hcju/proofwikiqa huggingface dataset:
Let $A, B \in \mathscr S$. We have that: {{begin-eqn}} {{eqn | l = T       | o = \subseteq       | r = A       | c = Definition of $\mathscr S$ }} {{eqn | l = T       | o = \subseteq       | r = B       | c = Definition of $\mathscr S$ }} {{eqn | n = 1       | ll= \leadsto       | l = T       | o = \subseteq       | r = A \cap B       | c = Intersection is Largest Subset }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = A       | o = \subseteq       | r = S       | c = {{Defof|Power Set}} }} {{eqn | l = B       | o = \subseteq       | r = S       | c = {{Defof|Power Set}} }} {{eqn | ll= \leadsto       | l = A \cap B       | o = \subseteq       | r = S       | c = Intersection is Subset }} {{eqn | n = 2       | ll= \leadsto       | l = A \cap B       | o = \in       | r = \powerset S       | c = {{Defof|Power Set}} }} {{end-eqn}} Thus we have: {{begin-eqn}} {{eqn | l = T       | o = \subseteq       | r = A \cap B       | c = from $(1)$ }} {{eqn | l = A \cap B       | o = \in       | r = \powerset S       | c = from $(2)$ }} {{eqn | ll= \leadsto       | l = A \cap B       | o = \in       | r = \mathscr S       | c = Definition of $\mathscr S$ }} {{end-eqn}} Hence the result by definition of closed algebraic structure. {{qed}}
For $n = 1$ the result follows trivially: there are no non-empty proper subsets of a singleton. Let $S = \set {s_1, s_2, \ldots, s_n}$. Let $\circ$ be defined on $S$ such that: :$\forall s_a \in S: s_a \circ s_a = \begin{cases} s_{a + 1} & : a < n \\ s_1 & : a = n \end{cases}$ For $a \ne b$ the operation $s_a \circ s_b$ can be arbitrary as long as $s_a \circ s_b \in S$. Let $T$ be such that $\O \subsetneq T \subsetneq S$. Then either: :$\exists s_k \in T: s_{k + 1} \notin T$ or: :$s_n \in T$ but $s_1 \notin T$ otherwise $T = S$. Thus either: :$s_k \circ s_k \notin T$ or: :$s_n \circ s_n \notin T$ In either case, $\circ$ is not closed in $T$. Thus by definition, $\struct {T, \circ}$ is not a submagma of $\struct {S, \circ}$. As $T$ is arbitrary, the result follows. {{qed}}
Let $x \in S$. Then: :$x \circ x^{-1} \in H$ and so $H \ne \O$. Now let $a, b \in H$. Then: :$a = x_a \circ y_a^{-1}$ and: :$b = x_b \circ y_b^{-1}$ for some $x_a, y_a, x_b, y_b \in S$. Thus: {{begin-eqn}} {{eqn | l = a \circ b^{-1}       | r = \paren {x_a \circ y_a^{-1} } \circ \paren {x_b \circ y_b^{-1} }^{-1}       | c =  }} {{eqn | r = \paren {x_a \circ y_a^{-1} } \circ \paren {\paren {y_b^{-1} }^{-1} \circ x_b^{-1} }       | c = Inverse of Group Product }} {{eqn | r = \paren {x_a \circ y_a^{-1} } \circ \paren {y_b \circ x_b^{-1} }       | c = Inverse of Group Inverse }} {{eqn | r = x_a \circ \paren {y_a^{-1} \circ y_b} \circ x_b^{-1}       | c = {{GroupAxiom|1}} }} {{eqn | r = x_a \circ \paren {y_b\circ y_a^{-1} } \circ x_b^{-1}       | c = {{Defof|Abelian Group}} }} {{eqn | r = \paren {x_a \circ y_b} \circ \paren {y_a^{-1} \circ x_b^{-1} }       | c = {{GroupAxiom|1}} }} {{eqn | r = \paren {x_a \circ y_b} \circ \paren {x_b \circ y_a}^{-1}       | c = Inverse of Group Product }} {{end-eqn}} As $\struct {S, \circ}$ is closed, both $x_a \circ y_b \in S$ and $x_b \circ y_a \in S$. Thus $a \circ b$ is in the form $x \circ y^{-1}$ for $x, y \in S$. Thus $a \circ b \in H$ Hence by the One-Step Subgroup Test, $H$ is a subgroup of $G$. {{qed}}
