hcju/proofwikiqa huggingface dataset:
Consider arbitrary $x_n \in N, x_h \in H$. Thus: :$x_n x_h \in N H$ We aim to show that: :$x_n x_h N K \paren {x_n x_h}^{-1} \subseteq N K$ thus demonstrating $N K \lhd N H$ by the Normal Subgroup Test. We have: {{begin-eqn}} {{eqn | l = x_n x_h N K \paren {x_n x_h}^{-1}       | r = x_n x_h N K {x_h}^{-1} {x_n}^{-1}       | c = Inverse of Group Product }} {{eqn | r = x_n N x_h K {x_h}^{-1} {x_n}^{-1}       | c = as $N \lhd G$ and $H \le G$ }} {{eqn | r = x_n N K {x_n}^{-1}       | c = as $K \lhd H$ }} {{eqn | r = N K {x_n}^{-1} }} {{eqn | r = K N {x_n}^{-1}       | c = as $N \lhd G$ and $K \le G$ }} {{eqn | r = K N }} {{eqn | r = N K       | c = as $N \lhd G$ and $K \le G$ }} {{end-eqn}} {{qed}}
By Groups of Order 8, the only group satisfying all three properties is $Q = \Dic 2$. {{qed}}
From Group Presentation of Dihedral Group, $D_4$ is generated by two elements of orders $4$ and $2$ respectively. Let these generators be $\alpha$ and $\beta$ where: :$\alpha^4 = e$ :$\beta^2 = e$ Hence $\alpha^2$ and $\beta$ are different elements of $D_4$ which both have order $2$. But the quaternion group: :$Q = \set {\mathbf 1, -\mathbf 1, \mathbf i, -\mathbf i, \mathbf j, -\mathbf j, \mathbf k, -\mathbf k}$ has only one element of order $2$, which is $-\mathbf 1$. The rest have order one or four. Thus $Q$ and $D_4$ cannot be isomorphic because there are not the same number of elements of order $2$ in both groups. {{qed}}
Let $G' = H K$. From Subset Product of Normal Subgroups is Normal, $G'$ is a normal subgroup of $G$. That is $G'$ is itself a group. So by the Internal Direct Product Theorem, $G'$ is the internal group direct product of $H$ and $K$. The result follows by definition of the internal group direct product. {{qed}}
Consider arbitrary $x_n \in N_1, x_h \in H_1 \cap H_2$. Thus: :$x_n x_h \in N_1 \paren {H_1 \cap H_2}$ Note that as $x_h \in H_1 \cap H_2$ it follows that $x_h \in H_1$. We aim to show that: :$x_n x_h N_1 \paren {H_1 \cap N_2} \paren {x_n x_h}^{-1} \subseteq N_1 \paren {H_1 \cap H_2}$ thus demonstrating $N_1 \paren {H_1 \cap N_2} \lhd N_1 \paren {H_1 \cap H_2}$ by the Normal Subgroup Test. We have: {{begin-eqn}} {{eqn | l = x_n x_h N_1 \paren {H_1 \cap N_2} \paren {x_n x_h}^{-1}       | r = x_n x_h N_1 \paren {H_1 \cap N_2} {x_h}^{-1} {x_n}^{-1}       | c = Inverse of Group Product }} {{eqn | r = x_n x_h N_1 {x_h}^{-1} x_h \paren {H_1 \cap N_2} {x_h}^{-1} {x_n}^{-1}       | c =  }} {{eqn | r = x_n N_1 x_h \paren {H_1 \cap N_2} {x_h}^{-1} {x_n}^{-1}       | c = as $N_1 \lhd H_1$ }} {{eqn | r = x_n N_1 \paren {H_1 \cap N_2} {x_n}^{-1}       | c = $H_1 \cap N_2$ is normal in $H_1$ from Intersection with Normal Subgroup is Normal }} {{eqn | r = N_1 \paren {H_1 \cap N_2} {x_n}^{-1} }} {{eqn | r = \paren {H_1 \cap N_2} N_1 {x_n}^{-1}       | c = Subset Product of Subgroups }} {{eqn | r = \paren {H_1 \cap N_2} N_1 }} {{eqn | r = N_1 \paren {H_1 \cap N_2}       | c = Subset Product of Subgroups }} {{end-eqn}} {{qed}} {{Proofread}}
