nyuuzyou/wikis huggingface dataset:
Subgroup of Integers is Ideal

== Theorem ==
Let $\struct {\Z, +}$ be the additive group of integers.
Every subgroup of $\struct {\Z, +}$ is an ideal of the ring $\struct {\Z, +, \times}$.
=== Corollary ===
== Proof ==
Let $H$ be a subgroup of $\struct {\Z, +}$.
Let $n \in \Z, h \in H$.
Then from the definition of cyclic group and Negative Index Law for Monoids:
:$n h = n \cdot h \in \gen h \subseteq H$
The result follows.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.2$
* : Chapter $5$: Rings: $\S 21$. Ideals: Example $34$
Category:Subgroups
Category:Integers
Category:Ideal Theory
Subgroup of Order 1 is Trivial

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Then $\struct {G, \circ}$ has exactly $1$ subgroup of order $1$: the trivial subgroup.
== Proof ==
From Trivial Subgroup is Subgroup, $\struct {\set e, \circ}$ is a subgroup of $\struct {G, \circ}$.
Suppose $\struct {\set g, \circ}$ is a subgroup of $\struct {G, \circ}$.
From Group is not Empty, $e \in \set g$.
Thus it follows trivially that $\struct {\set g, \circ} = \struct {\set e, \circ}$.
That is, $\struct {\set e, \circ}$ is the only subgroup of $\struct {G, \circ}$ of order $1$.
== Sources ==
* : $0$: Some Conventions and some Basic Facts
Category:Subgroups
Subgroup of Order p in Group of Order 2p is Normal

== Theorem ==
Let $p$ be an odd prime.
Let $G$ be a group of order $2 p$.
Let $a \in G$ be of order $p$.
Let $K = \gen a$ be the subgroup of $G$ generated by $a$.
Then $K$ is normal in $G$.
=== Corollary ===
== Proof ==
The result Non-Abelian Order 2p Group has Order p Element demonstrates that such an element $a$ exists in $G$.
By definition of generator of cyclic group, $K$ is the cyclic group $C_p$ of order $p$.
By Lagrange's Theorem, the index of $K$ is:
:$\index G K = \dfrac {\order G} {\order K} = \dfrac {2 p} p = 2$
The result follows from Subgroup of Index 2 is Normal.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $23$
Category:Groups of Order 2 p
Subgroup of Order p in Group of Order 2p is Normal/Corollary

== Theorem ==
Let $p$ be an odd prime.
Let $G$ be a group of order $2 p$ whose identity element is $e$.
Let $a \in G$ be of order $p$.
Let $K = \gen a$ be the subgroup of $G$ generated by $a$.
Let $G$ be non-abelian.
Every element of $G \setminus K$ is of order $2$, and:
:$\forall b \in G \setminus K: b a b^{-1} = a^{-1}$
== Proof ==
By Lagrange's Theorem, the elements of $G \setminus K$ can be of order $1$, $2$, $p$ or $2 p$.
$1$ is not possible because Identity is Only Group Element of Order 1.
Then we have that $G$ is non-abelian.
Hence from Cyclic Group is Abelian, $G$ is not cyclic.
Thus $\order b \ne 2 p$.
It remains to investigate $2$ and $p$.
Let $b \in G \setminus K$.
By Subgroup of Index 2 contains all Squares of Group Elements:
:$b^2 \in K$
 $b$ has order $p$.
Then by Intersection of Subgroups of Prime Order:
:$K \cap \gen b = e$
which contradicts $b^2 \in K$.
Thus by Proof by Contradiction $\order b \ne p$
Hence it must follow that $\order b = 2$.
We have that:
:$b a \in G \setminus K$
Thus:
:$\paren {b a}^2 = e$
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $23$
Category:Groups of Order 2 p
Subgroup of Real Numbers is Discrete or Dense

