hcju/proofwikiqa huggingface dataset:
For $n \in \Z_{>1}$, we have: {{begin-eqn}} {{eqn | l = 10 \times H_n + 9       | r = 40 \dfrac {n \left({5 n - 3}\right)} 2 + 9       | c = Closed Form for Heptagonal Numbers }} {{eqn | r = 20 n \left({5 n - 3}\right) + 9       | c =  }} {{eqn | r = 100 n^2 - 60 n + 9       | c =  }} {{eqn | r = \left({10 n - 3}\right)^2       | c =  }} {{eqn | r = \left({10 \left({n - 1}\right) + 7}\right)^2       | c =  }} {{end-eqn}} The positive integers of the form $10 \left({n - 1}\right) + 7$ for $n \in \Z_{>1}$ are precisely the positive integers which end in a $7$: :$7, 17, 27, 37, 47, \ldots$ {{qed}} Category:Heptagonal Numbers Category:Square Numbers hsge2ptz9w33aiomxuqcxf5gltagzc9
By Join Succeeds Operands: :$k \preceq p \vee k$ and $p \preceq p \vee k$ By definition of upper closure of element: :$p \vee k \in k^\succeq$ and $p \vee k \in p^\succeq$ By definition of intersection: :$p \vee k \in p^\succeq \cap k^\succeq$ By Maximal implies Difference equals Intersection: :$p^\succeq \setminus \set p = p^\succeq \cap k^\succeq$ By Infimum of Intersection of Upper Closures equals Join Operands :$\map \inf {p^\succeq \cap k^\succeq} = p \vee k$ By definition: :$p^\succeq \setminus \set p$ admits a minimum. Thus by definition: :$p$ is completely irreducible. {{qed}}
By Intersection of Upper Closures is Upper Closure of Join Operands: :$x^\succeq \cap y^\succeq = \left({x \vee y}\right)^\succeq$ Thus by Infimum of Upper Closure of Element: :$\inf\left({x^\succeq \cap y^\succeq}\right) = x \vee y$ {{qed}}
We will prove that $\left({x \vee y}\right)^\succeq \subseteq x^\succeq \cap y^\succeq$ Let $a \in \left({x \vee y}\right)^\succeq$ By definition of upper closure of element: :$x \vee y \preceq a$ By Join Succeeds Operands: :$x \preceq x \vee y$ and $y \preceq x \vee y$ By definition of transitivity: :$x \preceq a$ and $y \preceq a$ By definition of upper closure of element: :$a \in x^\succeq$ and $a \in y^\succeq$ Thus by definition of intersection: :$a \in x^\succeq \cap y^\succeq$ {{qed|lemma}} We will prove that :$x^\succeq \cap y^\succeq \subseteq \left({x \vee y}\right)^\succeq$ Let $a \in x^\succeq \cap y^\succeq$ Thus bt definition of intersection: :$a \in x^\succeq$ and $a \in y^\succeq$ By definition of upper closure of element: :$x \preceq a$ and $y \preceq a$ By definitions of supremum and Upper bound: :$x \vee y \preceq a$ Thus by definition of upper closure of element: :$a \in \left({x \vee y}\right)^\succeq$ {{qed|lemma}} Hence by definition of set equality: :$\left({x \vee y}\right)^\succeq = x^\succeq \cap y^\succeq$ {{qed}}
From the definition of join, it is required to prove that: :$\forall s \in S: p \preceq s \land u \preceq s \implies q \vee u \preceq s$ Let $s \in S$ be such that: :$p \preceq s$ and $u \preceq s$ We have: :$p \ne s$ By definition of strictly precede: :$p \prec s$ By assumption: :$q \preceq s$ Thus by definition of the join $q \vee u$: :$q \vee u \preceq s$ and hence, as desired: :$q \vee u = p \vee u$ {{qed}}