Let $\phi_1: G \to G / H$ and $\phi_2: G' \to G' / H'$ be the quotient epimorphisms with $H$ and $H'$ as their kernels, respectively. Now define a homomorphism $\phi: G \times G' \to \paren {G / H} \times \paren {G' / H'}$ by: :$\phi = \phi_1 \times \phi_2$ so: :$\map \phi {\tuple {x, x'} } = \tuple {\map {\phi_1} x, \map {\phi_1} {x'} }$ The kernel of $\phi$ is clearly $H \times H'$, and $\phi$ is surjective. So by the First Isomorphism Theorem for Groups: :$\paren {G / H} \times \paren {G' / H'} \cong \paren {G \times G'} / \paren {H \times H'}$ {{qed}}
The Identity Mapping is Group Automorphism. By definition, a group endomorphism is a group homomorphism from $G$ to itself. A group automorphism is a group isomorphism from $G$ to itself. As a group isomorphism is a group homomorphism which is also a bijection, the result follows by definition. {{qed}}
{{begin-eqn}} {{eqn | l = \norm {\sum_{i \mathop = 1}^n f_i}^2       | r = \innerprod {\sum_{i \mathop = 1}^n f_i} {\sum_{j \mathop = 1}^n f_j}       | c = {{Defof|Inner Product Norm}} }} {{eqn | r = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \innerprod {f_i} {f_j}       | c = Linearity of Inner Product }} {{eqn | r = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \begin {cases} \innerprod {f_i} {f_i} & i = j \\ 0 & i \ne j \end {cases}       | c = The $f_i$ are pairwise orthogonal }} {{eqn | r = \sum_{i \mathop = 1}^n \norm {f_i}^2       | c = {{Defof|Inner Product Norm}} }} {{end-eqn}} {{qed}}
=== Necessary Condition === Let $\phi: \struct {\Z, +} \to \struct {\Z, +}$ be an endomorphism. Let $k = \map \phi 1$. We have that $n = 1 + \cdots \paren n \cdots + 1$ for any positive integer $n$. Thus: {{begin-eqn}} {{eqn | l = \map \phi n       | r = \map \phi 1 + \cdots \paren n \cdots + \map \phi 1       | c =  }} {{eqn | r = k + \cdots \paren n \cdots + k       | c =  }} {{eqn | r = k n       | c =  }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \map \phi 1       | r = \map \phi {1 + 0}       | c =  }} {{eqn | r = \map \phi 1 + \map \phi 0       | c =  }} {{eqn | ll= \leadsto       | l = \map \phi 0       | r = 0       | c =  }} {{eqn | r = k \cdot 0       | c =  }} {{eqn | ll= \leadsto       | l = \map \phi {-1}       | r = -k       | c =  }} {{eqn | ll= \leadsto       | l = \map \phi {-n}       | r = -k n       | c = for all positive integers $n$ }} {{end-eqn}} Thus: :$\forall n \in \Z: \map \phi n = k n$ {{qed|lemma}} === Sufficient Condition === Let $k \in \Z$ such that: : $\forall n \in \Z: \map \phi n = k n$ Then: {{begin-eqn}} {{eqn | ll= \forall n, m \in \Z:       | l = \map \phi {n + m}       | r = k \paren {n + m}       | c =  }} {{eqn | r = k n + k m       | c = Integer Multiplication Distributes over Addition }} {{eqn | r = \map \phi n + \map \phi m       | c =  }} {{end-eqn}} Thus $\phi: \struct {\Z, +} \to \struct {\Z, +}$ is a group homomorphism from $\Z$ to $\Z$. Hence by definition $\phi$ is a group endomorphism. {{qed}}
