# Tag Info

5

So pointed out by Simon Henry, $\operatorname{PSL}(2,\mathbb{C})\cong \operatorname{SO}(3, \mathbb{R})$, not $\operatorname{SO}(3, \mathbb{C})$. Recall that $SO(3, \mathbb{R})$ is the set of orientation-preserving isometries of $\mathbb{R}^3$, and hence $S^2$ (which is homeomorphic to $\mathbb{P}^1$). The finite subgroups of $SO(3, \mathbb{R})$ have been ...

4

In practice, you should be able to tell that a table is a group table by matching it with a group that you already know. Assuming that you don't look at tables with more than six elements, here are the possible groups: Cyclic groups: $C_2, C_4,\dots,C_6$ The Klein four group $V_4$ The symmetric group $S_3$. So if you know what the multiplication tables ...

4

Any symmetry operations of the trirectangular tetrahedron will send vertices to vertices, edges to edges and faces to faces. Since the vertex with three right angles is distinguished, any symmetry operation will leave that vertex unchanged. If the lengths of the edges attached to that vertex are all different, any symmetry operation will leave them ...

2

Think of $\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$. It has order $4$ but all elements have order one or two.

2

1) You have $\alpha^{p^i}\in U_{i+1}-U_{i+2}$: a) For i=n-1 you get that $\alpha^{p^{n-1}}\in U_{n}$, therefore the image of $(\alpha_n)^{p^{n-1}}$ of $(\alpha)^{p^{n-1}}$ in the quotient $U_1/U_n$ is 1. b) Taking i=n-2, you get that $\alpha^{p^{n-2}}\notin U_n$ hence $\alpha_n^{p^{n-2}}\neq 1$. ($\alpha^{p^{n-2}}$ is not in the kernel $U_n$ of the ...

2

I think it is an open problem whether or not there exists a finitely presented group $G$ satisfying $G \simeq G \times G$. However, several such finitely generated groups are known. Probably the first example was given by Jones in Direct products and the Hopf property.

1

With the definition of sign in terms of transpositions, this is a one line proof. Namely, if $\phi$ is a product of an odd number of transpositions, then $\psi$ is the product of an even number of transpositions, and vice versa. Hence they have opposite signs. Less trivial is with the definition in terms of inversions (odd number of inversions means sign ...

1

If $(G_1,\phi_1)$,...,$(G_n,\phi_n)$ is a collection of $n$ groups with $n$ respective binary operations, we define their product as $(G_1\times...\times G_n, \phi_1\times...\times\phi_n)$ where $\phi_1\times...\times\phi_n:(G_1\times...\times G_n)\times(G_1\times...\times G_n)\rightarrow G_1\times...\times G_n$ by computing the group operation ...

1

As the comments have pointed it out, it is indeed modular addition. $\mathbb{Z}^n_2$ is $\underbrace{\mathbb{Z}_2 \times \dots \times \mathbb{Z}_2}_{n \text{ times}}$, so think about it as just dealing with $\mathbb{Z}_2$ (where the sum is modular) with $n$ separate components. So we have $(1, \dots, 1) + (1, \dots, 1) = (0, \dots, 0)$ since $1 + 1 \equiv ... 1 It is modular in each component. So $$(1,\ldots, 1)+(1,\ldots,1)=(0,\ldots,0)$$ and the set is closed under component-wise addition mod 2. 1$\alpha\in U_1\setminus U_2$is a$p$-adic integer of the form$1+p\beta$, where$v_p(\beta)=0$. Think of$p$-adic integers more or less as power series in$p$. The n$\alpha\in U_1\setminus U_2$does have a non-zero term in$p$. An element$a$lies in$U_n\setminus U_{n+1}$if its first term with positive valuation has valuation$n$, i.e. if it can be ... 1 The elements of$N(H)$aren't$gHg^{-1}$but$g \in G$such that$H$has that property. If$g, g' \in N(H)$then $$gg' H (gg')^{-1} = g(g' H g'^{-1})g^{-1} = gHg^{-1} = H$$ then$g\dot \,g' \in N(H)$If$g \in N(H)$then$g^{-1} H (g^{-1})^{-1} = g^{-1} H g = H$, because$gHg^{-1} = H$, then$g^{-1} \in N(H)$. And you're done. 1 Like you say, in order for a multiplication table for an operation to define a group structure, its rows and columns must each contain each element exactly once. (This is a consequence of divisibility: In a group$(G, \ast)$, if$a \ast b = a \ast c$then$b = c$.) This property of an operation$\star: X \times X \to X$, called the Latin square property, ... 1 The given definition of$\operatorname{P\Gamma L}(V)$is not the usual one: Typically$\operatorname{P\Gamma L}(V)$is not defined as$\operatorname{\Gamma L}(V)/Z(\operatorname{\Gamma L}(V))$, but as the quotient group$\operatorname{\Gamma L}(V)/Z(\operatorname{GL}(V))$. Now it is clear that$\operatorname{PGL}(V)$is a subgroup of$\operatorname{P\Gamma ...

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