nyuuzyou/wikis huggingface dataset:
Subgroup equals Conjugate iff Normal/Proof 2

== Theorem ==
== Proof ==
From Subgroup is Superset of Conjugate iff Normal, $N$ is normal in $G$ :
:$\forall g \in G: N \supseteq g \circ N \circ g^{-1}$
:$\forall g \in G: N \supseteq g^{-1} \circ N \circ g$
From Subgroup is Subset of Conjugate iff Normal, $N$ is normal in $G$ :
:$\forall g \in G: N \subseteq g \circ N \circ g^{-1}$
:$\forall g \in G: N \subseteq g^{-1} \circ N \circ g$
The result follows by definition of set equality.
Category:Subgroup equals Conjugate iff Normal
Subgroup is Closed iff Quotient is Hausdorff

== Theorem ==
Let $G$ be a topological group.
Let $H \le G$ be a subgroup.
Let $G / H$ be their quotient.
Then the following are equivalent:
:$H$ is closed in $G$
:$G / H$ is Hausdorff
== Proof ==
Category:Topological Groups
Category:Hausdorff Spaces
Subgroup is Normal Subgroup of Normalizer

== Theorem ==
Let $G$ be a group.
A subgroup $H \le G$ is a normal subgroup of its normalizer:
:$H \le G \implies H \lhd \map {N_G} H$
== Proof ==
From Subgroup is Subgroup of Normalizer we have that $H \le \map {N_G} H$.
It remains to show that $H$ is normal in $\map {N_G} H$.
Let $a \in H$ and $b \in \map {N_G} H$.
By the definition of normalizer:
:$b a b^{-1} \in H$
Thus $H$ is normal in $\map {N_G} H$.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $12 \ \text{(i)}$
Category:Normal Subgroups
Category:Normalizers
Category:Subgroups
Subgroup is Normal iff Contains Conjugate Elements

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ :
== Proof ==
By definition, a subgroup is normal in $G$ :
:$\forall g \in G: g \circ N = N \circ g$
=== Necessary Condition ===
Suppose that $g \circ N = N \circ g$, by definition 1 of normality in $G$.
Let $n \in N$.
Then:
=== Sufficient Condition ===
Suppose that:
: $\forall g \in G: \paren {n \in N \iff g \circ n \circ g^{-1} \in N}$
Let $g \circ n \circ g^{-1} \in N$.
Similarly:
As $g$ is arbitrary, then so is $g^{-1}$.
Thus:
:$N \circ g \subseteq g \circ N$
By definition of set equality:
:$g \circ N = N \circ g$
== Also see ==
* Equivalence of Definitions of Normal Subgroup
== Sources ==
* : $\S 6.6$. Normal subgroups
* : $\S 1.10$: Theorem $26$
* : $\text{II}$: Morphisms
* : $\S 49.3$ Normal subgroups
* : $\S 7$: Definition $7.3$
Category:Conjugacy
Category:Normal Subgroups
Subgroup is Normal iff Left Cosets are Right Cosets

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ (by definition 1) :
:Every right coset of $N$ in $G$ is a left coset
or equivalently:
:The right coset space of $N$ in $G$ equals its left coset space.
== Proof ==
=== Necessary Condition ===
Let $N$ be a normal subgroup of $G$ by Definition 1.
Then the equality of the coset spaces follows directly from definition of normal subgroup and coset.
=== Sufficient Condition ===
Suppose that every right coset of $N$ in $G$ is a left coset of $N$ in $G$.
Let $g \in G$.
Since every right coset of $N$ in $G$ is a left coset, there exists an $h \in G$ such that $N \circ g = h \circ N$.
By Element of Group is in its own Coset:
:$g \in N \circ g = h \circ N$
From Element in Left Coset iff Product with Inverse in Subgroup:
: $g^{-1} \circ h \in N$
Then:
Since this holds for all $g \in G$, $N$ is normal in $G$ (by definition 1).
== Also see ==
* Equivalence of Definitions of Normal Subgroup
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \alpha$
* : Chapter $7$: Normal subgroups and quotient groups: Proposition $7.4 \ \text{(e)}$
Category:Conjugacy
Category:Normal Subgroups
Subgroup is Normal iff Normal Subset

