nyuuzyou/wikis huggingface dataset:
Group Epimorphism Induces Bijection between Subgroups

== Theorem ==
Let $G_1$ and $G_2$ be groups whose identities are $e_{G_1}$ and $e_{G_2}$ respectively.
Let $\phi: G_1 \to G_2$ be a group epimorphism.
Let $K := \map \ker \phi$ be the kernel of $\phi$.
Let $\mathbb H_1 = \set {H \subseteq G_1: H \le G_1, K \subseteq H}$ be the set of subgroups of $G_1$ which contain $K$.
Let $\mathbb H_2 = \set {H \subseteq G_2: H \le G_2}$ be the set of subgroups of $G_2$.
Then there exists a bijection $Q: \mathbb H_1 \leftrightarrow \mathbb H_2$ such that:
:$\forall N \lhd G_1: \map Q N \lhd G_2$
:$\forall N \lhd G_2: \map {Q^{-1} } N \lhd G_1$
where $N \lhd G_1$ denotes that $N$ is a normal subgroup of $G_1$.
That is, normal subgroups map bijectively to normal subgroups under $Q$.
=== Corollary ===
== Proof ==
Let $Q$ be the mapping defined as:
:$\forall H \le \mathbb H_1: \map Q H = \set {\map \phi h: h \in H}$
Let $H$ be a subgroup of $G_1$ such that $K \subseteq H$.
From Group Homomorphism Preserves Subgroups, $\phi \sqbrk H$ is a subgroup of $G_2$.
This establishes that $Q$ is actually a mapping.
Let $N \lhd G_1$.
From Group Epimorphism Preserves Normal Subgroups, $\phi \sqbrk N$ is a normal subgroup of $G_2$.
This establishes that:
:$\forall N \lhd G_1: \map Q N \lhd G_2$
Next it is shown that $Q$ is a bijection.
=== Injective Nature of $Q$ ===
Let $H, J \in \mathbb H_1$.
Let $\map Q H = \map Q J$.
Let $h \in H$.
A similar argument shows that $J \subseteq H$.
So by definition of set equality:
:$H = J$
Thus:
:$\map Q H = \map Q J \implies H = J$
So by definition, $Q$ is injective.
=== Surjective Nature of $Q$ ===
Now let $N' \in \mathbb H_2$.
By definition of $\mathbb H_2$, $N'$ is a subgroup of $G_2$.
Let $N = \set {x: \map \phi x = N'}$.
We have from Identity of Subgroup that $e_{G_2} \in N'$.
Thus by definition of kernel, $K \subseteq N$.
Now suppose $\map \phi x, \map \phi y \in N'$.
Then:
So by the One-Step Subgroup Test, $N$ is a subgroup of $G_1$.
It has been established that $K \subseteq N$, and so $N \in \mathbb H_1$.
Thus it follows that for all $N' \in \mathbb H_2$ there exists $N \in H_1$ such that $\map Q N = N'$.
So $Q$ is a surjection.
So $Q$ has been shown to be both an injection and a surjection, and so by definition is a bijection.
Finally, it can then be shown that if $N'$ is normal in $G_2$, it follows that $N = \map {Q^{-1} } {N'}$ is normal in $G_1$.
This establishes that: 
:$\forall N \lhd G_2: \map {Q^{-1} } N \lhd G_1$
== Also see ==
* First Isomorphism Theorem
== Sources ==
* : $\S 1.10$: Theorem $29$
Category:Group Epimorphisms
Group Epimorphism Induces Bijection between Subgroups/Corollary

