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4

Conjugacy classes in the symmetric group correspond to cycle-types. Thus, for $H = \langle (1234)\rangle$, the conjugacy class of $(1234)$ is just $\{(1234),(1243),(1324),(1342),(1423),(1432)\}$. We see that the cycles $(1234),(1432)$ generate the same group. Similarly, so do the cycles $(1243),(1342)$, and also the cycles $(1324),(1423)$. Thus, the orbit ...


3

The quotient is actually $\mathbb{C}^n$ indeed. This is really an algebraic corollary and the geometric intuition here wouldn't be that helpful, mainly because we are talking about the homeomorphism type of the quotient, so we have quite a lot of flexibility. For that matter, we are in complex space, where the hyperplane of a reflection has topological ...


2

First, note that if $\min\{p, q\} = 0$, then by definition $C$ is empty and so has zero connected components. For other signatures, the count is given by the following: Proposition If $\min\{p, q\} > 1$, then $C$ has one connected component. If $\min\{p, q\} = 1$ but $\max\{p, q\} > 1$, then $C$ has two connected components. If $p = q = 1$, then ...


2

Your mistake comes from (in my opinion) a bad notation (but it is widely used unfortunately). There are two subsets at stake when you give some $x\in X$ : $$Gx=G.x:=\{g.x|g\in G\}\text{ which is the orbit of }x\text{ under the action of } G $$ $$\text{ and }$$ $$G_x=Stab_G(x):=\{g\in G|g.x=x\}\text{ the stabilizer of } x\text{ under the action of } G$$ ...


2

Notice that $G/Kernel$ isomorphic to a subgroup of $Sym(X).$ As $Sym(X)$ is finite we are done.


2

For each $x\in X$, the orbit of $x$ is the set $\mathcal O(x)=\{gx:g\in G\}$. Since $G$ is a $p$-group, $|\mathcal O(x)|$ is a power of $p$. Note that it can not be infinite because $X$ is finite. Note that the set orbits is a partition in $X$. Since the cardinal of each orbit is a power of $p$, $|X|$ must be also a multiple of $p$, unless some of these ...


2

Since $\phi(\beta^{-1})\alpha \in G_2(p)\subset \phi(G_1)$, there exists $\gamma \in G_1$ such that $\phi(\beta^{-1})\alpha = \phi(\gamma)$. So $\alpha = \phi(\beta\gamma) \in \phi(G_1)$.


1

I denote the symmetric group on the set $G$ with $S_G$ Consider the following homomorphism $\varphi$: $G\rightarrow S_G\rightarrow Z_2$, where the first homomorphism is the left regular action (the homomorphism that sends $g$ to the permutation $f$ on $G$ defined by $f(a)=ga$ ) and the second homomorphism is the one that sends each permutation to its ...


1

A hint would be to take that element of order $2$, and decompose $\phi(g)$ as a permutation into disjoint cycles. If you know how many cycles and the length of each cycle, you can use parity to determine if that permutation is even or odd.


1

Our setup consists of a space $X$, a group $G$ equipped with the discrete topology, and a continuous proper map $\rho: G\times X \to X \times X$ defined by $(g,x) \mapsto (g\cdot x, x)$. Claim. If $X$ (hence $X \times X$) is a locally compact and Hausdorff, the orbit $G_x$ is closed. Proof. The inclusion $\iota_x: X \to X \times X$ defined by $y \mapsto ...



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