== Theorem ==
Let $G$ be a subgroup of the additive group of real numbers.
Then one of the following holds:
:$G$ is dense in $\R$.
:$G$ is discrete and there exists $a \in \R$ such that $G = a \Z$, that is, $G$ is cyclic.
== Proof ==
If $G$ is trivial, then $G$ is discrete and cyclic.
Let $G$ be non-trivial.
Because $x \in G \iff -x \in G$, $G$ has a strictly positive element.
Thus $G \cap \R_{>0}$ is non-empty.
We have that $G \cap \R_{>0}$ is bounded below by $0$.
Hence, by the Continuum Property, $G \cap \R_{>0}$ admits an infimum.
So, let $a = \map \inf {G \cap \R_{>0} }$.
=== Case 1 ===
Suppose $a = 0$.
We will show that $G$ is dense in $\R$.
Let $x \in \R$ and $\epsilon > 0$.
As $a = 0$ is the infimum, by Characterizing Property of Infimum of Subset of Real Numbers:
:$\exists g \in G \cap \R_{>0}: 0 \le g < \epsilon$
Because $g \in \R_{>0}$:
:$0 < g < \epsilon$
Then:
:$y = g \cdot \floor {\dfrac x g} \in G$
where $\floor {\, \cdot \,}$ denotes the floor function.
We have:
Thus $G$ is dense in $\R$.
=== Case 2 ===
Suppose $a > 0$.
First we need to show that $a \in G$.
 $a \notin G$.
By Characterizing Property of Infimum of Subset of Real Numbers:
:$\forall \epsilon > 0: \exists g \in G: a \le g < a + \epsilon$
In particular, let $\epsilon = a$.
Then:
:$\exists g_2 \in G: a \le g_2 < 2 a$
Then let $\epsilon = g_2$.
Thus:
:$\exists g_1 \in G: a \le g_1 < g_2$
Thus we have that there exists $g_1, g_2 \in \R$ such that:
:$a \le g_1 < g_2 < 2 a$
Then:
but:
:$g_2 - g_1 \in G$
This contradicts $a$ being the infimum.
Therefore, by Proof by Contradiction:
:$a \in G$
Then we show that $G = a \Z$, from which it follows that $G$ is discrete.
Let $g \in G$.
Then:
:$g - a \cdot \floor {\dfrac g a} \in G$
By Real Number minus Floor, we have:
Because $a = \map \inf {G \cap \R_{>0} }$, we have:
:$g = a \cdot \floor {\dfrac g a}$
Thus:
:$g \in a \Z$
Category:Group Theory
Category:Topological Groups
Category:Real Numbers
Subgroup of Solvable Group is Solvable

== Theorem ==
Let $G$ be a solvable group.
Let $H$ be a subgroup of $G$.
Then $H$ is solvable.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 75 \alpha$
Category:Subgroups
Category:Solvable Groups
Category:Subgroup of Solvable Group is Solvable
Subgroup of Solvable Group is Solvable/Proof 1

== Theorem ==
== Proof ==
Let $\set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$ be a normal series for $G$ with abelian quotients.
For every $i = 1, 2, \ldots, n$ we have:
:$\paren {H \cap G_i} \cap G_{i - 1} = H \cap G_{i - 1}$
From the Second Isomorphism Theorem for Groups:
:$\dfrac {\paren {H \cap G_i} G_{i - 1} } {G_{i - 1} } \cong \dfrac {H \cap G_i} {\paren {H \cap G_i} \cap G_{i - 1} } = \dfrac {H \cap G_i} {H \cap G_{i - 1} }$
where $\cong$ denotes group isomorphism.
In particular, $H \cap G_{i - 1}$ is a normal subgroup of $H \cap G_i$.
We have that:
:$\paren {H \cap G_i} G_{i - 1} \subseteq G_i$
and so from the Correspondence Theorem:
:$\dfrac {\paren {H \cap G_i} G_{i - 1} } {G_{i - 1} } \le G_i / G_{i - 1}$
We have that $G_i / G_{i - 1}$ is abelian.
Thus from Subgroup of Abelian Group is Abelian:
:$\dfrac {\paren {H \cap G_i } G_{i - 1} } {G_{i - 1} }$ is abelian.
Hence $\dfrac {H \cap G_i} {H \cap G_{i - 1} }$ is abelian.
Therefore, the series :
:$\set e = H \cap G_0 \lhd H \cap G_1 \lhd \cdots \lhd H \cap G_n = H$
is a normal series with abelian factor groups for $H$.
Therefore $H$ is solvable.
Category:Subgroup of Solvable Group is Solvable
Subgroup of Solvable Group is Solvable/Proof 2