== Theorem ==
Let $G_1$ and $G_2$ be groups whose identities are $e_{G_1}$ and $e_{G_2}$ respectively.
Let $\phi: G_1 \to G_2$ be a group epimorphism.
Let $K := \map \ker \phi$ be the kernel of $\phi$.
Let $H \le G$ denote that $H$ is a subgroup of $G$.
Then:
:$\forall H \le G, K \subseteq H: \phi \sqbrk H \cong H / K$
where $H / K$ denotes the quotient group of $H$ by $K$.
== Proof ==
Let $H$ be a subgroup of $G$ such that $K \subseteq H$.
Consider the restriction of $\phi$ to $H$.
By Group Homomorphism Preserves Subgroups, $\phi_{\restriction_H} \sqbrk H$ is a group.
From Group Epimorphism Induces Bijection between Subgroups it follows that the First Isomorphism Theorem can be applied.
Hence the result.
== Sources ==
* : $\S 1.10$: Theorem $29$: Corollary
Category:Group Epimorphisms
Group Epimorphism Preserves Generator

== Theorem ==
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: G \to H$ be a group epimorphism.
Let $A$ be a generator for $\struct {G, \circ}$.
Then $\phi \sqbrk A$ is a generator for $\struct {H, *}$.
== Proof ==
By definition of generator:
:$A$ is the intersection of all subgroups of $G$ containing $A$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.15$
Category:Generated Subgroups
Category:Group Epimorphisms
Group Epimorphism Preserves Normal Subgroups

== Theorem ==
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: G \to H$ be a group epimorphism.
Let $N \lhd G$, where $\lhd$ denotes that $N$ is a normal subgroup of $G$.
Then $\phi \sqbrk N \lhd H$.
That is, the image under $\phi$ of a normal subgroup is itself normal.
== Proof ==
Let $N' := \phi \sqbrk N$.
From Group Homomorphism Preserves Subgroups, $N'$ is a subgroup of $H$.
It remains to show that $N'$ is normal in $H$.
Let $h \in H$ be arbitrary.
Let $n' \in N'$ be arbitrary.
Because $\phi$ is an epimorphism, it is by definition surjective.
Therefore:
:$\exists n \in N: \map \phi n = n'$
:$\exists g \in G: \map \phi g = h$
Then:
As $h$ and $n'$ were arbitrary, it follows that:
:$\forall h \in H, n' \in N': h * n' * h^{-1} \in N'$
and so $N'$ is a normal subgroup of $H$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.20 \ \text {(a)}$
* : $\S 1.10$: Theorem $29$
Category:Group Epimorphisms
Category:Normal Subgroups
Group Epimorphism Preserves Subgroups

== Theorem ==
Let $\struct {G_1, \circ}$ and $\struct {G_2, *}$ be groups.
Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ be a group epimorphism.
Then:
:$H \le G_1 \implies \phi \sqbrk H \le G_2$
where:
:$\phi \sqbrk H$ denotes the image of $H$ under $\phi$
:$\le$ denotes subgroup.
That is, group epimorphism preserves subgroups.
== Proof ==
By definition, $\phi$ is a group homomorphism.
The result then follows from Group Homomorphism Preserves Subgroups.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.20 \ \text {(a)}$
Category:Group Epimorphisms
Category:Subgroups
Group Epimorphism is Isomorphism iff Kernel is Trivial

== Theorem ==
Let $\struct {G, \oplus}$ and $\struct {H, \odot}$ be groups.
Let $\phi: \struct {G, \oplus} \to \struct {H, \odot}$ be a group epimorphism.
Let $e_G$ and $e_H$ be the identities of $G$ and $H$ respectively.
Let $K = \map \ker \phi$ be the kernel of $\phi$.
Then:
:the epimorphism $\phi$ is an isomorphism
:$K = \set {e_G}$
== Proof 1 ==
== Proof 2 ==
Category:Quotient Groups
Category:Group Epimorphisms
Category:Group Isomorphisms
Category:Group Epimorphism is Isomorphism iff Kernel is Trivial
Group Epimorphism is Isomorphism iff Kernel is Trivial/Proof 1