By definition of algebraic: :$\forall z \in S: z^\ll$ is directed and :$L$ satisfies axiom of approximation. By Axiom of Approximation in Up-Complete Semilattice: :$\exists k \in S: k \ll y \land k \npreceq x$ By Algebraic iff Continuous and For Every Way Below Exists Compact Between: :$\exists z \in K\left({L}\right): k \preceq z \preceq y$ By definition of transitivity: :$z \npreceq x$ By definition of upper closure of element: :$x \notin z^\succeq$ By definition of difference: :$x \in S \setminus z^\succeq$ By definition of relative complement: :$x \in \complement_S\left({z^\succeq}\right)$ By definition of compact subset: :$z$ is a compact element. By Upper Closure of Element is Way Below Open Filter iff Element is Compact: :$z^\succeq$ is a way below open filter on $L$. By Upper Way Below Open Subset Complement is Non Empty implies There Exists Maximal Element of Complement: :$\exists p \in S: x \preceq p \land p = \max\complement_S\left({z^\succeq}\right)$ Thus by Maximal implies Completely Irreducible: :$p$ is completely irreducible. Thus $x \preceq p$ Aiming for a contradiction suppose that :$y \preceq p$ By definition of transitivity: :$z \preceq p$ By definition of maximum element: :$p \in \complement_S\left({z^\succeq}\right)$ By definition of relative complement: :$p \notin z^\succeq$ Thus this contradicts $z \preceq p$ by definition of upper closure of element. {{qed}}
By Not Preceding implies Exists Completely Irreducible Element in Algebraic Lattice: :$\forall x, y \in S: y \npreceq x \implies \exists p \in \mathit{Irr}\left({L}\right): p \preceq x \land p \npreceq y$ Thus by Order Generating iff Not Preceding implies There Exists Element Preceding and Not Preceding: :$\mathit{Irr}\left({L}\right)$ is order generating. Let $X \subseteq S$ such that :$X$ is order generating. Let $x \in \mathit{Irr}\left({L}\right)$. By definition of $\mathit{Irr}$: :$x$ is completely irreducible. Thus by Order Generating Includes Completely Irreducible Elements: :$x \in X$ {{qed}}
Aiming for a contradiction suppose that :$p \notin X$ By Completely Irreducible Element iff Exists Element that Strictly Succeeds First Element: :$\exists q \in S: p \prec q \land \left({\forall s \in S: p \prec s \implies q \preceq s}\right) \land p^\succeq = \left\{ {p}\right\} \cup q^\succeq$ where $p^\succeq$ denotes the upper closure of $p$. We will prove that :$q$ is lower bound for $X$ Let $x \in X$. By definitions of infimum and lower bound :$p \preceq x$ By assumption: :$x \ne p$ By definition of strictly precede: :$p \prec x$ Thus by step condition: :$q \preceq x$ {{qed|lemma}} By definition of strictly precede: :$p \preceq q$ We will prove that :$\forall b \in S: b$ is lower bound for $X \implies b \preceq q$ Let $b \in S$ such that :$b$ is lower bound for $X$. By definition of infimum: :$b \preceq p$ Thus by definition of transitivity: :$b \preceq q$ {{qed|lemma}} By definition of infimum: :$q = \inf X = p$ Thus this contradicts $p \ne q$ by definition of strictly precede. {{qed}}
Let $x \in f^{-1} \sqbrk X$, $y \in S$ such that :$y \preceq x$ By definition of increasing mapping: :$\map f y \precsim \map f x$ By definition of preimage of set: :$\map f x \in X$ By definition of lower set: :$\map f y \in X$ Thus by definition of preimage of set: :$y \in f^{-1} \sqbrk X$ {{qed}}