== Theorem ==
Let $\left({G, \circ}\right)$ be a group.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ (by definition 1) it is a normal subset of $G$.
== Proof ==
=== Necessary Condition ===
Let $N$ be normal in $G$ (by definition 1):
Thus for each $g \in G$:
:$\forall g \in G: g \circ N = N \circ g$
where $g \circ N$ denotes the subset product of $g$ with $N$.
Thus $N$ is a normal subset of $G$ (by definition 1):
:$\forall g \in G: g \circ N = N \circ g$
=== Sufficient Condition ===
Let $N$ be a normal subset of $G$ (by definition 1):
:$\forall g \in G: g \circ N = N \circ g$
Since $N$ is a subgroup, $N$ is a normal subgroup of $G$ (by definition 1).
Category:Conjugacy
Category:Normal Subgroups
Subgroup is Normal iff it contains Product of Inverses

== Theorem ==
A subgroup $H$ of a group $G$ is normal :
:$\forall a, b \in G: a b \in H \implies a^{-1} b^{-1} \in H$
== Proof ==
=== Necessary Condition ===
Suppose $H$ is normal.
Let $a b \in H$.
As $H$ is a group, then $x \in H \iff x^{-1} \in H$.
So $b a \in H \iff \left({b a}\right)^{-1} \in H \iff a^{-1} b^{-1} \in H$.
Thus $\forall a, b \in G: a b \in H \implies a^{-1} b^{-1} \in H$.
=== Sufficient Condition ===
Suppose $a b \in H \implies a^{-1} b^{-1} \in H$.
Then (from the above) we have $\forall a, b \in H: a b \in H \implies b a \in H$.
Let $g \in G$, and let $h \in H$. Let $h = x g$ where $x \in G$.
Now $g h g^{-1} = g x g g^{-1} = g x$.
By hypothesis, $x g \in H \implies g x \in H$.
The result follows.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \beta$
Category:Normal Subgroups
Subgroup is Subgroup of Normalizer

== Theorem ==
Let $G$ be a group.
A subgroup $H \le G$ is a subgroup of its normalizer:
:$H \le G \implies H \le \map {N_G} H$
== Proof ==
=== Subset of Normalizer ===
First we show that $H$ is a subset of $\map {N_G} H$.
This follows directly from Left Coset Equals Subgroup iff Element in Subgroup:
:$x \in H \implies x H = H$
As $x \in H \implies x^{-1} \in H$ it also follows that $x \in H \implies H x^{-1} = H$.
Thus:
:$x \in H \implies x H x^{-1} = H^x = H$
and so:
:$x \in \map {N_G} H$
So:
:$H \subseteq \map {N_G} H$
as we wanted to show.
=== Subgroup of Normalizer ===
By hypothesis, $H$ is a subgroup of $G$.
Thus $H$ is itself a group.
So by definition of subgroup:
: $(1): \quad H \subseteq \map {N_G} H$
: $(2): \quad $ is a group
it follows that $H$ is a subgroup of $\map {N_G} H$.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 35 \gamma$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $12 \ \text{(i)}$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.10$
Category:Subgroups
Category:Normalizers
Subgroup is Subset of Conjugate iff Normal