== Theorem ==
== Proof ==
=== Necessary Condition ===
Let $\phi$ be an isomorphism.
Then by definition $\phi$ is a bijective homomorphism.
Thus by definition of bijection, $\phi$ is an injection.
By definition of injection, there exists exactly one element $x$ of $G$ such that $\map \phi x = e_H$.
From Epimorphism Preserves Identity, that element $x$ is $e_G$:
:$\map \phi {e_G} = e_H$
Thus by definition of kernel:
:$\map \ker \phi = \set {e_G}$
=== Sufficient Condition ===
Let $K := \map \ker \phi = \set {e_G}$.
From the Quotient Theorem for Epimorphisms:
:$\RR_\phi$ is compatible with $\oplus$
and thus from Kernel is Normal Subgroup of Domain:
:$K \lhd G$
From Congruence Relation induces Normal Subgroup, $\RR_\phi$ is the equivalence defined by $K$.
Let $\RR_K$ be the congruence modulo $K$ induced by $K$.
Suppose $\map \phi x = \map \phi y$.
Then:
:$x \mathop {\RR_K} y$
as $\RR_\phi = \RR_K$ from Congruence Modulo Subgroup is Equivalence Relation.
Thus by Congruence Class Modulo Subgroup is Coset:
:$x \oplus y^{-1} \in K$
Hence:
:$x \oplus y^{-1} = e_G$
and so:
:$x = y$
Thus $\phi$ is injective.
By definition, an injective epimorphism is a isomorphism.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.6$
Category:Group Epimorphism is Isomorphism iff Kernel is Trivial
Group Epimorphism is Isomorphism iff Kernel is Trivial/Proof 2

== Theorem ==
== Proof ==
From Kernel is Trivial iff Group Monomorphism, $\phi$ is a monomorphism $K = \set {e_G}$.
By definition, a group $G$ is an epimorphism is an isomorphism $G$ is also a monomorphism.
Hence the result.
Category:Group Epimorphism is Isomorphism iff Kernel is Trivial
Group Epimorphism preserves Central Subgroups

== Theorem ==
Let $G$ and $H$ be groups.
Let $\theta: G \to H$ be an epimorphism.
Let $Z \le G$ be a central subgroup of $G$.
Then $\theta \sqbrk Z$ is a central subgroup of $H$.
== Proof ==
By definition of central subgroup:
:$Z \subseteq \map Z G$
where $\map Z G$ denotes the center of $G$.
From Image under Epimorphism of Center is Subset of Center:
:$\theta \sqbrk {\map Z G} \subseteq \map Z H$
From Image of Subset under Mapping is Subset of Image it follows that:
:$\theta \sqbrk Z \subseteq \map Z H$
The result follows.
Category:Central Subgroups
Category:Group Epimorphisms
Group Example: x inv c y

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $c \in G$.
We define a new operation $*$ on $G$ as:
:$\forall x, y \in G: x * y = x \circ c^{-1} \circ y$
Then $\struct {G, *}$ is a group.
== Proof ==
=== ===
Let $x, y \in G$.
Then:
:$\forall x * y = x \circ c^{-1} \circ y \in G$ as $c^{-1} \in G$
thus demonstrating that $\struct {G, *}$ is closed.
=== ===
Let $x, y, z \in G$.
thus demonstrating that $\struct {G, *}$ is associative.
=== ===
Let $x \in G$.
So $c$ serves as the identity in $\struct {G, *}$.
=== ===
Let $x \in G$.
We need to find $y \in G$ such that $x * y = c \implies x \circ c^{-1} \circ y = c$.
Thus the inverse of $x$ under the operation $*$ is $c \circ x^{-1} \circ c$ where $x^{-1}$ is the inverse of $x$ under $\circ$.
All of the group axioms have been demonstrated to be fulfilled, and so $\struct {G, *}$ is a group.
== Also see ==
* Definition:Heap
== Sources ==
* : Chapter $3$: Elementary consequences of the definitions: Exercise $3$
Category:Examples of Groups
Group Generated by Normal Intersection is Normal