=== Necessary Condition === Let $\CC$ be a local basis. Let $B \in \BB$. Since $\BB$ is a local basis, by the definition of a local basis then $B$ is open. Since $\CC$ is a local basis, by the definition of a local basis then: :$\exists C \in \CC : C\subseteq B$. {{qed|lemma}} === Sufficient Condition === Let $\CC$ satisfy: :$\forall B \in \BB \implies \exists C \in \CC: C\subseteq B$ Let $U \in \tau$ and $x \in U$. By the definition of a local basis then: :$\exists B \in \BB : B\subseteq U$ By the assumption then: :$\exists C \in \CC: C\subseteq B \subseteq U$ By the definition of a local basis then $\CC$ is a local basis. {{qed}} Category:Local Bases 58qxh2muzay6vkkz4kn3r0hmnbx7er5
We have that Non-Zero Real Numbers under Multiplication form Group. We also have that the set of non-zero integers $\Z_{\ne 0}$ form a subset of $\R_{\ne 0}$. From Non-Zero Integers Closed under Multiplication: :$\forall a, b \in \Z_{\ne 0}: a \times b \in \Z_{\ne 0}$ We have that: :$\forall x \in \Z_{\ne 0}: 1 \times x= x = x \times 1$ and so $1$ is the identity of $\struct {\Z_{\ne 0}, \times}$. But for $x \in \Z_{\ne 0}: x \ne 1$, there exists no $y \in \Z_{\ne 0}: x \times y = 1$. Thus $\struct {\Z_{\ne 0}, \times}$ does not have inverses for all $x \in \Z_{\ne 0}$. Thus, by definition, $\struct {\Z_{\ne 0}, \times}$ is not a group. It follows that $\struct {\Z_{\ne 0}, \times}$ is not a subgroup of $\struct {\R_{\ne 0}, \times}$. {{qed}}
As $H$ is a subgroup of $G$, it is a group in its own right. Thus the group axioms all apply to $H$. Let $x, y \in H$. Then by {{GroupAxiom|0}}, $x y \in H$. Let $x, x y \in H$. As $x \in H$, it follows that $x^{-1} \in H$ by {{GroupAxiom|3}}. Thus by {{GroupAxiom|0}}, $x^{-1} \paren {x y} = y \in H$. Let $y, x y \in H$. As $y \in H$, it follows that $y^{-1} \in H$ by {{GroupAxiom|3}}. Thus by {{GroupAxiom|0}}, $\paren {x y} y^{-1} = x \in H$. {{qed}}
{{AimForCont}} neither $H \subseteq K$ nor $K \subseteq H$. Then from Union of Subgroups it follows that $H \cup K$ is not a subgroup of $G$. The result follows by Proof by Contradiction. {{qed}}
Let $H = \gen S$ denote the subgroup generated by $S$. Let $a, b \in H$. Then: :$a = s_1$ :$b = s_2$ for some words $s_1, s_2$ of the set of words $\map W S$ of $S$. Then: {{begin-eqn}} {{eqn | l = a \circ b       | r = s_1 \circ s_2       | c =  }} {{eqn | r = s_2 \circ s_1       | c = as all elements of $S$ commute with each other }} {{eqn | r = b \circ a       | c =  }} {{end-eqn}} Hence the result, by definition of abelian group. {{qed}}
=== $(1)$ is equivalent to $(2)$ === Let $H$ be the smallest subgroup containing $S$. Let $\mathbb S$ be the set of subgroups containing $S$. To show the equivalence of the two definitions, we need to show that $H = \bigcap \mathbb S$. Since $H$ is a subgroup containing $S$: :$H \in \mathbb S$ By Intersection is Subset: :$\bigcap \mathbb S \subseteq H$ On the other hand, by Intersection of Subgroups is Subgroup: :$\bigcap \mathbb S$ is a subgroup containing $S$. Since $H$ be the smallest subgroup containing $S$: :$H \subseteq \bigcap \mathbb S$ By definition of set equality: :$H = \bigcap \mathbb S$ Hence the result. {{qed|lemma}} === $(1)$ is equivalent to $(3)$ === This is shown in Set of Words Generates Group. {{qed}}