== Theorem ==
== Proof ==
Firstly we, know that a group is solvable if and only if its derived series:
$\map D G = \sqbrk {G, G} \ , \ \map {D^i} G = \sqbrk {\map {D^{i - 1} } G, \map {D^{i - 1} } G}$
becomes trivial after finite iteration.
Meaning:
$\map {D^j} G = \set 1$ 
for some finite $j$.
Now it is trivial that:
:$\map D H \le \map D G$
since $H$ is smaller than $G$.
Further since $\map {D^i} H$ is dominated by $\map {D^i} G$, it too has to become trivial after a finite amount of steps.
Category:Subgroup of Solvable Group is Solvable
Subgroup of Solvable Group is Solvable/Proof 3

== Theorem ==
== Proof ==
Let $H \le G$ and $G$ be solvable with normal series:
:$\set e = G_0 \lhd G_1 \lhd \dots \lhd G_m = G$
such that $G_{i + 1} / G_i$ is abelian for all $i$.
Define $N_i = G_i \cap H$.
We show that these $N_i$ will form a normal series with abelian factors.
;Normality:
Let $x \in N_i$ and $y \in N_{i + 1}$.
Then $y x y^{-1} \in N$ since $N$ as a group is closed.
We also have $y x y^{-1} \in G_i$ since $G_i$ is normal in $G_{i + 1}$.
Hence $N_i$ is invariant under conjugation and therefore normal.
;Abelian Factors: 
Note that:
Thus:
so the quotient $\dfrac {N_{i + 1}} {N_i}$ is isomorphic to a subgroup of the abelian group $\dfrac{G_{i + 1}} {G_i}$.
Hence it is abelian, proving the theorem.
Category:Subgroup of Solvable Group is Solvable
Subgroup of Subgroup with Prime Index

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$.
Let $K$ be a subgroup of $H$.
Let:
:$\index G K = p$
where:
:$p$ denotes a prime number
:$\index G K$ denotes the index of $K$ in $G$.
Then either:
:$H = K$
or:
:$H = G$
=== Corollary ===
== Proof ==
From the Tower Law for Subgroups:
:$\index G K = \index G H \index H K$
As $\index G K = p$ is prime, either $\index G H = p$ or $\index H K = p$.
Thus either $\index G H = 1$ or $\index H K = 1$.
The result follows from Index is One iff Subgroup equals Group.
== Sources ==
* : $0$: Some Conventions and some Basic Facts: Exercise $11 \ \text{(ii)}$
Category:Index of Subgroups
Subgroup of Subgroup with Prime Index/Corollary

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H$ and $K$ be subgroups of $G$.
Let $K \subsetneq H$.
Let:
:$\index G K = p$
where:
:$p$ denotes a prime number
:$\index G K$ denotes the index of $K$ in $G$.
Then:
:$H = G$
== Proof ==
As $K \subsetneq H$ and $K$ is a subgroups of $G$, it follows that $K$ is a proper subgroup of $H$.
That is, $K \ne H$
Hence from Subgroup of Subgroup with Prime Index:
:$H = G$
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $13$
Category:Index of Subgroups
Subgroup of Symmetric Group that Fixes n

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $H$ denote the subgroup of $S_n$ which consists of all $\pi \in S_n$ such that:
:$\map \pi n = n$
Then:
:$H = S_{n - 1}$
and the index of $H$ in $S_n$ is given by:
:$\index {S_n} H = n$
== Proof ==
We have that $S_{n - 1}$ is the symmetric group on $n - 1$ letters.
Let $\pi \in S_{n - 1}$.
Then $\pi$ is a permutation on $n - 1$ letters.
Hence $\pi$ is also a permutation on $n$ letters which fixes $n$.
So $S_{n - 1} \subseteq H$.
Now let $\pi \in H$.
Then $\pi$ is a permutation on $n$ letters which fixes $n$.
That is, $\pi$ is a permutation on $n - 1$ letters.
Thus $\pi \in S_{n - 1}$.
So we have that $H = S_{n - 1}$.
We also have that $S_{n - 1}$ is a group, and:
:$\forall \rho \in S_{n - 1}: \rho \in S_n$
So $S_{n - 1}$ is a subset of $S_n$ which is a group.
Hence $S_{n - 1}$ is a subgroup of $S_n$ by definition.
Then we have:
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 39 \alpha$
Category:Symmetric Groups
Subgroups of Additive Group of Integers