== Theorem ==
Let $I$ be an indexing set, and $\set {N_i: i \in I}$ be a set of normal subgroups of the group $G$.
Then $\family {N_i: i \in I}$ is a normal subgroup of $G$.
== Proof ==
By definition, $\family {N_i: i \in I}$ is the intersection of all the subgroups of $G$ which contain every $N_i$.
For each $H \le G$, the conjugate $g H g^{-1}$ contains each $g N_i g^{-1}$.
Since each $N_i \lhd G$, it follows that:
:$N_i \subseteq g H g^{-1}$
Thus it follows that:
:$\family {N_i: i \in I} \lhd G$
Category:Normal Subgroups
Group Generated by Reciprocal of z and 1 minus z

== Theorem ==
Let $\struct {S, \circ}$ denote the group generated by $\dfrac 1 z$ and $1 - z$.
Then $\struct {S, \circ}$ is a finite group of order $6$.
== Proof ==
By definition:
:$S = \set {f_1, f_2, f_3, f_4, f_5, f_6}$
where $f_1, f_2, \ldots, f_6$ are complex functions defined for all $z \in \C \setminus \set {0, 1}$ as:
Also by definition, $\circ$ denotes composition of functions.
First it is necessary to establish the Cayley table for $\struct {S, \circ}$.
First note that it is apparent by definition of composition of functions that:
:$\forall n: f_1 \circ f_n = f_n = f_n \circ f_1$
It remains to establish the rest of the compositions
Hence the Cayley table is established:
Taking the group axioms in turn:
=== ===
By inspection it can be seen that $\struct {S, \circ}$ is closed.
=== ===
From Composition of Mappings is Associative, it follows that $\struct {S, \circ}$ is associative.
=== ===
We have that:
:$\forall n: f_1 \circ f_n = f_n = f_n \circ f_1$
Thus $\struct {S, \circ}$ has an identity element.
=== ===
From the above analysis:
Thus every element of $\struct {S, \circ}$ has an inverse.
All the group axioms are thus seen to be fulfilled, and so $\struct {S, \circ}$ is a group.
$\struct {S, \circ}$ has $6$ elements and so is a finite group of order $6$.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(ii)}$
* : Introduction
Category:Groups of Order 6
Category:Group Generated by Reciprocal of z and 1 minus z
Group Generated by Reciprocal of z and 1 minus z/Cayley Table

== Cayley Table for Group Generated by Reciprocal of $1 / z$ and $1 - z$ ==
We have:
Hence from Group Generated by Reciprocal of z and 1 minus z:
:$\begin{array}{r|rrrrrr}
\circ & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 \\
\hline
f_1 & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 \\
f_2 & f_2 & f_3 & f_1 & f_6 & f_4 & f_5 \\
f_3 & f_3 & f_1 & f_2 & f_5 & f_6 & f_4 \\
f_4 & f_4 & f_5 & f_6 & f_1 & f_2 & f_3 \\
f_5 & f_5 & f_6 & f_4 & f_3 & f_1 & f_2 \\
f_6 & f_6 & f_4 & f_5 & f_2 & f_3 & f_1 \\
\end{array}$
Expressing the elements in full:
:$\begin{array}{c|cccccc}
\circ & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} \\
\hline
z & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} \\
\dfrac 1 {1 - z} & \dfrac 1 {1 - z} & \dfrac {z - 1} z & z & \dfrac z {z - 1} & \dfrac 1 z & 1 - z \\
\dfrac {z - 1} z & \dfrac {z - 1} z & z & \dfrac 1 {1 - z} & 1 - z & \dfrac z {z - 1} & \dfrac 1 z \\
\dfrac 1 z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z \\
1 - z & 1 - z & \dfrac z {z - 1} & \dfrac 1 z & \dfrac {z - 1} z & z & \dfrac 1 {1 - z} \\
\dfrac z {z - 1} & \dfrac z {z - 1} & \dfrac 1 z & 1 - z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & z \\
\end{array}$
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(ii)}$
* : Introduction
Category:Examples of Cayley Tables
Group Generated by Reciprocal of z and Minus z