The group action axioms are investigated in turn. Let $\pi, \rho \in S_n$. Let $S \in B_r$. Thus: {{begin-eqn}} {{eqn | l = \pi * \paren {\rho * S}       | r = \pi * \rho \sqbrk S       | c = Definition of $*$ }} {{eqn | r = \pi \sqbrk {\rho \sqbrk S}       | c = Definition of $*$ }} {{eqn | r = \paren {\pi \circ \rho} \sqbrk S       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \paren {\pi \circ \rho} * n       | c = Definition of $*$ }} {{end-eqn}} demonstrating that {{GroupActionAxiom|1}} holds. Then: {{begin-eqn}} {{eqn | l = I_{\N_n} * S       | r = I_{\N_n} \sqbrk S       | c = where $I_{\N_n}$ is the identity mapping on $\N_n$ }} {{eqn | r = S       | c = {{Defof|Identity Mapping}} }} {{end-eqn}} demonstrating that {{GroupActionAxiom|2}} holds. {{qed}}
From Group Action of Symmetric Group on Subset it is established that $*$ is a group action. Let $U = \set {u_1, u_2, \ldots, u_r}$ and $V = \set {v_1, v_2, \ldots, v_r}$ be elements of $B_r$. Then there exists a permutation $\rho \in S_n$ such that: :$\map \rho {u_k} = v_k$ for all $k \in \set {1, 2, \ldots, r}$. Thus: :$\rho \sqbrk U = V$ Thus $B_r$ is the orbit of all $U \in B_r$. Hence the result by definition of transitive group action. {{qed}}
Proof by Counterexample: Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as: {{:Symmetric Group on 3 Letters/Cayley Table}} Consider the subgroups of $S_3$: {{:Symmetric Group on 3 Letters/Subgroups}} Let $H = \set {e, \tuple {12} }$. From Normal Subgroups in Symmetric Group on 3 Letters, $H$ is not normal. Let $A = \set {\tuple {123}, \tuple {23} }$. We have: {{begin-eqn}} {{eqn | l = \tuple {123} H        | r = \set {\tuple {123}, \tuple {23} }       | c =  }} {{eqn | l = \tuple {23} H        | r = \set {\tuple {123}, \tuple {23} }       | c =  }} {{end-eqn}} and so $A$ is the left coset of $H$ by both $\tuple {123}$ and $\tuple {23}$. Let $B = \set {\tuple {132}, \tuple {13} }$. {{begin-eqn}} {{eqn | l = \tuple {132} H        | r = \set {\tuple {132}, \tuple {13} }       | c =  }} {{eqn | l = \tuple {13} H        | r = \set {\tuple {132}, \tuple {13} }       | c =  }} {{end-eqn}} and so $B$ is the left coset of $H$ by both $\tuple {132}$ and $\tuple {13}$. Now consider: {{begin-eqn}} {{eqn | l = A \circ B       | r = \tuple {123} H \tuple {132} H       | c =  }} {{eqn | r = \tuple {123} \tuple {132} H       | c =  }} {{eqn | r = e H       | c =  }} {{eqn | r = H       | c =  }} {{end-eqn}} But: {{begin-eqn}} {{eqn | l = A \circ B       | r = \tuple {23} H \tuple {13} H       | c =  }} {{eqn | r = \tuple {23} \tuple {13} H       | c =  }} {{eqn | r = \tuple {132} H       | c =  }} {{eqn | r = B       | c =  }} {{eqn | o = \ne       | r = H       | c =  }} {{end-eqn}} So two different evaluations of the coset product give two different result. Hence by definition the coset product is not well-defined on $H$. The result follows. {{qed}}
{{ProofWanted|Straightforward but tedious, unless someone has a short cut better than testing all the products}} :$\index {A_4} V = 3$ follows from Lagrange's Theorem.