Let $x \in f^{-1} \sqbrk X$, $y \in S$ such that :$x \preceq y$ By definition of increasing mapping: :$\map f x \precsim \map f y$ By definition of preimage of set: :$\map f x \in X$ By definition of upper set: :$\map f y \in X$ Thus by definition of preimage of subset: :$y \in f^{-1} \sqbrk X$ {{qed}}
Let $x, y \in S_1$ such that :$x \preceq_1 y$ {{AimForCont}} that :$\map f x \npreceq_2 \map f y$ By definition of lower closure of element: :$\map f x \notin \paren {\map f y}^\preceq$ By definition of relative complement: :$\map f x \in \relcomp {S_2} {\paren {\map f y}^\preceq}$ By definition of reflexivity: :$\map f y \preceq_2 \map f y$ By definition of lower closure of element: :$\map f y \in \paren {\map f y}^{\preceq_2}$ By Closure of Singleton is Lower Closure of Element in Scott Topological Lattice: :$\set {\map f y}^- = \paren {\map f y}^{\preceq_2}$ By definition of closure: :$\paren {\map f y}^{\preceq_2}$ is a closed set. By definition of closed set: :$\relcomp {S_2} {\paren {\map f y}^{\preceq_2} }$ is a open set. By definition of continuous: :$f^{-1} \sqbrk {\relcomp {S_2} {\paren {\map f y}^{\preceq_2} } }$ is an open set. By definition of Definition:Scott Topology: :$f^{-1} \sqbrk {\relcomp {S_2} {\paren {\map f y}^{\preceq_2} } }$ is an upper set. By definition of preimage of set: :$x \in f^{-1} \sqbrk {\relcomp {S_2} {\paren {\map f y}^{\preceq_2} } }$ By definition of upper set: :$y \in f^{-1} \sqbrk {\relcomp {S_2} {\paren {\map f y}^{\preceq_2} } }$ By definition of preimage of set: :$\map f y \in \relcomp {S_2} {\paren {\map f y}^{\preceq_2} }$ Thus this contradicts $\map f y \in \paren {\map f y}^{\preceq_2}$ {{qed}}
By definition of continuous: :$x^\ll$ is directed and :$\left({S, \vee, \wedge, \preceq}\right)$ is up-complete and :$\left({S, \vee, \wedge, \preceq}\right)$ satisfies axiom of approximation. By definition of mapping preserves directed suprema: :$f$ preserves the supremum of $x^\ll$. By definition of up-complete: :$x^\ll$ admits a supremum. Thus {{begin-eqn}} {{eqn | l = f\left({x}\right)       | r = f\left({\sup \left({x^\ll}\right)}\right)       | c = Axiom of Approximation }} {{eqn | r = \sup \left({f \left[{x^\ll}\right]}\right)       | c = {{Defof|Mapping Preserves Supremum/Subset|Mapping Preserves Supremum: Subset}} }} {{eqn | r = \sup \left({f \left[{\left\{ {w \in S: w \ll x}\right\} }\right]}\right)       | c = {{Defof|Way Below Closure}} }} {{eqn | r = \sup \left\{ {f \left({w}\right): w \in S \land w \ll x}\right\}       | c = {{Defof|Image of Subset under Mapping}} }} {{end-eqn}} {{qed}}
Let $z \in x^\ll$. By definition of way below closure: :$z \ll x$ By Preceding and Way Below implies Way Below and definition of reflexivity: :$z \ll y$ Thus by definition of way below closure: :$z \in y^\ll$ {{qed}}