== Definition ==
Let:
:$S = \set {f_1, f_2, f_3, f_4}$
where $f_1, f_2, f_3, f_4$ are complex functions defined for all $z \in \C \setminus \set 0$ as:
Let $\circ$ denote composition of functions.
Then $\struct {S, \circ}$ is the group generated by $\dfrac 1 z$ and $-z$.
=== Cayley Table ===
== Also see ==
* Group Generated by Reciprocal of z and Minus z is Klein Four-Group, which demonstrates that this is a (finite) group.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 7$: Isomorphic Groups: Example $2 \ \text{(a)}$
Category:Klein Four-Group
Group Generated by Reciprocal of z and Minus z/Cayley Table

== Cayley Table for Group Generated by Reciprocal of $1 / z$ and $-z$ ==
We have:
:$\map {f_1} z = z$
:$\map {f_2} z = -z$
:$\map {f_3} z = \dfrac 1 z$
:$\map {f_4} z = -\dfrac 1 z$
Hence from Group Generated by Reciprocal of z and Minus z:
$\quad \begin {array} {r|rrrr}
\circ & f_1 & f_2 & f_3 & f_4 \\
\hline
f_1 & f_1 & f_2 & f_3 & f_4 \\
f_2 & f_2 & f_1 & f_4 & f_3 \\
f_3 & f_3 & f_4 & f_1 & f_2 \\
f_4 & f_4 & f_3 & f_2 & f_1 \\
\end{array}$
Expressing the elements in full:
$\quad \begin {array} {c|cccc}
\circ & z & -z & \dfrac 1 z & -\dfrac 1 z \\
\hline
z & z & -z & \dfrac 1 z & -\dfrac 1 z \\
-z & -z & z & -\dfrac 1 z & \dfrac 1 z \\
\dfrac 1 z & \dfrac 1 z & -\dfrac 1 z & z & -z \\
-\dfrac 1 z & -\dfrac 1 z & \dfrac 1 z & -z & z \\
\end{array}$
Category:Examples of Cayley Tables
Category:Group Generated by Reciprocal of z and Minus z
Group Generated by Reciprocal of z and Minus z is Klein Four-Group

== Theorem ==
Let $K_4$ denote the Klein $4$-group.
Let $S$ be the group generated by $1 / z$ and $-z$.
Then $K_4$ and $S$ are isomorphic algebraic structures.
== Proof ==
Establish the mapping $\phi: K_4 \to S$ as follows:
From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertain that $\phi$ is an isomorphism:
=== Cayley Table of Klein $4$-Group ===
</onlyinclude>
=== Group Generated by $1 / z$ and $-z$ ===
The Cayley table for $S$ is as follows:
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 7$: Isomorphic Groups: Example $2$
Category:Examples of Group Isomorphisms/Order 4
Category:Klein Four-Group
Category:Group Generated by Reciprocal of z and Minus z
Group Homomorphism/Examples/Mapping from D3 to Parity Group