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ (by definition 1) :
:$\forall g \in G: N \subseteq g \circ N \circ g^{-1}$
:$\forall g \in G: N \subseteq g^{-1} \circ N \circ g$
== Proof ==
By definition, a subgroup is normal in $G$ :
:$\forall g \in G: g \circ N = N \circ g$
First note that:
:$(1): \quad \paren {\forall g \in G: N \subseteq g \circ N \circ g^{-1} } \iff \paren {\forall g \in G: N \subseteq g^{-1} \circ N \circ g}$
which is shown by, for example, setting $h := g^{-1}$ and substituting.
=== Necessary Condition ===
Suppose that $N$ is normal in $G$.
Then:
Then by $(1)$ above:
:$\paren {\forall g \in G: N \subseteq g \circ N \circ g^{-1} } \implies \paren {\forall g \in G: N \subseteq g^{-1} \circ N \circ g}$
=== Sufficient Condition ===
Let $N$ be a subgroup of $G$ such that:
: $\forall g \in G: N \subseteq g \circ N \circ g^{-1}$
and so from $(1)$ above:
: $\forall g \in G: N \subseteq g^{-1} \circ N \circ g$
Then:
Similarly:
Thus we have:
:$N \circ g \subseteq g \circ N$
:$g \circ N \subseteq N \circ g$
By definition of set equality:
:$g \circ N = N \circ g$
Hence the result.
== Also see ==
* Equivalence of Definitions of Normal Subgroup
* Subgroup is Superset of Conjugate iff Normal
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.2: \ 4^\circ, 5^\circ$
Category:Conjugacy
Category:Normal Subgroups
Subgroup is Superset of Conjugate iff Normal

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ (by definition 1) :
:$\forall g \in G: g \circ N \circ g^{-1} \subseteq N$
:$\forall g \in G: g^{-1} \circ N \circ g \subseteq N$
== Proof ==
By definition, a subgroup is normal in $G$ :
:$\forall g \in G: g \circ N = N \circ g$
First note that:
:$(1): \quad \paren {\forall g \in G: g \circ N \circ g^{-1} \subseteq N} \iff \paren {\forall g \in G: g^{-1} \circ N \circ g \subseteq N}$
which is shown by, for example, setting $h := g^{-1}$ and substituting.
=== Necessary Condition ===
Suppose that $N$ is normal in $G$.
Then:
Similarly:
=== Sufficient Condition ===
Let $N$ be a subgroup of $G$ such that:
:$\forall g \in G: g \circ N \circ g^{-1} \subseteq N$
and so from $(1)$ above:
:$\forall g \in G: g^{-1} \circ N \circ g \subseteq N$
Then:
Similarly:
Thus we have:
:$N \circ g \subseteq g \circ N$
:$g \circ N \subseteq N \circ g$
By definition of set equality:
:$g \circ N = N \circ g$
Hence the result.
== Also known as ==
Because of its convenience as a technique for determining whether a subgroup is normal or not, this result is often seen referred to as the Normal Subgroup Test.
== Also see ==
* Equivalence of Definitions of Normal Subgroup
* Subgroup is Subset of Conjugate iff Normal
== Sources ==
* : $\S 6.6$. Normal subgroups
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.2: \ 2^\circ, 3^\circ$
* : Chapter $7$: Normal subgroups and quotient groups: Proposition $7.4 \ \text{(b)}$
Category:Conjugacy
Category:Normal Subgroups
Subgroup of Abelian Group is Normal

== Theorem ==
Every subgroup of an abelian group is normal.
== Proof ==
Let $G$ be an abelian group.
Let $H \le G$ be a subgroup of $G$.
Then for all $a \in G$:
== Sources ==
* : $\S 6.6$. Normal subgroups
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures
* : $\S 1.10$: Example $36$
* : Chapter $\text{II}$: Groups: Morphisms
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46$
* : $\S 5$: Groups $\text{I}$: Subgroups
* : $\S 49.6$ Normal subgroups
* : Chapter $7$: Normal subgroups and quotient groups: Example $7.5$
Category:Normal Subgroups
Category:Abelian Groups
Subgroup of Additive Group Modulo m is Ideal of Ring