Let $x, y \in S$ such that :$x \preceq y$ By Preceding implies Way Below Closure is Subset of Way Below Closure: :$x^\ll \subseteq y^\ll$ By definitions of image of set and way below closure: :$f\left[{x^\ll}\right] = \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$ and :$f\left[{y^\ll}\right] = \left\{ {f\left({w}\right): w \in S \land w \ll y}\right\}$ where $f\left[{x^\ll}\right]$ denotes the image of $x^\ll$ under $f$. By Image of Subset under Relation is Subset of Image/Corollary 2: :$f\left[{x^\ll}\right] \subseteq f\left[{y^\ll}\right]$ Thus {{begin-eqn}} {{eqn | l = f\left({x}\right) | r = \sup \left({f\left[{x^\ll}\right]}\right) | c = assumption }} {{eqn | o = \precsim | r =  \sup \left({f\left[{y^\ll}\right]}\right) | c = Supremum of Subset }} {{eqn | r = f\left({y}\right) | c = assumption }} {{end-eqn}} {{qed}}
Let $n$ be semiperfect. Then by definition, the sum of the aliquot parts of $n$ is not less than $n$. The result follows by definition of deficient. {{qed}} Category:Semiperfect Numbers Category:Deficient Numbers hc85iiuau9aa2yjhlxau95ge4lnn53b
Let $n$ be perfect. From Divisor of Perfect Number is Deficient, all divisors of $n$ are deficient. But from Semiperfect Number is not Deficient, it follows that the divisors of $n$ cannot be semiperfect. Hence the result, by definition of primitive semiperfect number. {{qed}} Category:Perfect Numbers Category:Primitive Semiperfect Numbers m2rwpvgnzeynsz9mte4tobkvu5pwm2i
By Mapping at Element is Supremum implies Mapping is Increasing: :$f$ is an increasing mapping. === Sufficient Condition === Let $y \ll f\left({x}\right)$ By Way Below Closure is Directed in Bounded Below Join Semilattice: :$x^\ll$ is directed where $x^\ll$ denotes the way below closure of $x$. By Image of Directed Subset under Increasing Mapping is Directed: :$f\left[{x^\ll}\right]$ is directed where $f\left[{x^\ll}\right]$ denotes the image of $x^\ll$ under $f$. By assumption: :$f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$ By definitions of image of set and way below closure: :$f\left[{x^\ll}\right] = \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$ By Way Below iff Second Operand Preceding Supremum of Directed Set There Exists Element of Directed Set First Operand Way Below Element: :$\exists z \in f\left[{x^\ll}\right]: y \ll z$ By definition of image of set: :$\exists w \in x^\ll: f\left({w}\right) = z$ Thus by definition of way below closure: :$w \ll x$ Thus $y \ll f\left({w}\right)$ {{qed|lemma}} === Necessary Condition === Let :$\exists w \in S: w \ll x \land y \ll f\left({w}\right)$ By Way Below implies Preceding: :$w \preceq x$ By definition of increasing mapping: :$f\left({w}\right) \precsim f\left({x}\right)$ Thus by Preceding and Way Below implies Way Below: :$y \ll f\left({x}\right)$ {{qed}}
By definition of supremum: :$\sup D$ is upper bound for $D$. By Increasing Mapping Preserves Upper Bounds: :$f\left({\sup D}\right)$ is upper bound for $f\left[{D}\right]$. Thus by definition of supremum: :$\sup \left({f\left[{D}\right]}\right) \precsim f\left({\sup D}\right)$ {{qed}}
Let $g \in G$. Because $N$ is normal, conjugation by $g$ is an automorphism of $N$. Because $H$ is characteristic in $N$, $g H g^{-1} = H$. Thus $H$ is normal in $G$. {{qed}}