=== Necessary Condition === Let $\struct {G, \circ}$ be a cyclic group of order $n$. From List of Elements in Finite Cyclic Group: :$G = \set {a^0, a^1, a^2, \ldots, a^n}$ where $a^0 = e, a^1 = a$. From the definition of integers modulo $n$, $\Z_n$ can be expressed as: :$\Z_n = \set {\eqclass 0 n, \eqclass 1 n, \ldots, \eqclass {n - 1} n}$ where $\eqclass x n$ is the residue class of $x$ modulo $n$. Let $\phi: G \to \Z_n$ be the mapping defined as: :$\forall k \in \set {0, 1, \ldots, n - 1}: \map \phi {a^k} = \eqclass k n$ By its definition it is clear that $\phi$ is a bijection. Also: {{begin-eqn}} {{eqn | l = \map \phi {a^r \circ a^s}       | r = \map \phi {a^{r + s} }       | c = Powers of Group Elements: Sum of Indices }} {{eqn | r = \eqclass {r + s} n       | c = Definition of $\phi$ }} {{eqn | r = \eqclass r n +_n \eqclass s n       | c = {{Defof|Modulo Addition}} }} {{eqn | r = \map \phi {a^r} +_n \map \phi {a^s}       | c = Definition of $\phi$ }} {{end-eqn}} Thus the morphism property of $\phi$ is demonstrated, and $\phi$ is thus a group homomorphism. By definition, a group isomorphism is a group homomorphism which is also a bijection. {{qed|lemma}} === Sufficient Condition === Now suppose $G$ is a group such that $\phi: \Z_n \to G$ is a group isomorphism. Let $a = \map \phi {\eqclass 1 n}$. Let $g \in G$. Then $g = \map \phi {\eqclass k n}$ for some $\eqclass k n \in \Z_n$. Therefore: {{begin-eqn}} {{eqn | l = g       | r = \map \phi {\eqclass k n}       | c =  }} {{eqn | r = \map \phi {\eqclass 1 n +_n \cdots (k) \cdots +_n \eqclass 1 n}       | c =  }} {{eqn | r = a^k       | c =  }} {{end-eqn}} So every element of $G$ is a power of $a$. So by definition $G$ is cyclic. {{qed}}
{{begin-eqn}} {{eqn | l = \norm {f + g}^2 + \norm {f - g}^2       | r = \innerprod {f + g} {f + g} + \innerprod {f - g} {f - g}       | c = {{Defof|Inner Product Norm}} }} {{eqn | r = \innerprod f f + \innerprod f g + \innerprod g f + \innerprod g g + \innerprod f f - \innerprod f g - \innerprod g f + \innerprod g g       | c = Linearity of Inner Product }} {{eqn | r = 2 \innerprod f f + 2 \innerprod g g }} {{eqn | r = 2 \paren {\norm f^2 + \norm g^2}       | c = {{Defof|Inner Product Norm}} }} {{end-eqn}} {{qed}}
Let $\mathbf 0_H$ be the zero of $H$. Since for every $k \in K$, we have: :$\map d {h, k} = \norm {h - k} = \map d {\mathbf 0_H, k - h}$ it follows that: :$\map d {h, K} = \map d {\mathbf 0_H, K - h}$ {{WLOG}}, we may therefore assume that $h = \mathbf 0_H$. The problem has therefore reduced to finding $k_0 \in K$ such that: :$\norm {k_0} = \map d {\mathbf 0_H, K} = \inf \set {\norm k : k \in K}$ Let $d = \map d {\mathbf 0_H, K}$. By definition of infimum, there exists a sequence $\sequence {k_n}_{n \mathop \in \N}$ such that: :$\displaystyle \lim_{n \mathop \to \infty} \norm {k_n} = d$ By the Parallelogram Law, we have that for all $m, n \in \N$: :$(1): \quad \norm {\dfrac {k_n - k_m} 2 } = \dfrac 1 2 \paren {\norm {k_n}^2 + \norm {k_m}^2} - \norm {\dfrac {k_n + k_m} 2 }^2$ Since $K$ is convex, $\dfrac {k_n + k_m} 2 \in K$. Hence: :$\norm {\dfrac {k_n + k_m} 2 }^2 \ge d^2$ Now given $\epsilon > 