== Theorem ==
Let $m \in \Z: m > 1$.
Let $\struct {\Z_m, +_m}$ be the additive group of integers modulo $m$.
Then every subgroup of $\struct {\Z_m, +_m}$ is an ideal of the ring of integers modulo $m$ $\struct {\Z_m, +_m, \times_m}$.
== Proof ==
Let $H$ be a subgroup of $\struct {\Z_m, +_m}$
Suppose:
: $(1): \quad h + \ideal m \in H$, where $\ideal m$ is a principal ideal of $\struct {\Z_m, +_m, \times_m}$
and 
: $(2): \quad n \in \N_{>0}$.
Then by definition of multiplication on integers and Homomorphism of Powers as applied to integers:
But:
:$n \cdot \map {q_m} h \in \gen {\map {q_m} h}$
where $\gen {\map {q_m} h}$ is the group generated by $\map {q_m} h$.
Hence by Epimorphism from Integers to Cyclic Group, $n \cdot \map {q_m} h \in H$.
The result follows.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.4$
Category:Subgroups
Category:Ideal Theory
Category:Additive Groups of Integers Modulo m
Category:Modulo Addition
Subgroup of Additive Group of Integers Generated by Two Integers

== Theorem ==
Let $m, n \in \Z_{> 0}$ be (strictly) positive integers.
Let $\struct {\Z, +}$ denote the additive group of integers.
Let $\gen {m, n}$ be the subgroup of $\struct {\Z, +}$ generated by $m$ and $n$.
Then:
:$\gen {m, n} = \struct {\gcd \set {m, n} \Z, +}$
That is, the additive groups of integer multiples of $\gcd \set {m, n}$, where $\gcd \set {m, n}$ is the greatest common divisor of $m$ and $n$.
== Proof ==
By definition:
:$\gen {m, n} = \set {x \in \Z: \gcd \set {m, n} \divides x}$
Hence the result.
== Sources ==
* : Chapter $5$: Subgroups: Exercise $10$
* : Chapter $2$: Subgroups and Cosets: $\S 36 \beta$
Category:Additive Groups of Integer Multiples
Subgroup of Circle Group Generated by Distinct Roots of Unity

== Theorem ==
Let $K$ be the circle group.
Let $m, n \in \Z_{>0}$ be (strictly) positive integers.
Let $d = \lcm \set {m, n}$ be the least common multiple of $m$ and $n$.
Let $\alpha$ be a primitive $n$th root of unity.
Let $\beta$ be a primitive $m$th root of unity.
Let $\gamma$ be a primitive $d$th root of unity.
Let $H = \gen {\alpha, \beta}$ be the subgroup of $K$ generated by $\alpha, \beta$.
Then $H = \gen \gamma$.
== Proof ==
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 44 \beta$
Category:Multiplicative Groups of Complex Roots of Unity
Category:Circle Group
Subgroup of Cyclic Group is Cyclic

== Theorem ==
Let $G$ be a cyclic group.
Let $H$ be a subgroup of $G$.
Then $H$ is cyclic.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text {Y}$
* 
* 
Category:Cyclic Groups
Category:Subgroups
Category:Subgroup of Cyclic Group is Cyclic
Subgroup of Cyclic Group is Cyclic/Proof 1