Let $g \in G$. We show that $g H$ contains exactly $1$ element of $T$. Since $S$ is a right transversal: :$\exists ! x \in S: x \in H g^{-1}$ By Right Cosets are Equal iff Element in Other Right Coset: :$H x = H g^{-1}$ By Right Cosets are Equal iff Left Cosets by Inverse are Equal: :$x^{-1} H = g H$ We have from definition $x^{-1} \in T$. The result follows from uniqueness of $x$. {{qed}}
From {{Defof|Right Transversal}}, $S$ is a right transversal for $H$ in $G$ {{iff}} every right coset of $H$ contains exactly one element of $S$. Since there are $n$ right cosets of $H$  and $S$ has cardinality $n$, if $S$ is not a right transversal for $H$ in $G$, at least one right coset of $H$ contain more than one element of $S$. Thus the contrapositive of the theorem is as follows: :At least one right coset of $H$ contain more than one element of $S$ {{iff}}: ::$\exists x, y \in S: x \ne y \land x y^{-1} \in H$ This is a consequence of Elements in Same Right Coset iff Product with Inverse in Subgroup. {{qed}}
It is established in Action of Group on Coset Space is Group Action that $*$ is a group action. It remains to be shown that: :$\forall g' H \in G / H: \Orb {g' H} = G / H$ where $\Orb {g' H}$ denotes the orbit of $g' H \in G / H$ under $*$. Let $a H, b H \in G / H$ such that $a H \ne b H$. We have that: :$\exists x \in G: x a = b$ and so: :$x * a H = \paren {x a} H = b H$ and so: :$b H \in \Orb {a H}$ As both $a$ and $b$ are arbitrary, the result follows. {{qed}}
It is established in Action of Group on Coset Space is Group Action that $*$ is a group action. Then: {{begin-eqn}} {{eqn | l = \Stab {a H}       | r = \set {g \in G: g * a H = a H}       | c = {{Defof|Stabilizer}} }} {{eqn | r = \set {g \in G: \paren {g a} H = a H}       | c =  }} {{eqn | r = \set {g \in G: g H = a H a^{-1} }       | c =  }} {{eqn | r = a H a^{-1}       | c =  }} {{end-eqn}} {{qed}}
Proof by Counterexample: Let $D_4$ denote the dihedral group $D_4$. Let $D_4$ be presented in matrix representation: {{:Dihedral Group D4/Matrix Representation/Formulation 1}} Its Cayley table is given as: {{:Dihedral Group D4/Matrix Representation/Formulation 1/Cayley Table}} Consider the subgroup $H$ whose underlying set is: :$H = \set {\mathbf I, \mathbf A, -\mathbf I, -\mathbf A}$ From Subgroup of Index 2 is Normal, $H$ is normal in $D_4$. Consider the subgroup $H$ whose underlying set is: :$K = \set {\mathbf I, \mathbf A} = \gen {\mathbf A}$ From Subgroup of Index 2 is Normal, $K$ is normal in $H$. It remains to be demonstrated that $K$ is not normal in $D_4$. From the Cayley table: :$\mathbf C \mathbf A \mathbf C^{-1} = \mathbf B \mathbf C = -\mathbf A \notin K$ Hence $K$ is not normal in $D_4$. {{qed}}
=== Necessary Condition === Let $\Stab x$ be normal in $G$. {{AimForCont}} $y \in \Orb x$ such that $\Stab x \ne \Stab y$. Then: {{ProofWanted}} === Sufficient Condition === Let $\Stab x$ be the stabilizer of every element in $\Orb x$. {{ProofWanted}}
From Quotient Group is Group, the operation: :$\forall a, b \in G: \paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$ is the group operation in the quotient group $\struct {G / N, \circ}$. The result follows directly by definition of power of group element. {{qed}}
Let $\paren {a N}^n$ be the identity of $G / N$. Then: {{begin-eqn}} {{eqn | l = \paren {a N}^n       | r = N       | c = Quotient Group is Group: {{GroupAxiom|2}} }} {{eqn | ll= \leadstoandfrom       | l = \paren {a^n} N       | r = N       | c = Power of Coset Product is Coset of Power }} {{eqn | ll= \leadstoandfrom       | l = a^n       | o = \in       | r = N       | c = Coset Equals Subgroup iff Element in Subgroup }} {{end-eqn}} {{qed}}