0$, choose $N$ such that for all $n \ge N$: :$\norm {k_n}^2 < d^2 + \epsilon$ From $(1)$, it follows that: :$\norm {\dfrac {k_n - k_m} 2} < d^2 + \epsilon - d^2 = \epsilon$ and hence that $\sequence {k_n}_{n \mathop \in \N}$ is a Cauchy sequence. Since $H$ is a Hilbert space and $K$ is closed, it follows that there is a $k_0 \in K$ such that: :$\displaystyle \lim_{n \mathop \to \infty} k_n = k_0$ From Norm is Continuous, we infer that $\norm {k_0} = d$. This demonstrates existence of $k_0$. For uniqueness, suppose that $h_0 \in K$ has $\norm {h_0} = d$. Since $K$ is convex, it follows that $\dfrac {h_0 + k_0} 2 \in K$. This implies that $\norm {\dfrac {h_0 + k_0} 2} \ge d$. Now from the Triangle Inequality: :$\norm {\dfrac {h_0 + k_0} 2} \le \dfrac {\norm {h_0} + \norm {k_0} } 2 = d$ meaning that $\norm {\dfrac {h_0 + k_0} 2} = d$. Thus, the Parallelogram Law implies that: :$d^2 = \norm {\dfrac {h_0 + k_0} 2}^2 = d^2 - \norm {\dfrac {h_0 - k_0} 2}^2$ from which we conclude that $h_0 = k_0$. {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac 1 {\sqrt {a^2 - x^2} } \rd x       | r = \int \frac {\rd x} {\sqrt {a^2 \paren {1 - \frac {x^2} {a^2} } } }        | c = factor $a^2$ out of the radicand }} {{eqn | r = \int \frac {\rd x} {\sqrt{a^2} \sqrt {1 - \paren {\frac x a}^2} }       | c =  }} {{eqn | r = \frac 1 a \int \frac {\rd x} {\sqrt {1 - \paren {\frac x a}^2} } }} {{end-eqn}} Substitute: :$\sin \theta = \dfrac x a \iff x = a \sin \theta$ for $\theta \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$. From Real Sine Function is Bounded and Shape of Sine Function, this substitution is valid for all $x / a \in \openint {-1} 1$. By hypothesis: {{begin-eqn}} {{eqn | l = a^2       | o = >        | r = x^2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = 1       | o = >       | r = \frac {x^2} {a^2}       | c = dividing both terms by $a^2$ }} {{eqn | ll= \leadstoandfrom       | l = 1       | o = >       | r = \paren {\frac x a}^2       | c = Powers of Group Elements }} {{eqn | ll= \leadstoandfrom       | l = 1       | o = >       | r = \size {\frac x a}       | c = taking the square root of both terms }} {{eqn | ll= \leadstoandfrom       | l = -1       | o = <       | r = \paren {\frac x a} < 1       | c = Negative of Absolute Value }} {{end-eqn}} so this substitution will not change the domain of the integrand. Then: {{begin-eqn}} {{eqn | l = x        | r = a \sin \theta       | c = from above }} {{eqn | ll= \leadsto       | l = 1       | r = a \cos \theta \frac {\rd \theta} {\rd x}       | c = differentiating {{WRT|Differentiation}} $x$, Derivative of Sine Function, Chain Rule for Derivatives }} {{eqn | l = \frac 1 a \int \frac 1 {\sqrt {1 - \paren {\frac x a}^2 } } \rd x       | r = \frac 1 a \int \frac {a \cos \theta} {\sqrt {1 - \sin^2 \theta} } \frac {\rd \theta} {\rd x} \rd x       | c = from above }} {{eqn | r = \frac a a \int \frac {\cos \theta} {\sqrt {1 - \sin^2 \theta} } \rd \theta       | c = Integration by Substitution }} {{eqn | r = \int \frac {\cos \theta} {\sqrt {\cos^2 \theta} } \rd \theta       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \int \frac {\cos \theta} {\size {\cos \theta} } \rd \theta       | c =  }} {{end-eqn}} We have defined $\theta$ to be in the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$. From Sine and Cosine are Periodic on Reals, $\cos \theta > 0$ for the entire interval. Therefore the absolute value is unnecessary, and the integral simplifies to: {{begin-eqn}} {{eqn | l = \int \rd \theta       | r = \theta + C }} {{end-eqn}} As $\theta$ was stipulated to be in the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$: :$\sin \theta = \dfrac x a \iff \theta = \arcsin \dfrac x a$ The answer in terms of $x$, then, is: :$\displaystyle \int \frac 1 {\sqrt {a^2 - x^2} } \rd x = \arcsin \frac x a + C$ {{qed}}
By definition, $\kappa_x: G \to G$ is a mapping defined as: :$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$ We need to show that $\kappa_x$ is an automorphism. First we show $\kappa_x$ is a homomorphism. {{begin-eqn}} {{eqn | ll= \forall g, h \in G:       | l = \map {\kappa_x} g \map {\kappa_x} h       | r = \paren {x g x^{-1} } \paren {x h x^{-1} }       | c = Definition of $\kappa_x$ }} {{eqn | r = x g \paren {x^{-1} x} h x^{-1}       | c = {{GroupAxiom|1}} }} {{eqn | r = x \paren {g e h} x^{-1}       | c = {{GroupAxiom|3}} }} {{eqn | r = x \paren {g h} x^{-1}       | c = {{GroupAxiom|2}} }} {{eqn | r = \map {\kappa_x} {g h}       | c = Definition of $\kappa_x$ }} {{end-eqn}} Thus the morphism property is demonstrated. Next we show that $\kappa_x$ is injective. {{begin-eqn}} {{eqn | l = \map {\kappa_x} g       | r = \map {\kappa_x} h }} {{eqn | ll= \leadsto       | l = x g x^{-1}       | r = x h x^{-1}       | c = Definition of $\kappa_x$ }} {{eqn | ll= \leadsto       | l = g       | r = h       | c = Cancellation Laws }} {{end-eqn}} So $\kappa_x$ is injective. Finally we show that $\kappa_x$ is surjective. Note that $\forall h \in G: x^{-1} h x \in G$ from fact that $G$ is a group and therefore closed. So: {{begin-eqn}} {{eqn | ll= \forall h \in G:       | l = \map {\kappa_x} {x^{-1} h x}       | r = x \paren {x^{-1} h x} x^{-1}       | c = Definition of $\kappa_x$ }} {{eqn | r = \paren {x x^{-1} } h \paren {x x^{-1} }       | c = {{GroupAxiom|1}} }} {{eqn | r = e h e       | c = {{GroupAxiom|3}} }} {{eqn | r = h       | c = {{GroupAxiom|2}} }} {{end-eqn}} Thus every element of $G$ is the image of some element of $G$ under $\kappa_x$ (that is, of $x^{-1} h x$), and surjectivity is proved. {{Qed}}
There are $3$ cases, and for each case we find an automorphism that is not the identity automorphism. === Case $1$: $G$ is non-Abelian === There exists some elements $g, h \in G$ such that $g h \ne h g$. Then the inner automorphism: :$x \mapsto g x g^{-1}$ gives an automorphism that is not the identity automorphism, since: :$g h g^{-1} \ne h$ Hence $\order {\Aut G} > 1$. {{qed|lemma}} === Case $2$: $G$ has an element with order greater than $2$ === Suppose $g \in G$ has order $k$, where $k > 2$. Then $g \ne g^{-1}$. Thus