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- https://math.libretexts.org/Workbench/Group_Theory_4e_(Milne)/zz%3A_Back_Matter/21%3A_Detailed_Licensing
- https://math.libretexts.org/Workbench/Group_Theory_4e_(Milne)/zz%3A_Back_Matter/20%3A_GlossaryExample and Directions Words (or words that have the same definition) The definition is case sensitive (Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pag...Example and Directions Words (or words that have the same definition) The definition is case sensitive (Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages] (Optional) Caption for Image (Optional) External or Internal Link (Optional) Source for Definition "Genetic, Hereditary, DNA ...") (Eg. "Relating to genes or heredity") The infamous double helix CC-BY-SA; Delmar Larsen Glossary Entries Definition Image Sample Word 1 Sample Definition 1
- https://math.libretexts.org/Workbench/Group_Theory_4e_(Milne)/zz%3A_Back_Matter/10%3A_Index
- https://math.libretexts.org/Workbench/Group_Theory_4e_(Milne)/01%3A_Basic_Definitions_and_Results
- https://math.libretexts.org/Workbench/Group_Theory_4e_(Milne)/04%3A_Groups_Acting_on_Sets
- https://math.libretexts.org/Workbench/Group_Theory_4e_(Milne)/00%3A_Front_Matter/05%3A_Preface_and_IntroductionWe use the standard (Bourbaki) notation: \(\mathbb{N}=\{0,1,2,\ldots\}\); \(\mathbb{Z}\) is the ring of integers; \(\mathbb{Q}{}\) is the field of rational numbers; \(\mathbb{R}{}\) is the field of re...We use the standard (Bourbaki) notation: \(\mathbb{N}=\{0,1,2,\ldots\}\); \(\mathbb{Z}\) is the ring of integers; \(\mathbb{Q}{}\) is the field of rational numbers; \(\mathbb{R}{}\) is the field of real numbers; \(\mathbb{C}{}\) is the field of complex numbers; \(\mathbb{F}_{q}\) is a finite field with \(q\) elements, where \(q\) is a power of a prime number.
- https://math.libretexts.org/Workbench/Group_Theory_4e_(Milne)/07%3A_Representations_of_Finite_Groups/7.02%3A_Roots_of_1_in_fieldsOtherwise, \(X^{n}-1\) has distinct roots (a multiple root would have to be a root of its derivative \(nX^{n-1}\)), and we can always arrange that \(|\mu_{n}(F)|=n\) by extending \(F\), for example, b...Otherwise, \(X^{n}-1\) has distinct roots (a multiple root would have to be a root of its derivative \(nX^{n-1}\)), and we can always arrange that \(|\mu_{n}(F)|=n\) by extending \(F\), for example, by replacing a subfield \(F\) of \(\mathbb{C}{}\) with \(F[\zeta]\), where \(\zeta=e^{2\pi i/n}\), or by replacing \(F\) with \(F[X]/(g(X))\), where \(g(X)\) is an irreducible factor of \(X^{n}-1\) not dividing \(X^{m}-1\) for any proper divisor \(m\) of \(n.\)
- https://math.libretexts.org/Workbench/Group_Theory_4e_(Milne)/05%3A_The_Sylow_Theorems_Application/5.04%3A_Exercises[x81]Show that a finite group (not necessarily commutative) is cyclic if, for each \(n>0\), it contains at most \(n\) elements of order dividing \(n\).
- https://math.libretexts.org/Workbench/Group_Theory_4e_(Milne)/07%3A_Representations_of_Finite_GroupsLet \(\{e_{1},\ldots,e_{n}\}\) be a basis for \(A\) as an \(F\)-vector space; then \(e_{i}e_{j}=\sum\nolimits_{k}a_{ij}^{k}e_{k}\)for some \(a_{ij}^{k}\in F\), called the structure constants of \(A\) ...Let \(\{e_{1},\ldots,e_{n}\}\) be a basis for \(A\) as an \(F\)-vector space; then \(e_{i}e_{j}=\sum\nolimits_{k}a_{ij}^{k}e_{k}\)for some \(a_{ij}^{k}\in F\), called the structure constants of \(A\) relative to the basis; once a basis has been chosen, the algebra \(A\) is uniquely determined by its structure constants.
- https://math.libretexts.org/Workbench/Group_Theory_4e_(Milne)/04%3A_Groups_Acting_on_Sets/4.04%3A_Primitive_actions[ga37] The group \(G\) acts imprimitively if and only if there is a proper subset \(A\) of \(X\) with at least \(2\) elements such that, \[\text{for each }g\in G\text{, either }gA=A\text{ or }gA\cap A...[ga37] The group \(G\) acts imprimitively if and only if there is a proper subset \(A\) of \(X\) with at least \(2\) elements such that, \[\text{for each }g\in G\text{, either }gA=A\text{ or }gA\cap A=\emptyset. \label{e18}% \] Conversely, suppose that there exists an \(x\) in \(X\) and a subgroup \(H\) such that \[\Stab(x)\subsetneqq H\subsetneqq G\text{.}% \nonumber \] Then I claim that \(A=Hx\) is a block \(\neq X\) with at least two elements.
- https://math.libretexts.org/Workbench/Group_Theory_4e_(Milne)/03%3A_Automorphisms_and_Extensions/3.01%3A_Automorphisms_of_groupsThe map \(\alpha\mapsto m\) defines an isomorphism \[\Aut(C_{n})\rightarrow(\mathbb{Z}/n\mathbb{Z})^{\times}, \nonumber \] where \[(\mathbb{Z}/n\mathbb{Z})^{\times}=\{\text{units in the ring }\mathbb{...The map \(\alpha\mapsto m\) defines an isomorphism \[\Aut(C_{n})\rightarrow(\mathbb{Z}/n\mathbb{Z})^{\times}, \nonumber \] where \[(\mathbb{Z}/n\mathbb{Z})^{\times}=\{\text{units in the ring }\mathbb{Z}% /n\mathbb{Z}\}=\{m+n\mathbb{Z}\mid\gcd(m,n)=1\}. \nonumber \] This isomorphism is independent of the choice of a generator \(a\) for \(G\): if \(\alpha(a)=a^{m}\), then for any other element \(b=a^{i}\) of \(G\), \[\alpha(b)=\alpha(a^{i})=\alpha(a)^{i}=a^{mi}=(a^{i})^{m}=(b)^{m}. \nonumber \]