== Theorem ==
== Proof ==
Let $G$ be a cyclic group generated by $a$.
Let $H$ be a subgroup of $G$.
If $H = \set e$, then $H$ is a cyclic group subgroup generated by $e$.
Let $H \ne \set e$.
By definition of cyclic group, every element of $G$ has the form $a^n$.
Then as $H$ is a subgroup of $G$, $a^n \in H$ for some $n \in \Z$.
Let $m$ be the smallest (strictly) positive integer such that $a^m \in H$.
Consider an arbitrary element $b$ of $H$.
As $H$ is a subgroup of $G$, $b = a^n$ for some $n$.
By the Division Theorem, it is possible to find integers $q$ and $r$ such that $n = m q + r$ with $0 \le r < m$.
It follows that:
:$a^n = a^{m q + r} = \paren {a^m}^q a^r$
and hence:
:$a^r = \paren {a^m}^{-q} a^n$
Since $a^m \in H$ so is its inverse $\paren {a^m}^{-1}$
By , so are all powers of its inverse.
Now $a^n$ and $\paren {a^m}^{-q}$ are both in $H$, thus so is their product $a^r$, again by .
However:
:$m$ was the smallest (strictly) positive integer such that $a^m \in H$
and:
:$0 \le r < m$
Therefore it follows that:
:$r = 0$
Therefore:
:$n = q m$
and:
:$b = a^n = \paren {a^m}^q$.
We conclude that any arbitrary element $b = a^n$ of $H$ is a power of $a^m$.
So, by definition, $H = \gen {a^m}$ is cyclic.
== Sources ==
* : $\S 1.9$: Theorem $14$
* : Chapter $2$: Subgroups and Cosets: $\S 43$
* : $\S 39.4$ Cyclic Groups
* : $\S 3.4$: Cyclic groups: Theorem $1$
* : Chapter $4$: Subgroups: Proposition $4.13$
* : $\S 1.14$: Exercise $19 \ \text{(b)}$
Category:Subgroup of Cyclic Group is Cyclic
Subgroup of Cyclic Group is Cyclic/Proof 2

== Theorem ==
== Proof ==
Let $G$ be a cyclic group generated by $a$.
=== Finite Group ===
Let $G$ be finite.
By Bijection from Divisors to Subgroups of Cyclic Group there are exactly as many subgroups of $G$ as divisors of the order of $G$.
As each one of these is cyclic by Subgroup of Finite Cyclic Group is Determined by Order‎, the result follows.
=== Infinite Group ===
By Morphism from Integers to Group, an infinite cyclic group is isomorphic to $\struct {\Z, +}$.
So all we need to do is show that any subgroup of $\struct {\Z, +}$ is cyclic.
Suppose $H$ is a subgroup of $\struct {\Z, +}$.
From Subgroup of Integers is Ideal and Ring of Integers is Principal Ideal Domain, we have that:
:$\exists m \in \Z_{> 0}: H = \ideal m$
where $\ideal m$ is the principal ideal of $\struct {\Z, +, \times}$ generated by $m$.
But $m$ is also a generator of the subgroup $\ideal m$ of $\struct {\Z, +}$, as:
:$n \in \Z: n \circ m = n \cdot m \in \gen m$
Hence the result.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.8$ Corollary
Category:Subgroup of Cyclic Group is Cyclic
Subgroup of Cyclic Group is Cyclic/Proof 3

== Theorem ==
== Proof ==
Let $G$ be a cyclic group generated by $a$.
Let $H$ be a subgroup of $G$.
By Cyclic Group is Abelian, $G$ is abelian.
By Subgroup of Abelian Group is Normal, $H$ is normal in $G$.
Let $G / H$ be the quotient group of $G$ by $H$.
Let $q_H: G \to G / H$ be the quotient epimorphism from $G$ to $G / H$:
:$\forall x \in G: \map {q_H} x = x H$
Then from Quotient Group Epimorphism is Epimorphism, $H$ is the kernel of $q_H$.
From Kernel of Homomorphism on Cyclic Group, $H$ is a cyclic group.
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $9$
Category:Subgroup of Cyclic Group is Cyclic
Subgroup of Direct Product is not necessarily Direct Product of Subgroups