the inverse automorphism: :$x \mapsto x^{-1}$ is not the identity automorphism. Hence $\order {\Aut G} > 1$. {{qed|lemma}} === Case $3$: $G$ is Abelian and has no element with order greater than $2$ === By Fundamental Theorem of Finite Abelian Groups, write: :$G = C_{a_1} \times \dots \times C_{a_k}$ where $C_{a_i}$ are cyclic groups of prime power order. Since all elements of $G$ has order not more than $2$, $a_i = 2$ for all $i$, that is: :$G = C_2^k$ As $\order G = 2^k > 2$: :$k > 1$ Writing each element of $G$ as $\tuple {g_1, \dots, g_k}$ where $g_i = 0$ or $1$, the swapping automorphism: :$\tuple {g_1, g_2, \dots, g_k} \mapsto \tuple {g_2, g_1, \dots, g_k}$ is an automorphism that is not the identity automorphism.  Hence $\order {\Aut G} > 1$. {{qed}}
Let $\phi: G \to G$ be a group automorphism. Since $H$ is a characteristic subgroup of $G$, we have: :$\phi \sqbrk H = H$ Thus, from Group Homomorphism Preserves Subgroups, $\phi {\restriction_H}$, the restriction of $\phi$ to $H$, is an automorphism of $H$. Now since $K$ is a characteristic subgroup of $H$, we have that: :$\phi {\restriction_H} \sqbrk K = K$ but this immediately implies that: :$\phi \sqbrk K = K$ by definition of the restriction $\phi {\restriction_H}$. That is, $K$ is a characteristic subgroup of $G$ as well. {{qed}}
=== Monomorphism === The mapping $\alpha$ is the inclusion mapping defined as $i: \Img \phi \to G'$: :$\forall x \in \Img \phi: \map i x = x$ From Inclusion Mapping is Monomorphism, it follows that $\alpha$ is a monomorphism. {{qed|lemma}} === Isomorphism === From the First Isomorphism Theorem for Groups: :$\Img \phi \cong G / \map \ker \phi$ for any homomorphism $\phi$. That is, the image of $\phi$ is isomorphic to the quotient group of $G$ by the kernel of $\phi$. Thus $\beta$ is such an isomorphism. {{qed|lemma}} === Epimorphism === The mapping $\gamma: G \to G / \map \ker \phi$ is given by: :$\forall a \in G: \map \gamma a = a \, \map \ker \phi$ where $a \, \map \ker \phi$ is the left coset of $\map \ker \phi$ by $a$. This is justified by Kernel is Normal Subgroup of Domain. By definition, each (left) coset of $\map \ker \phi$ is the (left) coset of each of its elements. So $\gamma$ is a surjection. $\gamma$ is shown to be a homomorphism thus: {{begin-eqn}} {{eqn | l = \map \gamma {a b}       | r = \paren {a b} \, \map \ker \phi       | c =  }} {{eqn | r = \paren {a \, \map \ker \phi} \paren {b \, \map \ker \phi}       | c = {{Defof|Coset Product}} }} {{eqn | r = \map \gamma a \map \gamma b       | c =  }} {{end-eqn}} thus demonstrating the morphism property. A homomorphism which is surjective is an epimorphism. {{qed|lemma}} Hence the result. {{qed}} This theorem can be illustrated by means of the following commutative diagram: ::$\begin{xy}\xymatrix@L+2mu@+1em{  G \ar@{-->}[r]^*{\phi}      \ar[d]^*{\gamma} &  G' \\   G / \map \ker \phi \ar[r]_*{\beta}^*{\cong} &  \Img \phi \ar[u]^*{\alpha} }\end{xy}$