== Examples of Group Homomorphisms ==
Let $D_3$ denote the symmetry group of the equilateral triangle:
:
Let $G$ denote the parity group, defined as:
:$\struct {\set {1, -1}, \times}$
where $\times$ denotes conventional multiplication.
Let $\theta: D_3 \to G$ be the mapping defined as:
:$\forall x \in D_3: \map \theta x = \begin{cases} 1 & : \text{$x$ is a rotation} \\ -1 & : \text{$x$ is a reflection} \end{cases}$
Then $\theta$ is a (group) homomorphism, where:
== Proof ==
Let $x, y \in D_3$.
Let $P$ denote the subset of $D_3$ consisting of the rotations:
:$P := \set {e, p, q}$
Let $R$ denote the subset of $D_3$ consisting of the reflections:
:$R := \set {r, s, t}$
It is noted that $e$ is classified as a rotation of zero angle.
The Cayley table for $D_3$ is:
Direct reference to this Cayley table gives us:
Thus we have:
By definition, $\map \ker \phi$ is the set of all elements of $D_3$ which map to $1$.
Hence by definition:
:$\map \ker \phi = \set {e, p, q}$
Also by definition, $\Img \phi$ is the set of all elements of $G$ which are mappped to by $\phi$.
Hence by definition:
:$\Img \phi = \set {1, -1}$
Hence the result.
== Sources ==
* : Chapter $8$: The Homomorphism Theorem: Exercise $2$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval

== Examples of Group Homomorphisms ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ be the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
From Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group we have that $G$ is indeed a group.
=== Direct Sum ===
==== Example $1$ ====
==== Example $2$ ====
==== Example $3$ ====
==== Example $4$ ====
==== Example $5$ ====
==== Example $6$ ====
==== Example $7$ ====
==== Example $8$ ====
==== Example $9$ ====
==== Example $10$ ====
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Direct Sum

== Example of Group Homomorphisms ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ be the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be a homomorphism from $\struct {\map {\mathscr C} J, +}$ to $\struct {\R, +}$.
Let $\phi$ satisfy the condition:
:$\forall c \in \R: \map \phi {f_c} = c$
where $f_c$ is the constant mapping on $\R$ defined as:
:$\forall x \in \R: \map {f_c} x = c$
Then $\struct {\map {\mathscr C} J, +}$ is the internal group direct product of $\map \ker \phi$ and the subgroup of constant mappings on $\R$.
== Proof ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 1

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ be the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
From Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group we have that $G$ is indeed a group.
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the mapping defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \map f 1$
Then $\phi$ is a homomorphism.
=== Kernel ===
== Proof ==
Let $f, g \in \map {\mathscr C} J$ be arbitrary.
We have:
Thus $\phi$ is a homomorphism by definition.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(a)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 1/Kernel

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ denote the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
Let $I_J$ denote the identity mapping on $J$:
:$\forall x \in J: \map {I_J} x = x$
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the homomorphism defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \map f 1$
The kernel of $\phi$ is given by:
:$\map \ker \phi = I_J - f_m$
where:
:$f_m: \R \to \R$ denotes the constant mapping on $\R$
:$m = 1$
:$I_J$ denotes the identity mapping on $J$.
== Proof ==
From Group Homomorphism: Example 1, we have that $\phi$ is indeed a homomorphism.
For all $c \in \R$, let $f_c: \R \to \R$ be the constant mapping:
:$\forall x \in \R: \map {f_c} x = c$
First we show that:
:$\forall c \in \R: \map \phi {f_c} = c$
Let $c \in \R$ be arbitrary.
We have:
Then we show that there exists a unique $m \in \R$ such that:
:$\map \phi {I_J - f_m} = 0$
where in this case:
:$m = 1$
We have:
Hence the result by definition of kernel.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(a)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 10

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ be the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
From Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group we have that $G$ is indeed a group.
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the mapping defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = -\map f 0 + \ds \int_{-2}^0 \map f {e^x} \rd x$
Then $\phi$ is a homomorphism.
=== Kernel ===
== Proof ==
Let $f, g \in \map {\mathscr C} J$ be arbitrary.
We note that:
:when $x = 0$ we have that $e^x = 1$
:when $x = -2$ we have that $0 < e^x < 1$
Therefore the limits of integration of the given definition of $\phi$ are within $\closedint 0 1$.
As both $f$ and $g$ are continuous real functions on $J$, they are integrable on $\closedint {e^{-2} } 1$.
We have:
Thus $\phi$ is a homomorphism by definition.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(j)}$
Category:Examples of Group Homomorphisms