Let $\phi$ be an automorphism of $G$. Let $x \in \map Z G, y \in G$. Then: {{begin-eqn}} {{eqn | l = \map \phi x y       | r = \map \phi x \map \phi {\map {\phi^{-1} } y}       | c = automorphisms are bijections }} {{eqn | r = \map \phi {x \map {\phi^{-1} } y}       | c = {{Defof|Group Homomorphism}} }} {{eqn | r = \map \phi {\map {\phi^{-1} } y x}       | c = {{Defof|Center of Group}} }} {{eqn | r = \map \phi {\map {\phi^{-1} } y} \map \phi x       | c = {{Defof|Group Homomorphism}} }} {{eqn | r = y \map \phi x }} {{end-eqn}} Hence $\map \phi x \in \map Z G$. So we have $\phi \sqbrk {\map Z G} \subseteq \map Z G$. Since $\phi^{-1}$ is also an automorphism: :$\phi^{-1} \sqbrk {\map Z G} \subseteq \map Z G$ Since $\phi$ is a bijection: :$\map Z G = \phi \sqbrk {\phi^{-1} \sqbrk {\map Z G}} \subseteq \phi \sqbrk {\map Z G}$ Therefore we conclude that $\phi \sqbrk {\map Z G} = \map Z G$. Hence $\map Z G$ is characteristic in $G$. {{qed}} Category:Characteristic Subgroups Category:Centers of Groups 7j9sccn0viksp9srk03nwwlmiikg2gc
=== Associativity === Let $(n_1,h_1),(n_2,h_2),(n_3,h_3)\in N\times H$. Then {{begin-eqn}} {{eqn | l = ((n_1, h_1) \circ (n_2, h_2)) \circ (n_3, h_3)       | r = ((n_1\cdot \phi_{h_1}(n_2), h_1\cdot h_2)) \circ (n_3,h_3) }} {{eqn | l =        | r = (n_1\cdot \phi_{h_1}(n_2) \cdot \phi_{h_1h_2}(n_3), h_1\cdot h_2\cdot h_3) }} {{eqn | l =        | r = (n_1\cdot \phi_{h_1}(n_2) \cdot \phi_{h_1}(\phi_{h_2}(n_3)), h_1\cdot h_2\cdot h_3)       | c = {{Defof|Group Action}} }} {{eqn | l =        | r = (n_1\cdot \phi_{h_1}(n_2\phi_{h_2}(n_3)), h_1\cdot h_2\cdot h_3)       | c = $H$ acts by automorphisms }} {{eqn | l =        | r = (n_1, h_1) \circ ((n_2, h_2) \circ (n_3, h_3))       | c =  }} {{end-eqn}} {{qed|lemma}} === Identity Element === Let $e = (e_N,e_H)$ and $(n,h)\in N\times H$. Then  {{begin-eqn}} {{eqn | l = e \circ (n,h)       | r = (e_N\cdot \phi_{e_H}(n), e_H\cdot h) }} {{eqn | l =        | r = (n, h)       | c = {{Defof|Group Action}} }} {{end-eqn}} and {{begin-eqn}} {{eqn | l = (n,h) \circ e       | r = (n\cdot \phi_{h}(e_N), h\cdot e_H) }} {{eqn | l =        | r = (n, h)       | c = $\phi_h(e_N)=e_N$ because $H$ acts by automorphisms }} {{end-eqn}} Thus $e$ is an identity element for $\circ$. {{qed|lemma}} === Existence of inverse element === Let $(n,h)\in N\times H$. We are looking for $(m,g)$ with $(n,h)\circ(m,g) = (e_N,e_H) = (m,g)\circ(n,h)$, that is: :$\begin{eqnarray}    n \phi_h(m) &= e_N \\    hg &= e_H \\    m \phi_g(n) &= e_N \\    gh &= e_H \end{eqnarray}$ Thus we take $g=h^{-1}$. From the first equation we have $m=\phi_h^{-1}(n^{-1})$; from the third $m=\phi_{h^{-1}}(n)^{-1}$. These are equal because $H$ acts by automorphisms. The inverse of $(n,h)$ is thus $(\phi_{h^{-1}}(n^{-1}),h^{-1})$. {{qed|lemma}} {{qed}} Category:Semidirect Products 1u8lm7p5m9mp1f6iwtyhm10i6d3ytrl
Follows from Semidirect Product of Groups is Group. The alternatives follow from the fact that $H$ acts by automorphisms. {{qed}} {{finish|Expand the proof by demonstrating how it works.}} Category:Group Theory Category:Semidirect Products 9mlhlpyesu2zbhormoglwgj83gye5uv