== Theorem ==
Let $\struct {\Z, +}$ be the additive group of integers.
Let $n \Z$ be the additive group of integer multiples of $n$.
Every non-trivial subgroup of $\struct {\Z, +}$ has the form $n \Z$.
== Proof ==
First we note that, from Integer Multiples under Addition form Infinite Cyclic Group, $\struct {n \Z, +}$ is an infinite cyclic group.
From Cyclic Group is Abelian, it follows that $\struct {n \Z, +}$ is an infinite abelian group.
Let $H$ be a non-trivial subgroup of $\struct {\Z, +}$.
Because $H$ is non-trivial:
:$\exists m \in \Z: m \in H: m \ne 0$
Because $H$ is itself a group:
:$-m \in H$
So either $m$ or $-m$ is positive and therefore in $\Z_{>0}$.
Thus:
:$H \cap \Z_{>0} \ne \O$
From the Well-Ordering Principle, $H \cap \Z_{>0}$ has a smallest element, which we can call $n$.
It follows from Subgroup of Infinite Cyclic Group is Infinite Cyclic Group that:
: $\forall a \in \Z: a n \in H$
Thus:
: $n \Z \subseteq H$
:
:$\exists m \in \Z: m \in H \setminus n \Z$
Then $m \ne 0$, and also $-m \in H \setminus n \Z$.
Assume $m > 0$, otherwise we consider $-m$.
By the Division Theorem:
:$m = q n + r$
If $r = 0$, then $m = q n \in n \Z$, so $0 \le r < n$.
Now this means $r = m - q n \in H$ and $0 \le r < n$.
This would mean $n$ was not the smallest element of $H \cap \Z$.
Hence, by Proof by Contradiction, there can be no such $m \in H \setminus n \Z$.
Thus:
:$H \setminus n \Z = \O$
Thus from Set Difference with Superset is Empty Set‎:
:$H \subseteq n \Z$
Thus we have $n \Z \subseteq H$ and $H \subseteq n \Z$.
Hence:
:$H = n \Z$
== Examples ==
== Sources ==
* : Chapter $\text{II}$: Groups: A Little Number Theory
* : Chapter $2$: Subgroups and Cosets: $\S 36$
* : $\S 1.2$
Category:Subgroups
Category:Additive Group of Integers
Category:Additive Groups of Integer Multiples
Category:Subgroups of Additive Group of Integers
Subgroups of Additive Group of Integers/Examples/Even Integers

== Example of Subgroup of Additive Group of Integers ==
Let $2 \Z$ denote the set of even integers.
Let $\struct {2 \Z, +}$ be the algebraic structure formed from $2 \Z$ with the operation of integer addition.
Then $\struct {2 \Z, +}$ is a subgroup of the additive group of integers $\struct {\Z, +}$.
== Proof ==
The set of even integers is the underlying set of the additive group of (integer) multiples of $2$.
Thus from Subgroups of Additive Group of Integers, $\struct {2 \Z, +}$ is a subgroup of $\struct {\Z, +}$.
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{E ii}$
* : Chapter $4$: Subgroups: Exercise $1 \ \text{(a)}$
* 
* 
Category:Subgroups of Additive Group of Integers
Category:Even Integers
Subgroups of Additive Group of Integers/Examples/Multiples of 4