== Theorem ==
Let $G$ and $H$ be groups.
Let $G \times H$ denote the direct product of $G$ and $H$.
Let $K$ be a subgroup of $G \times H$.
Then it is not necessarily the case that $K$ is of the form:
:$G' \times H'$
where:
:$G'$ is a subgroup of $G$
:$H'$ is a subgroup of $H$.
== Proof ==
Let $G = H = C_2$, the cyclic group of order $2$.
Let $G = \gen x$ and $H = \gen y$, so that:
:$G = \set {e_G, x}$
:$H = \set {e_H, y}$
where $e_G$ and $e_H$ are the identity elements of $G$ and $H$ respectively.
Consider the element $\tuple {x, y} \in G \times H$.
We have that:
:$\gen {\tuple {x, y} } =\set {\tuple {e_G, e_H}, \tuple {x, y} }$
but this is not the direct product of a subgroup of $G$ with a subgroup of $H$.
== Sources ==
* : Chapter $13$: Direct products: Remark $2$
Category:Group Direct Products
Subgroup of Elements whose Order Divides Integer

== Theorem ==
Let $A$ be an abelian group.
Let $k \in \Z$ and $B$ be a set of the form:
:$\left\{{x \in A : x^k = e}\right\}$
Then $B$ is a subgroup of $A$.
== Proof ==
First note that the identity $e$ satisfies $e^k = e$ and so $B$ is non-empty.
Now assume that $a, b \in B$.
Then:
Hence, by the One-Step Subgroup Test, $B$ is a subgroup of $A$.
Category:Abelian Groups
Subgroup of Finite Cyclic Group is Determined by Order

== Theorem ==
Let $G = \gen g$ be a cyclic group whose order is $n$ and whose identity is $e$.
Let $d \divides n$, where $\divides$ denotes divisibility.
Then there exists exactly one subgroup $G_d = \gen {g^{n / d} }$ of $G$ with $d$ elements.
== Proof ==
Let $G$ be generated by $g$, such that $\order g = n$.
From Number of Powers of Cyclic Group Element, $g^{n/d}$ has $d$ distinct powers.
Thus $\gen {g^{n / d} }$ has $d$ elements.
Now suppose $H$ is another subgroup of $G$ of order $d$.
Then by Subgroup of Cyclic Group is Cyclic, $H$ is cyclic.
Let $H = \gen y$ where $y \in G$.
Thus $\order y = d$.
Thus $\exists r \in \Z: y = g^r$.
Since $\order y = d$, it follows that $y^d = g^{r d} = e$.
From Equal Powers of Finite Order Element:
: $n \divides r d$
Thus:
Thus $y$ is a power of $g^{n / d}$.
Hence $H$ is a subgroup of $\gen {g^{n / d} }$.
Since both $H$ and $\gen {g^{n / d} }$ have order $d$, they must be equal.
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $9$
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.8$
* : $\S 1.9$: Theorem $16$
* : $\S 39$. Cyclic Groups: Worked Example
* : $\S 3.4$: Cyclic groups: Theorem $1$
* : Chapter $4$: Subgroups: Proposition $4.14$
Category:Finite Cyclic Groups
Category:Subgroups
Subgroup of Free Group is Free Group

== Theorem ==
Let $G$ be a free group.
Let $H \le G$ be a subgroup of $G$ such that $H$ is not the trivial group.
Then $H$ is also a free group.
== Proof ==
Let $x \in H$ such that $x$ is not the identity element.
Because $H$ is not the trivial group, such an $x$ always exists.
As $H$ is a subgroup of $G$:
:$x \in G$
where, , $G$ is a free group.
By Element of Free Group can be Expressed Uniquely as Finite Product, $x$ can be expressed uniquely in the form: 
:$a^\alpha b^\beta \dotsm r^\sigma$
where:
:adjacent elements $a, b, \ldots, r$ of $a^\alpha b^\beta \dotsm r^\sigma$ are distinct elements of $G$
:$\alpha, \beta, \ldots, \sigma$ are non-zero integers.
Hence, by definition, $H$ is a free group.
== Sources ==
* 
* 
Category:Free Groups
Subgroup of Index 2 contains all Squares of Group Elements

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$ whose index is $2$.
Then:
:$\forall x \in G: x^2 \in H$
== Proof ==
By Subgroup of Index 2 is Normal, $H$ is normal in $G$.
Hence the quotient group $G / H$ exists.
Then we have:
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $18 \ \text {(i)}$
Category:Subgroups
Subgroup of Index 2 is Normal