Pick arbitrary $\tuple {n_1, h_1}, \tuple {n_2, h_2} \in N \rtimes_\phi H$. {{begin-eqn}} {{eqn | l = \tuple {n_1, h_1} \tuple {n_2, h_2}       | r = \tuple {n_1 \cdot \map \phi {h_1} \paren {n_2}, h_1 h_2}       | c = {{Defof|Semidirect Product}} }} {{eqn | r = \tuple {n_1 \cdot \map {I_N} {n_2}, h_1 h_2}       | c = Definition of $\phi$ }} {{eqn | r = \tuple {n_1 n_2, h_1 h_2}       | c =  }} {{end-eqn}} which meets the definition of direct product. {{qed}} Category:Semidirect Products rukq7wpt4k5wgd159gbyr0un0kx53rm
=== 1 implies 2 === Let $n\in N$, $h\in H$. From $(n,e)(e,h) = (e,h)(n,e)$ we have $n\phi_e(e) = e\phi_h(n)$. Thus $H$ acts trivially. By Semidirect Product with Trivial Action is Direct Product, $N\rtimes_\phi H = N\times H$. By External Direct Product of Abelian Groups is Abelian Group, $N$ and $H$ are abelian. {{qed|lemma}} === 2 implies 1 === By Semidirect Product with Trivial Action is Direct Product, $N\rtimes_\phi H = N\times H$. By External Direct Product of Abelian Groups is Abelian Group, $N\rtimes_\phi H$ is abelian. {{qed|lemma}} {{qed}} Category:Semidirect Products 0cxwo51lkiba91x6nn93qaiug5et4rg
{{begin-eqn}} {{eqn | l = 38^2       | r = 1444 }} {{eqn | l = 462^2       | r = 213 \, 444 }} {{eqn | l = 538^2       | r = 289 \, 444 }} {{eqn | l = 962^2       | r = 925 \, 444 }} {{eqn | l = 1038^2       | r = 1 \, 077 \, 444 }} {{eqn | l = 1462^2       | r = 2 \, 137 \, 444 }} {{eqn | l = 1538^2       | r = 2 \, 365 \, 444 }} {{eqn | l = 1962^2       | r = 3 \, 849 \, 444 }} {{eqn | l = 2038^2       | r = 4 \, 153 \, 444 }} {{eqn | l = 2462^2       | r = 6 \, 061 \, 444 }} {{eqn | l = 2538^2       | r = 6 \, 441 \, 444 }} {{eqn | l = 2962^2       | r = 8 \, 773 \, 444 }} {{eqn | l = 3038^2       | r = 9 \, 229 \, 444 }} {{eqn | l = 3462^2       | r = 11 \, 985 \, 444 }} {{end-eqn}} All such $n$ are of the form $500 m + 38$ or $500 m - 38$: {{begin-eqn}} {{eqn | l = \paren {500 m + 38}^2       | r = 250 \, 000 m^2 + 38 \, 000 m + 1444       | c =  }} {{eqn | r = \paren {250 m^2 + 38 m + 1} + 444       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \paren {500 m - 38}^2       | r = 250 \, 000 m^2 - 38 \, 000 m + 1444       | c =  }} {{eqn | r = \paren {250 m^2 - 38 m + 1} + 444       | c =  }} {{end-eqn}} and it is seen that all such numbers end in $444$. Now we show that all such numbers are so expressed. In Squares Ending in Repeated Digits, we have shown the only numbers with squares ending in $44$ ends in: :$12, 38, 62, 88$ hence any number with square ending in $444$ must also end in those numbers. Suppose $\sqbrk {axy}^2 \equiv 444 \pmod {1000}$, where $a < 10$ and $\sqbrk {xy}$ is in the above list. For $\sqbrk {xy} = 12$: {{begin-eqn}} {{eqn | l = 444       | r = \paren {100 a + 12}^2       | rr = \pmod {200} }} {{eqn | r = 10000 a^2 + 2400 a + 144       | rr = \pmod {200} }} {{eqn | r = 144       | rr = \pmod {200} }} {{end-eqn}} This is a contradiction. Similarly for $\sqbrk {xy} = 88$: {{begin-eqn}} {{eqn | l = 444       | r = \paren {100 a + 88}^2       | rr = \pmod {200} }} {{eqn | r = 10000 a^2 + 17600 a + 7744       | rr = \pmod {200} }} {{eqn | r = 144       | rr = \pmod {200} }} {{end-eqn}} Again, a contradiction. For $\sqbrk {xy} = 38$: {{begin-eqn}} {{eqn | l = 444       | r = \paren {100 a + 38}^2       | rr = \pmod {1000} }} {{eqn | r = 10000 a^2 + 7600 a + 