We have that the inclusion mapping is an injection. Now let $x, y \in T$: {{begin-eqn}} {{eqn | l=\iota \left({x \circ y}\right)       | r=x \circ y       | c=Definition of inclusion mapping }} {{eqn | r=\iota \left({x}\right) \circ \iota \left({y}\right)       | c=Definition of inclusion mapping }} {{end-eqn}} demonstrating that $\iota$ has the morphism property. So $\iota$ is a homomorphism which is also an injection. Thus by definition, $\iota$ is a monomorphism. {{qed}} Category:Monomorphisms Category:Inclusion Mappings 5r56ehrqxt3wyhwk48szr9ve4rktbtf
First note that we have: {{begin-eqn}} {{eqn | l = y       | r = f \left({x}\right)       | c =  }} {{eqn | ll = \implies       | l = x       | r = f^{-1} \left({y}\right)       | c = Image of Element under Inverse Mapping }} {{eqn | ll = \implies       | l = -x       | r = -f^{-1} \left({y}\right)       | c = Multiply both sides by $-1$ }} {{eqn-intertext|Then:}} {{eqn | l = -y        | r = f \left({-x}\right)       | c = {{Defof|Odd Function}} }} {{eqn | ll = \implies       | l = f^{-1} \left({-y}\right)       | r = f^{-1} \circ f \left({-x}\right)       | c = because $f^{-1}$ is a mapping  }} {{eqn | r = -x       | c = Composite of Bijection with Inverse is Identity Mapping }} {{eqn | r = -f^{-1} \left({y}\right)       | c = from above }} {{end-eqn}} The result follows by definition of an odd function.  {{qed}} Category:Inverse Mappings Category:Odd Functions s7y3uva37y1vsh20h9fu1t2cy1s7mfz
Suppose that: :$(1): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_{n - 1} \lhd G_n = G$ and: :$(2): \quad \set e = H_0 \lhd H_1 \lhd \cdots \lhd H_{m - 1} \lhd H_m = G$ are two normal series for $G$. Let a new series be formed: :$(3): \quad \set e = \hat G_0 \subseteq \hat G_1 \subseteq \cdots \subseteq \hat G_{n m - 1} \subseteq \hat G_{n m} = G$ such that: :$\hat G_k = G_q \paren {G_{q + 1} \cap H_r}$ for $k = q m + r$, where $0 \le q < n$ and $0 \le r \le m$. For completeness, define $\hat G_{n m} = G_n$. Note that, by the above construction: :$\hat G_{q m} = G_q \paren {G_{q + 1} \cap H_0} = G_q \paren {\set e} = G_q$ This implies that each group in $(1)$ also appears in $(3)$. It needs to be demonstrated that $\hat G_k$ is well-defined. Consider the ambiguity of $k = q m + r$ in the following cases: :When $m \nmid k$, there is none. :When $m \divides k$, we have: :::$k = q m + m = \paren {q + 1} m$ For $(3)$ to be well-defined, we require: :$G_q \paren {G_{q + 1} \cap H_m} = G_{q + 1} \paren {G_{q + 2} \cap H_0}$ Note that the {{RHS}} is just: :$G_{q + 1} \paren {G_{q + 2} \cap \set e} = G_{q + 1} \paren {\set e} = G_{q + 1}$ The {{LHS}} is: :$G_q \paren {G_{q + 1} \cap H_m} = G_q \paren {G_{q + 1} \cap G} = G_q G_{q + 1} = G_{q + 1}$ Thus the construction is verified. It next needs to be demonstrated that: :$\hat G_k \lhd \hat G_{k + 1}$ Choose the representation of $k$ in which $r < m$. It follows that: :$\hat G_{k + 1} = G_q \paren {G_{q + 1} \cap H_{r + 1} }$ Since $H_{r + 1} \rhd H_r$: :$\hat G_{k + 1} = G_q \paren {G_{q + 1} \cap H_{r + 1} } \rhd G_q \paren {G_{q + 1} \cap H_r} = \hat G_k$ That is, $\hat G_k$ is a normal subgroup