From Additive Group of Integers is Subgroup of Reals, $\struct {\Z, +}$ is a normal subgroup of $\struct {\R, +}$. As the additive group of real numbers is abelian, from Subgroup of Abelian Group is Normal it follows that $\struct {\Z, +} \lhd \struct {\R, +}$. {{qed}}
From Additive Group of Integers is Normal Subgroup of Reals, we have that $\struct {\Z, +}$ is a normal subgroup of $\struct {\R, +}$. Hence $\R / \Z$ is indeed a quotient group. By definition of rational number, what is to be proved is: :$x + \Z$ is of finite order {{iff}}: :$x = \dfrac m n$ for some $m \in \Z, n \in \Z_{> 0}$. Let $x + \Z$ be of finite order in $\R / \Z$. Then: {{begin-eqn}} {{eqn | lo= \exists n \in \Z_{\ge 0}:       | l = \paren {x + \Z}^n       | r = \Z       | c = {{Defof|Quotient Group}}: {{GroupAxiom|2}} }} {{eqn | ll= \leadstoandfrom       | l = n x       | o = \in       | r = \Z       | c = Condition for Power of Element of Quotient Group to be Identity }} {{eqn | ll= \leadstoandfrom       | l = n x       | r = m       | c = for some $m \in \Z$ }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \dfrac m n       | c =  }} {{end-eqn}} {{qed}}
From Additive Group of Integers is Subgroup of Rationals, $\struct {\Z, +}$ is a subgroup of $\struct {\Q, +}$. From Rational Numbers under Addition form Abelian Group, $\struct {\Q, +}$ is an abelian group. From Subgroup of Abelian Group is Normal it follows that $\struct {\Z, +}$ is a normal subgroup of $\struct {\Q, +}$. {{qed}} Category:Additive Group of Integers Category:Additive Group of Rational Numbers Category:Examples of Normal Subgroups 7jc6hb49k1u71jvloy78brhxaj4hu5e
Let $k, l \in \Z$. {{begin-eqn}} {{eqn | l = \map {\phi_g} {k + l}       | r = a^{k + l}       | c =  }} {{eqn | r = a^k a^l       | c =  }} {{eqn | r = \map {\phi_g} k \circ \map {\phi_g} l       | c =  }} {{end-eqn}} thus proving that $\phi_g$ is a homomorphism as required. {{qed}}
An inner automorphism is a permutation on $G$ by definition. From Inner Automorphisms form Subgroup of Automorphism Group: :$\Inn G \le \Aut G$ where $\Aut G$ denotes the set of automorphisms of $G$. From Automorphism Group is Subgroup of Symmetric Group: :$\Aut G \le \struct {\map \Gamma G, \circ}$ Thus $\Inn G \le \struct {\map \Gamma G, \circ}$ as required. {{qed}}
By definition, an isomorphism is a bijection. By definition, the order of a group is the cardinality of its underlying set. The result follows by definition of set equivalence. {{qed}}
Let $x, y \in \C$ such that $x = x_1 + 0 i, y = y_1 + 0 i$. As $x$ and $y$ are wholly real, we have that $x, y \in \R$. Then $x + y = \paren {x_1 + y_1} + \paren {0 + 0} i$ which is also wholly real. Also, the inverse of $x$ is $-x = -x_1 + 0 i$ which is also wholly real. Thus by the Two-Step Subgroup Test, $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$. {{Qed}}
Note that both $C_6$ and $S_3$ are of order $6$. From Cyclic Group is Abelian, $C_6$ is abelian. From Symmetric Group is not Abelian, $S_6$ is not abelian. From Isomorphism of Abelian Groups, if two groups are isomorphic, they are either both abelian or both not abelian. Hence $C_6$ and $S_3$ are not isomorphic. {{qed}}
Let $S_n$ denote the symmetric group on $n$ letters. From Alternating Group is Normal Subgroup of Symmetric Group: :$\index {S_n} {A_n} = 2$ where $\index {S_n} {A_n}$ denotes the index of $A_n$ in $S_n$. From Order of Symmetric Group: :$\order {S_n} = n!$ The result follows from Lagrange's Theorem. {{qed}}