== Theorem ==
A subgroup of index $2$ is always normal.
== Proof ==
Suppose $H \le G$ such that $\index G H = 2$.
Thus $H$ has two left cosets (and two right cosets) in $G$.
If $g \in H$, then $g H = H = H g$.
If $g \notin H$, then $g H = G \setminus H$ as there are only two cosets and the cosets partition $G$.
For the same reason, $g \notin H \implies H g = G \setminus H$.
That is, $g H = H g$.
The result follows from the definition of normal subgroup.
== Also see ==
* Subgroup of Index Least Prime Divisor is Normal
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.9$
* : $\S 1.10$: Theorem $23$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \gamma$
* : $\S 49.2$ Normal subgroups
* : Chapter $7$: Normal subgroups and quotient groups: Example $7.6$
Category:Normal Subgroups
Subgroup of Index 3 does not necessarily contain all Cubes of Group Elements

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$ whose index is $3$.
Then it is not necessarily the case that:
:$\forall x \in G: x^3 \in H$
== Proof ==
Proof by Counterexample:
Consider $S_3$, the symmetric group on $3$ letters.
From Subgroups of Symmetric Group on 3 Letters, the subgroups of $S_3$ are:
subsets of $S_3$ which form subgroups of $S_3$ are:
One such subgroup of $G$ whose index is $3$ is $\set {e, \tuple {12} }$
But $\set {e, \tuple {12} }$ does not contain $\tuple {123}$ or $\tuple {132}$, both of which are of order $3$.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $18 \ \text {(ii)}$
Category:Subgroups
Subgroup of Infinite Cyclic Group is Infinite Cyclic Group

== Theorem ==
Let $G = \gen a$ be an infinite cyclic group generated by $a$, whose identity is $e$.
Let $g \in G, g \ne e: \exists k \in \Z, k \ne 0: g = a^k$.
Let $H = \gen g$.
Then $H \le G$ and $H \cong G$.
Thus, all non-trivial subgroups of an infinite cyclic group are themselves infinite cyclic groups.
A subgroup of $G = \gen a$ is denoted as follows:
:$n G := \gen {a^n}$
This notation is usually used in the context of $\struct {\Z, +}$, where $n \Z$ is (informally) understood as the set of integer multiples of $n$.
== Proof ==
The fact that $H \le G$ follows from the definition of subgroup generator.
By Infinite Cyclic Group is Isomorphic to Integers:
:$G \cong \struct {\Z, +}$
Now we show that $H$ is of infinite order.
Suppose $\exists h \in H, h \ne e: \exists r \in \Z, r > 0: h^r = e$.
But:
:$h \in H \implies \exists s \in \Z, s > 0: h = g^s$
where $g = a^k$.
Thus:
:$e = h^r = \paren {g^s}^r = \paren {\paren {a^k}^s}^r = a^{k s r}$
and thus $a$ is of finite order.
This would mean that $G$ was also of finite order.
So $H$ must be of infinite order.
From Subgroup of Cyclic Group is Cyclic, as $G$ is cyclic, then $H$ must also be cyclic.
From Infinite Cyclic Group is Isomorphic to Integers:
:$H \cong \struct {\Z, +}$
Therefore, as $G \cong \struct {\Z, +}$:
:$H \cong G$
== Comment ==
The interesting thing to note here is that a non-trivial subgroup of an infinite group is itself isomorphic to the group of which it is a subgroup.
This can be compared with the result Infinite Set is Equivalent to Proper Subset.
== Sources ==
* : $\S 1.9$: Theorem $15$
* : $\S 3.4$: Cyclic groups: Theorem $1$
* : Chapter $5$: Cosets and Lagrange's Theorem: Example $5.7$
Category:Subgroups
Category:Infinite Cyclic Group