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3

The modular group $SL_2 \mathbb Z$ is presented by $$ \langle T,P | T^4 = P^3 = 1 \rangle, $$ where we mean specific matrices $$ T = \left( \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right) $$ and $$ P = \left( \begin{array}{rr} 0 & 1 \\ -1 & -1 \end{array} \right) $$ This is from page 9 in Binary Quadratic Forms by Duncan A. Buell. ...


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Define $\sigma:(x_1,x_2,\dots)\mapsto(y_1,y_2,\dots)$ so that $y_n=x_n\Leftrightarrow x_1+x_2+\cdots+x_{n-1}=0$. In words, flip everything after the first $1$.


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It is a bit too long for a comment, it provides details for Will's answer: Consider a group $G$ which splits as a free product $G_1 * G_2$. Theorem. Every finite order element is conjugate either into $G_1$ or into $G_2$. Proof. The trick is to think of this not as an algebraic but a geometric problem. The free product decomposition of $G$ corresponds to ...


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Using your first multiplication $\mathrm{AGL}(V)$ acts on $V$ from the left via: $(v,A)*z:=Az+v$ This is an action since $$(w,B)*((v,A)*z)=(w,B)*(Az+v)=BAz+Bv+w=(Bv+w,BA)*z=((w,B)(v,A))*z$$ and feels fairly natural to me.


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There's probably a better way that this, but my answer would be "by guesswork". Let's look at $D_3$ as an example. I'd start by guessing "in polar coordinates, a wedge $E$ from $0$ to $2\pi/3$." Clearly the order-3 subgroup $\{e, a, a^2\}$of $D_3$ sends this to the entire plane (i.e., $E \cup aE \cup a^2E = R^2$). If we say that the order-two subgroup $\{e, ...


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The relevant fact here is that if a set $Y$ is equipped with a right $G$-action, then it is also automatically equipped with a left $G$-action by defining $g \cdot y := y \cdot g^{-1}$ (the axioms are easy to check, this is basically what's done in Michael Joyce's answer). Therefore, in your case we view $Y$ as a set with a left $G$-action, and then the ...


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If $G$ is a topological group acting on a topological space $M$, the usual definition is that the action is proper if the map $G\times M\to M\times M$ defined by $(g,x)\mapsto (g\cdot x,x)$ is a proper map, which means that the preimage of every compact set is compact. This definition works both in the topological category and in the smooth category. For ...


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For the stabilizer calculation, you need as well to use that, by virtue of the matrix being in $SL(2, \mathbb{R})$, you have $$ad - bc = 1.$$ It may also be helpful to note that because, e.g., $a^2 + b^2 = 1$, there is some angle $\theta$ such that $$a = \cos \theta, b = -\sin \theta.$$ To show that the action of a group $G$ on a space $X$ is transitive, ...


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Let's use polar coordinates on the $(x,y)$ plane and catesian on the $z$ axis. Then the application $$\varphi : S^2 \to S^2, \quad (\rho,\theta,z) \mapsto (\rho,2\theta,z)$$ is a ramified covering whose fibers are the orbits of your second involution. More generally, in higher dimensions as you say you can classify the involution by the spectrum of the ...


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The key property you need for the definition to be well-defined is that $f((gh) \cdot x) = f(g \cdot (h \cdot x))$, since $(gh) \cdot x = g \cdot (h \cdot x)$ by the definition of a $G$-action on $X$. Using the definition given, we see that $$f(g \cdot (h \cdot x)) = f(h \cdot x) \cdot g^{-1} = (f(x) \cdot h^{-1}) \cdot g^{-1} \\ = f(x) \cdot (h^{-1} ...


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Observing what happens when a stabilizer is a normal subgroup is one of my favorite pieces of math. A priori, the conjugation action and normal subgroups do not seem all that important especially when you first begin studying group theory. However, then you are introduced to quotient groups, the Sylow Theorems, etc., and you see that they are indeed ...


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The quotient space topology in the space of orbits $X/G$ is induced by the topology in $X$ via the projection $\pi \colon X \to X/G$ where $\pi(x) = Gx$. More specifically a subset $U \subset X/G$ is open if and only if $\pi^{-1} U$ is open. Another way of saying this is that a set $\mathcal{O}$ of orbits is open if and only if $\bigcup \mathcal{O}$ is open ...


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The open sets of $X/G$ are precisely images of $G$-invariant open subsets of $X$, or if you want, images of saturated open subsets of $X$. This works for any quotient topology where the map $f\colon X \to Y$ is surjective. Note however that in the case of spaces of orbits $X/G$ the map $X\to X/G$ is moreover open, that is, the image of an open subset of $X$ ...


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For $GL_n$ you have the Gauss decomposition $G' = L \cdot D \cdot U$, where $L$ are the lower triangular matrices with $1$ on the diagonal, $D$ the diagonal matrices and $U$ the upper triangular matrices with $1$ on the diagonal. $G'$ is the open subset of $GL_n$ consisting of all matrices with nonzero leading principal minors $M_1$, $\ldots $ , $M_n \ne 0$. ...


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Consider the homeomorphism from $\{0,2,3\}^{\mathbb{N}}$ to $\{0,1\}^{\mathbb{N}}$ mapping a sequence $(a_n)$ to the concatenation of binary writings of the $a_n$'s.


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$\newcommand\N{\mathbb{N}}$Let $p=3$ for simplicity (what follows could work for any $p$, not that we really need it). First we define a homeomorphism $f : \{0,1,2\}^\N \to \{0,1\}^\N$. $f(x)$ is defined by concatenating the $\varphi(x_n)$ where: $$\varphi(u) = \begin{cases} (0) & u = 0 \\ (1,0) & u = 1 \\ (1,1) & u = 2 \end{cases}$$ This is ...


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Recall that a group $G$ is said to act on a set $X$ if each $g \in G$ can be thought of as a bijection $X \to X$ (and we need that the identity $e \in G$ is the trivial bijection and that these maps satisfy associativity $(gh)(x) = g(h(x))$ for $x \in X$). Any bijection of a set can be thought of as a rearrangement or a permutation of the set $X$, so what ...



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