== Example of Subgroup of Additive Group of Integers ==
Let $4 \Z$ denote the set of integers which are divisible by $4$.
Let $\struct {4 \Z, +}$ be the algebraic structure formed from $4 \Z$ with the operation of integer addition.
Then $\struct {4 \Z, +}$ is a subgroup of the additive group of integers $\struct {\Z, +}$.
== Proof ==
The set of integers which are divisible by $4$ is the underlying set of the additive group of (integer) multiples of $4$.
Thus from Subgroups of Additive Group of Integers, $\struct {4 \Z, +}$ is a subgroup of $\struct {\Z, +}$.
== Sources ==
* : Chapter $\text{II}$: Groups: Subgroups
Category:Subgroups of Additive Group of Integers
Subgroups of Additive Group of Integers Modulo m

== Theorem ==
Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
Let $\struct {\Z_m, +_m}$ denote the additive group of integers modulo $m$.
The subgroups of $\struct {\Z_m, +_m}$ are the additive groups of integers modulo $k$ where:
:$k \divides m$
== Proof ==
From Integers Modulo m under Addition form Cyclic Group, $\struct {\Z_m, +_m}$ is cyclic.
Let $H$ be a subgroup of $\struct {\Z_m, +_m}$
From Subgroup of Cyclic Group is Cyclic, $H$ is of the form $\struct {\Z_k, +_k}$ for some $k \in \Z$.
From Lagrange's Theorem, it follows that $k \divides m$.
Hence the result.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 36 \gamma$
Category:Additive Groups of Integers Modulo m
Subgroups of Cartesian Product of Additive Group of Integers

== Theorem ==
Let $\struct {\Z, +}$ denote the additive group of integers.
Let $m, n \in \Z_{> 0}$ be (strictly) positive integers.
Let $\struct {\Z \times \Z, +}$ denote the Cartesian product of $\struct {\Z, +}$ with itself.
The subgroups of $\struct {\Z \times \Z, +}$ are not all of the form:
:$\struct {m \Z, +} \times \struct {n \Z, +}$
where $\struct {m \Z, +}$ denotes the additive group of integer multiples of $m$.
== Proof ==
Consider the map $\phi: \struct {m \Z, +} \times \struct {n \Z, +} \mapsto \struct {\Z, +} \times \struct {\Z, +}$ defined by:
:$\forall c, d \in \Z: \map \phi {m c, n d} = \tuple {c, d}$
which is a group isomorphism.
Hence, $\struct {m \Z, +} \times \struct {n \Z, +}$ is a free abelian group of rank $2$.
Therefore, any subgroup generated by a singleton, for example, $\set {\tuple {x, 0}: x \in \Z}$ is a subgroup not in the form $\struct {m \Z, +} \times \struct {n \Z, +}$.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 36 \delta$
Category:Additive Group of Integers
Category:Additive Groups of Integer Multiples
Subgroups of Symmetric Group Isomorphic to Product of Subgroups

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $k \in \closedint 1 n$.
Then there are $\dbinom n k$ subgroups of $S_n$ which are isomorphic to $S_k \times S_{n - k}$, where $\dbinom n k$ denotes the binomial coefficient.
All of these $\dbinom n k$ subgroups are conjugate.
== Proof ==
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 79 \delta$
Category:Symmetric Groups
Subgroups of Symmetry Group of Regular Hexagon

== Theorem ==
Let $\HH = ABCDEF$ be a regular hexagon.
Let $D_6$ denote the symmetry group of $\HH$.
:
Let $e$ denote the identity mapping
Let $\alpha$ denote rotation of $\HH$ anticlockwise through $\dfrac \pi 3$ radians ($60 \degrees$)
Let $\beta$ denote reflection of $\HH$ in the $AD$ axis.
The subsets of $D_6$ which form its subgroups are as follows:
;Order $1$:
:$\set e$
;Order $2$:
:$\set {e, \alpha^3}$
:$\set {e, \beta}$
:$\set {e, \alpha \beta}$
:$\set {e, \alpha^2 \beta}$
:$\set {e, \alpha^3 \beta}$
:$\set {e, \alpha^4 \beta}$
:$\set {e, \alpha^5 \beta}$
;Order $3$:
:$\set {e, \alpha^2, \alpha^4}$
;Order $4$:
:$\set {e, \alpha^3, \beta, \alpha^3 \beta}$
:$\set {e, \alpha^3, \alpha \beta, \alpha^4 \beta}$
:$\set {e, \alpha^3, \alpha^2 \beta, \alpha^5 \beta}$
;Order $6$:
:$\set {e, \alpha, \alpha^2, \alpha^3, \alpha^4, \alpha^5}$
:$\set {e, \alpha^2, \alpha^4, \beta, \alpha^2 \beta, \alpha^4 \beta}$
:$\set {e, \alpha^2, \alpha^4, \alpha \beta, \alpha^3 \beta, \alpha^5 \beta}$
;Order $12$:
:$D_6$ itself.
== Proof ==
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 38 \beta$
Category:Symmetry Group of Regular Hexagon
Submartingale Composed with Increasing Convex Function is Submartingale

