit, so if there are only two in all, $G\setminus\{e\}$ must be an orbit. This implies immediately that all elements of $G$ which are not $e$ are of the same order $n$. The number must be prime, for otherwise $G$ contains cyclic groups of order $n$ which contain elements which are not of order $n$. It follows ...


2

Firstly, it is relatively trivial that every group acts transitively on itself, so we don't really restrict ourselves to talking about a group acting on itself when we define transitivity. In general, we can speak of transitive actions of a group $G$ on any set $X$. The fact that $G$ is transitive on a set $X$ does not imply $G\cong {\rm Sym}(X)$. Indeed, ...


2

Not quite. Especially, your solution $aH$ fails to be a group in general. $$gaH=aH\iff a^{-1}gaH= H\iff a^{-1}ga\in H\iff g\in aHa^{-1}$$ Alternatively: "Clearly" the stabilizer of $H$ is $H$. And for any action with $a\cdot x_1=x_2$ we have $G_{x_2}=aG_{x_1}a^{-1}$


1

$gaH=aH$ if and only if $ga$ is in $aH$ that is if $ga=ah$ for some $h\in H$. That is if $g=aha^{-1}\iff g\in aHa^{-1}$


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Your answers are correct. ${}{}{}{}{}$


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You are missing a hypothesis they are assuming, which is that $H_*(M)[\pi^{-1}]$ (which is just $H_*(M)[m_1^{-1}]$ in this case) can be constructed by right fractions. This implies that $H_*(M_\infty)=H_*(M)[m_1^{-1}]$, as the colimit that computes $H_*(M_\infty)$ is exactly the right fractions for $H_*(M)[m_1^{-1}]$. Since element of $M$ is homotopic to ...


1

The answer depends on which topology you put on the orbit $Gx$. Namely, there are two possibilities: (1) Since the orbit $G.x$ is the coset space $G/G_x$, where $G_x$ is the isotropy subgroup at $x$ you can put on the orbit $Gx$ the quotient topology. Then, by definition, the projection map $\pi : G \to G/G_x$ is open hence the map $\phi_x : G \to Gx$ is ...


1

Let $\varphi:G\rightarrow S_X$ be a homomorphism. Then we can define a group action of $G$ on $X$ by $g.x:=\varphi(g)(x)$. Conversely, given a group action of $G$ on $X$, we can define a homomorphism $\psi:G\rightarrow S_X$ by $\psi(g)(x):=g.x$ (technically, it should be checked that these define group actions and homomorphisms but this isn't a hard check ...



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