=== Necessary Condition === Let $\struct {G, \circ}$ be an abelian group. Let $a, b \in G$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = \map \phi {a \circ b}       | r = \paren {a \circ b}^n       | c = Definition of $\phi$ }} {{eqn | r = a^n \circ b^n       | c = Power of Product of Commutative Elements in Group }} {{eqn | r = \map \phi a \circ \map \phi b       | c = Definition of $\phi$ }} {{end-eqn}} As $a$ and $b$ are arbitrary, the above holds for all $a, b \in G$. Thus $\phi$ is a group homomorphism from $G$ to $G$. So by definition, $\phi$ is a group endomorphism. {{qed|lemma}} === Sufficient Condition === Let $\phi: G \to G$ as defined above be a group endomorphism. Then: {{begin-eqn}} {{eqn | lo= \forall a, b \in G:       | l = \map \phi {a \circ b}       | r = \map \phi a \circ \map \phi b       | c = {{Defof|Group Endomorphism}} }} {{eqn | ll= \leadsto       | lo= \forall a, b \in G:       | l = \paren {a \circ b}^n       | r = a^n \circ b^n       | c = Definition of $\phi$ }} {{end-eqn}} From Power of Product of Commutative Elements in Group it follows that $G$ is an abelian group. {{qed}}
We have that: :Infinite Cyclic Group is Isomorphic to Integers. :Integer Multiples under Addition form Infinite Cyclic Group. :Infinite Cyclic Group is Unique up to Isomorphism Hence the result. {{qed}}
Let $G / N$ be the quotient group of $G$ by $N$. Let $q_N: G \to G / N$ be the quotient epimorphism from $G$ to $G / N$: :$\forall x \in G: \map {q_N} x = x N$ Then from Quotient Group Epimorphism is Epimorphism, $N$ is the kernel of $q_n$ Thus $q_N$ is that group homomorphism of which $N$ is the kernel. {{qed}}
Let $y \in \Img G$. Then $\exists x \in G: y = \map \phi x$. As $G$ be a cyclic group with generator $g$, $x = g^n$ for some $n \in \Z$. Thus by Homomorphism of Power of Group Element: :$y = \paren {\map \phi g}^n$ and so is a power of $\map \phi g$. As $y$ is arbitrary, it follows that all elements of $\Img G$ are powers of $\map \phi g$. Hence the result, by definition of cyclic group. {{qed}}
{{begin-eqn}} {{eqn | l = x^n       | o = \in       | r = \map \ker \phi       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \map \phi {x^n}       | r = e_H       | c = {{Defof|Kernel of Group Homomorphism}} }} {{eqn | ll= \leadstoandfrom       | l = \paren {\map \phi x}^n       | r = e_H       | c = Homomorphism of Power of Group Element }} {{end-eqn}} {{qed}}
From Kernel of Group Homomorphism is Subgroup and Subgroup of Cyclic Group is Cyclic: :$\exists m \in \N: \map \ker \phi = \gen {g^m}$ From Homomorphic Image of Cyclic Group is Cyclic Group: :$\Img \phi$ is a cyclic group generated by $\map \phi g$. === Case $1$: $\Img \phi$ is infinite === {{AimForCont}} $m \ne 0$. Then $g^m$ is not the identity. Thus: {{begin-eqn}} {{eqn | l = \map \phi {g^m}       | r = \paren {\map \phi g}^m       | c = Homomorphism of Power of Group Element }} {{eqn | o = \ne       | r = e_H       | c = {{Defof|Infinite Cyclic Group}} }} {{end-eqn}} However this contradicts $g^m \in \gen {g^m} = \map \ker \phi$. Hence we must have $m = 0$. {{qed|lemma}} === Case $2$: $\Img \phi$ is finite === From Order of Cyclic Group equals Order of Generator, the order of $\map \phi g$ is $\order {\Img \phi}$. Let $n \in \Z$. By Division Theorem: :$\exists q, r \in \Z: 0 \le r < \order {\Img \phi}: n = q \order {\Img \phi} + r$ We have: {{begin-eqn}} {{eqn | l = \map \phi {g^n}       | r = \paren {\map \phi g}^n       | c = Homomorphism of Power of Group Element }} {{eqn | r = \paren {\map \phi g}^{q \order {\Img \phi} + r} }} {{eqn | r = \paren {\paren {\map \phi g}^{\order {\Img \phi} } }^q \paren {\map \phi g}^r }} {{eqn | r = {e_H}^q \paren {\map \phi g}^r       | c = as $\paren {\map \phi g}^{\order {\Img \phi} } = e_H$ }} {{eqn | r = \paren {\map \phi g}^r       | c = Power of Identity is Identity }} {{end-eqn}} From definition of order of group element: :$0 < r < \order {\Img \phi} \implies \paren {\map \phi g}^r \ne e_H$ Hence: {{begin-eqn}} {{eqn | l = \map \phi {g^n} = e_H       | o = \iff       | r = r = 0 }} {{eqn | ll = \leadstoandfrom       | l = g^n \in \ker \phi       | o = \iff       | r = n = q \order {\Img \phi}       | c = {{Defof|Kernel of Group Homomorphism}} }} {{eqn | ll = \leadstoandfrom       | l = g^n \in \gen {g^m}       | o = \iff       | r = g^n \in \gen {g^\order {\Img \phi} }       | c = Group Generated by Singleton }} {{eqn | ll = \leadstoandfrom       | l = \gen {g^m}       | r = \gen {g^\order {\Img \phi} }       | c = {{Defof|Set Equality}} }} {{end-eqn}} and we see that $m = \order {\Img \phi}$ satisfies the above relation. {{qed}}
Let $x, y \in G$. By definition of automorphism group, we have that: :$\map \kappa x \map \kappa y = \kappa_x \circ \kappa_y$ where $\circ$ denotes composition of mappings. Then $\forall g \in G$: {{begin-eqn}} {{eqn | l = \map {\kappa_x \circ \kappa_y} g       | r = \map {\kappa_x} {\map {\kappa_y} g}       | c = for all $g \in G$ }} {{eqn | r = \map {\kappa_x} {y g y^{-1} }       | c = Definition of $\kappa_y$ }} {{eqn | r = x \paren {y g y^{-1} } x^{-1}       | c = Definition of $\kappa_x$ }} {{eqn | r = \paren {x y} g \paren {y^{-1} x^{-1} }       | c = {{GroupAxiom|1}} }} {{eqn | r = \paren {x y} g \paren {x y}^{-1}       | c = Inverse of Group Product }} {{eqn | r = \map {\kappa_{x y} } g       | c = Definition of $\kappa_{x y}$ }} {{end-eqn}} And so: :$\forall g \in G: \map {\kappa_{x y} } g = \map {\kappa_x \circ \kappa_y} g$ Thus by definition of $\kappa$: :$\map \kappa x \map \kappa y = \map \kappa{x y}$ demonstrating that $\kappa$ is a homomorphism. {{qed}}
Let $\Inn G$ denote the inner automorphism group of $G$. For all $x \in G$, $\map \kappa x = \kappa_x \in \Inn G$. Hence $\Img \kappa \subseteq \Inn G$.  Let $\phi \in \Inn G$. Then: :$\exists y \in G: \forall g \in G: \map \phi g = y g y^{-1}$ Then $\map \kappa y = \phi$. Hence $\Inn G \subseteq \Img \kappa$.  Therefore $\Img \kappa = \Inn G$. {{qed}}
By definition of monomorphism, $\phi$ is a homomorphism which is also an injection. From Order of Homomorphic Image of Group Element: :$\forall g \in G: \order {\map \phi g} \divides \order g$ {{begin-eqn}} {{eqn | l = \map \phi {g^m}       | r = \paren {\map \phi g}^m       | c = Homomorphism of Power of Group Element }} {{eqn | r = e_H       | c =  }} {{eqn | r = \map \phi {e_G}       | c = Homomorphism to Group Preserves Identity }} {{end-eqn}} So $g^m = e$, as $\phi$ is injective. From the definition of order of group element, that means $n \le m$ since $n$ is the ''smallest'' such power. Thus $m = n$ and the result holds. {{Qed}}
Let $y \in \theta \sqbrk {\map Z G}$. Let $t \in H$. We have that: :$y = \map \theta z$ for some $z \in \map Z G$ As $\theta$ is an epimorphism, it is by definition surjective. Then: :$t = \map \theta s$ for some $s \in G$. Hence: {{begin-eqn}} {{eqn | l = y t       | r = \map \theta z \map \theta s       | c =  }} {{eqn | r = \map \theta {z s}       | c = {{Defof|Group Epimorphism}} }} {{eqn | r = \map \theta {s z}       | c = {{Defof|Center of Group}} }} {{eqn | r = \map \theta s \map \theta z       | c = {{Defof|Group Epimorphism}} }} {{eqn | r = t y       | c = {{Defof|Group Epimorphism}} }} {{end-eqn}} As $t$ is arbitrary, it follows that $y \in \map Z H$. The result follows. {{qed}}