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $\sequence {X_t}_{t \ge 0}$ be a $\sequence {\FF_t}_{t \ge 0}$-submartingale.
Let $f : \R \to \R$ be an increasing convex function such that $\map f {X_t}$ is integrable for each $t \in \hointr 0 \infty$. 
Then $\sequence {\map f {X_t} }_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-submartingale. 
== Proof ==
Since $\sequence {X_t}_{t \ge 0}$ is a martingale, we have: 
:$X_t$ is $\FF_t$-measurable
for each $t \in \hointr 0 \infty$. 
From Convex Real Function is Measurable and Composition of Measurable Mappings is Measurable:
:$\map f {X_t}$ is $\FF_t$-measurable
for each $t \in \hointr 0 \infty$.
So $\sequence {\map f {X_t} }_{t \ge 0}$ is $\sequence {\FF_t}_{t \ge 0}$-adapted. 
Now let $s, t \in \hointr 0 \infty$ be such that $0 \le s < t$. 
Let $\expect {\map f {X_t} \mid \FF_s}$ be a version of the conditional expectation of $\map f {X_t}$ given $\FF_s$.
Since $\sequence {X_t}_{t \ge 0}$ is a submartingale, we have: 
:$\expect {X_t \mid \FF_s} \ge X_s$ almost surely. 
So, since $f$ is increasing, we have: 
:$\map f {\expect {X_t \mid \FF_s} } \ge \map f {X_s}$ almost surely.
We then have: 
So $\sequence {\map f {X_t} }_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-submartingale.
== Sources ==
* : Proposition $3.12$
Category:Submartingales
Category:Convex Real Functions
Submodule Test

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring with unity $1_R$.
Let $\struct {G, +, \circ}_R$ be a unitary $R$-module.
Let $H$ be a non-empty subset of $G$.
Then $\struct {H, +, \circ}_R$ is a submodule of $G$ :
== Proof ==
=== Necessary Condition ===
Let $\struct {H, +, \circ}_R$ fulfil the conditions $\text {SM} 1$ and $\text {SM} 2$.
We have that $H \subseteq G$ such that $H \ne \O$.
We have from $\text {SM} 1$:
:$\forall x, y \in H: x + y \in H$
We have that $\struct {R, +_R, \times_R}$ is a ring with unity whose unity is $1_R$.
Hence:
It follows from the Two-Step Subgroup Test that $H$ is a subgroup of $\struct {G, +}$.
It remains to be shown that $\struct {H, +, \circ}_R$ fulfils the module axioms as follows:
;
Hence $\struct {H, +, \circ}_R$ fulfils .
;
Hence $\struct {H, +, \circ}_R$ fulfils .
;
Hence $\struct {H, +, \circ}_R$ fulfils .
=== Sufficient Condition ===
Let $\struct {H, +, \circ}_R$ be a submodule of $G$.
As $H$ is an $R$-module, $\struct {H, +}$ is an abelian group.
As $\struct {H, +}$ is a group, it is closed under $+$ it follows that
:$\forall x, y \in H: x + y \in H$
which is $\text {SM} 1$.
As $H$ is an $R$-module, it is closed for scalar product:
:$\forall \lambda \in R, x \in H: \lambda \circ x \in H$
which is $\text {SM} 2$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.1$
Category:Submodules
Submodule of Module of Polynomial Functions