of $\hat G_{k + 1}$. $(A): \quad$ But note that $(3)$ may not actually be a normal series as it is possible that $\hat G_k = \hat G_{k + 1}$ for some $k$. Having created series $(3)$, we use the same procedure to form: :$(4): \quad \set e = \hat H_0 \subseteq \hat H_1 \subseteq \cdots \subseteq \hat H_{n m - 1} \subseteq \hat H_{n m} = G$ such that: :$\hat H_k = H_q \paren {H_{q + 1} \cap G_r}$ for $k = q n + r$, where $0 \le q < m$ and $0 \le r \le n$. The same statements that were made about $(3)$ also hold about $(4)$. Now let $k = u m + v$ and $l = v n + u$. It follows that: {{begin-eqn}} {{eqn | l = \frac {\hat G_{k + 1} } {\hat G_k}       | r = \frac {G_u \paren {G_{u + 1} \cap H_{v + 1} } } {G_u \paren {G_{u + 1} \cap H_v} }       | c = from above }} {{eqn | o = \cong       | r = \frac {H_v \paren {H_{v + 1} \cap G_{u + 1} } } {H_v \paren {H_{v + 1} \cap G_u} }       | c = Zassenhaus Lemma }} {{eqn | r = \frac {\hat H_{l + 1} } {\hat H_l}       | c = from above }} {{end-eqn}} So $(3)$ and $(4)$ have isomorphic factors. In $(A)$ it was remarked that some of the factors in these isomorphic series may have redundant elements where $\hat G_k = \hat G_{k + 1}$ and similarly for $\hat H_l = \hat H_{l + 1}$. As the series consist of isomorphic elements, all we now have to do is remove these redundant elements from both (they will occur at the same place). Hence we end up with two refinements of equal length whose factors are isomorphic. {{qed}}
By the Schreier-Zassenhaus Theorem, two normal series have refinements of equal length whose factors are isomorphic. But from the definition of composition series, $\HH_1$ and $\HH_2$ have no proper refinements. Hence any such refinements must be identical to $\HH_1$ and $\HH_2$ themselves. {{qed}} {{Namedfor|Marie Ennemond Camille Jordan|name2 = Otto Ludwig Hölder|cat = Jordan|cat2 = Hölder}}
Let $\sequence {x_n} \subset A^\perp$ be a convergent sequence. Let $x$ be the limit of $\sequence {x_n}$. Then, by the definition of orthocomplement: :$\forall n \in \N, y \in A: \innerprod {x_n} y = 0$ {{LinkWanted|The notation $\innerprod {x_n} y$ is to be specified}} Passing to the limit, from Inner Product is Continuous we have: :$\forall y \in A: \innerprod x y = 0$ So: :$x \in A^\perp$ This shows that $A^\perp$ is closed. {{qed}} Category:Hilbert Spaces 7kd5zfnewigrdywd9swhwm13j4dw7qr
=== Sufficient Condition === Assume that $K$ is everywhere dense and $x \in K^\perp$. Then: {{begin-eqn}} {{eqn | o =        | r = x \in K^\perp }} {{eqn | ll = \leadsto       | lo = \forall y \in K :       | o =        | r = x \perp y }} {{eqn | ll = \leadsto       | lo = \forall y \in H :       | o =        | r = x \perp y       | c = $K$ is everywhere dense, and Inner Product is Continuous }} {{eqn | ll = \leadsto       | lo =        | o =        | r = x = 0 }} {{end-eqn}} {{qed|lemma}} === Necessary Condition === Assume $K^\perp = 0$. Then by Double Orthocomplement is Closed Linear Span, $\vee K=\paren{0}^\perp=H$. Hence $K$ is everywhere dense. {{qed|lemma}} {{qed}}