1444       | rr = \pmod {1000} }} {{eqn | r = 600 a + 444       | rr = \pmod {1000} }} {{end-eqn}} The solutions to $600 a \equiv 0 \pmod {1000}$ are $a = 0$ or $5$. Hence: :$\paren {500 n + 38}^2 \equiv 444 \pmod {1000}$ Similarly, for $\sqbrk {xy} = 62$: {{begin-eqn}} {{eqn | l = 444       | r = \paren {100 a + 62}^2       | rr = \pmod {1000} }} {{eqn | r = 10000 a^2 + 12400 a + 3844       | rr = \pmod {1000} }} {{eqn | r = 400 a + 844       | rr = \pmod {1000} }} {{end-eqn}} The solutions to $400 a + 400 \equiv 0 \pmod {1000}$ are $a = 4$ or $9$. Hence: :$\paren {500 n + 462}^2 \equiv \paren {500 n - 38}^2 \equiv 444 \pmod {1000}$ {{qed}}
By definition of infimum: :$\inf D$ is lower bound for $D$. By Increasing Mapping Preserves Lower Bounds: :$\map f {\inf D}$ is a lower bound for $f \sqbrk D$. Thus by definition of infimum: :$\map f {\inf D} \precsim \map \inf {f \sqbrk D}$ {{qed}}
=== 1 implies 2 === Note that by Finite Field Extension has Finite Galois Group, $G = \Aut {L / K}$ is finite. Let $\alpha\in L$. Then its orbit under $G$ is finite. By: :Minimal Polynomial of Element with Finite Orbit under Group of Automorphisms over Fixed Field in terms of Orbit its minimal polynomial over $K$ splits completely in $L$. Thus $L/K$ is normal. By: :Finite Orbit under Group of Automorphisms of Field implies Separable over Fixed Field $L / K$ is separable. {{qed|lemma}} === 2 implies 3 === Since $L / K$ is Galois, it is separable. Thus, by the Primitive Element Theorem, there exists an $\alpha \in L$ such that $L = \map K \alpha$. Let $m_\alpha \in K \sqbrk x$ be the minimal polynomial of $\alpha$ over $K$. Then: :$\index L K = \map \deg {m_\alpha}$ Suppose $\sigma \in \Gal {L / K}$. Then: :$\map {m_\alpha} {\map \sigma \alpha} = \map \sigma {\map {m_\alpha} \alpha} = \map \sigma 0 = 0$ since $m_\alpha$ has coefficients in $K$. Therefore $\map \sigma \alpha$ must be a root of $m_\alpha$. Every element of $\Gal {L / K}$ is determined by its value at $\alpha$ by our assumption that $L = \map K \alpha$.  Therefore: :$\order {\Gal {L / K} } \le \map \deg {m_\alpha} = \index L K$ Next, suppose $\beta$ is a root of $m_\alpha$. By the normality of $L / K$, we must have $\beta \in L$. Then, by Abstract Model of Algebraic Extensions: :$\map K \alpha \cong K \sqbrk x / \gen {m_\alpha} \cong \map K \beta$ Composing isomorphisms we have an automorphism of $L$ for each $\beta$. Thus: :$\order {\Gal {L / K} } \ge \map \deg {m_\alpha} = \index L K$ from which our result follows. {{qed|lemma}} {{ProofWanted}} Category:Galois Theory nw8ll9jqppiha21abbselt53sgqx2n9
Because $E / F$ is finite, it is finitely generated. Let $\alpha_1, \ldots, \alpha_n \in E$ with $E = \map F {\alpha_1, \ldots, \alpha_n}$. {{explain|notation $\map F {\alpha_1, \ldots, \alpha_n}$}} By Finite Field Extension is Algebraic, $\alpha_1, \ldots, \alpha_n$ are algebraic over $F$. Let $f_1, \ldots, f_n$ be their minimal polynomials. Let $f = f_1\dots f_n$. By Galois Group Acts Faithfully on Generating Set, $\Gal {E / F}$ acts faithfully on the roots of $f$. Thus $\Gal {E / F}$ is finite. {{qed}} Category:Galois Theory 2eckmkkpx5u20bf3ukk9lt60uc2mndv