== Theorem ==
Let $K$ be a commutative ring with unity.
Let $\map P K$ be the set of all polynomial functions on $K$.
Consider the set $\map {P_m} K$ of all the polynomial functions:
:$\ds \sum_{k \mathop = 0}^{m - 1} \alpha_k {I_K}^k$
for some $m \in \N^*$ where:
:$\sequence {\alpha_k}_{k \in \closedint 0 {m - 1} }$
is any sequence of $m$ terms of $K$.
Then $\map {P_m} K$ is a submodule of $\map P K$.
== Proof ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Example $27.4$
Category:Module Theory
Category:Polynomial Theory
Subobject Class in Category of Sets

== Theorem ==
Let $\mathbf {Set}$ be the category of sets.
Let $S$ be a set.
Let $\map {\overline {\mathbf {Sub} }_{\mathbf {Set} } } S$ be the category of subobject classes of $S$.
Let $\powerset S$ be the order category on the power set of $S$ induced by Subset Relation on Power Set is Partial Ordering.
Then $\map {\overline {\mathbf {Sub} }_{\mathbf {Set} } } S \cong \powerset S$.
== Proof ==
== Sources ==
* : $\S 5.1$: Remark $5.2$
Category:Categories of Subobjects
Category:Category of Sets
Subring Containing Ring Unity has Unity

== Definition ==
Let $\struct{R, +, *}$ be a ring with unity $1_R$.
Let $\struct{S, +_S, *_S}$ be a subring of $\struct{R, +, *}$.
Let $1_R \in S$.
Then $1_R$ is the unity of $\struct{S, +_S, *_S}$.
== Proof ==
We have:
== Also see ==
* Unity of Subring is not necessarily Unity of Ring
Category:Rings with Unity
Subring Generated by Unity of Ring with Unity

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let the mapping $g: \Z \to R$ be defined as $\forall n \in \Z: \map g n = n 1_R$, where $n 1_R$ the $n$th power of $1_R$.
Let $\ideal x$ be the principal ideal of $\struct {R, +, \circ}$ generated by $x$.
Then $g$ is an epimorphism from $\Z$ onto the subring $S$ of $R$ generated by $1_R$.
If $R$ has no proper zero divisors, then $g$ is the only nonzero homomorphism from $\Z$ into $R$.
The kernel of $g$ is either:
:$(1): \quad \ideal {0_R}$, in which case $g$ is an isomorphism from $\Z$ onto $S$
or:
:$(2): \quad \ideal p$ for some prime $p$, in which case $S$ is isomorphic to the field $\Z_p$.
== Proof ==
By the Index Law for Sum of Indices and Integral Multiple of Ring Element, we have:
:$\paren {n 1_R} \paren {m 1_R} = n \paren {m 1_R} = \paren {n m} 1_R$
Thus $g$ is an epimorphism from $\Z$ onto $S$.
 $R$ has no proper zero divisors.
By Kernel of Ring Epimorphism is Ideal, the kernel of $g$ is an ideal of $\Z$.
By Ring of Integers is Principal Ideal Domain, the kernel of $g$ is $\ideal p$ for some $p \in \Z_{>0}$.
By Kernel of Ring Epimorphism is Ideal (don't think this is the correct reference - check it), $S$ is isomorphic to $\Z_p$ and also has no proper zero divisors.
So from Integral Domain of Prime Order is Field either $p = 0$ or $p$ is prime.
Now we need to show that $g$ is unique.
Let $h$ be a non-zero (ring) homomorphism from $\Z$ into $R$.
As $\map h 1 = \map h {1^2} = \paren {\map h 1}^2$, either $\map h 1 = 1_R$ or $\map h 1 = 0_R$ by Idempotent Elements of Ring with No Proper Zero Divisors.
But, by Homomorphism of Powers: Integers, $\forall n \in \Z: \map h n = \map h {n 1} = n \map h 1$
So if $\map h 1 = 0_R$, then $\forall n \in \Z: \map h n = n 0_R = 0_R$.
Hence $h$ would be a zero homomorphism, which contradicts our stipulation that it is not.
So $\map h 1 = 1_R$, and thus $\forall n \in \Z: \map h n = n 1 = \map g n$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.7$
Category:Ideal Theory