nyuuzyou/wikis huggingface dataset:
Quotient Epimorphism Condition for Normal Subgroup Product to be Internal Group Direct Product/Sufficient Condition

== Theorem ==
Let $\struct {G, \odot}$ be a group
Let $\struct {H, \odot}$ and $\struct {K, \odot}$ be normal subgroups of $\struct {G, \odot}$.
Let $\struct {G, \odot}$ be the internal group direct product of $\struct {H, \odot}$ and $\struct {K, \odot}$.
Then:
:the restriction of the quotient epimorphism $q_H$ to $K$ is an isomorphism from $K$ onto the quotient group $G / H$
and:
:the restriction of the quotient epimorphism $q_K$ to $H$ is an isomorphism from $H$ onto the quotient group $G / K$.
== Proof ==
Recall that the quotient epimorphism $q_H: G \to G / H$ is defined as:
:$\forall x \in G: \map {q_H} x = x \odot H$
From Restriction of Homomorphism is Homomorphism:
:the restriction of $q_H$ to $K$ is a homomorphism.
From Lagrange's Theorem (Group Theory):
:$\card {G / H} = \card K$
Then we have:
But by definition of internal group direct product, a representation of $g \in G$ in the form $h \odot k$ where $h \in H$ and $k \in K$ is unique.
That is:
Hence the restriction of $q_H$ to $K$ is an injection.
Hence by Injection to Equivalent Finite Set is Bijection, it follows that:
:$q_H$ is a bijection.
A bijective homomorphism is an isomorphism by definition.
Similarly, the quotient epimorphism $q_K: G \to G / K$ is defined as:
:$\forall x \in G: \map {q_K} x = x \odot K = K \odot x$
as $K$ is a normal subgroup of $G$
From Restriction of Homomorphism is Homomorphism:
:the restriction of $q_K$ to $H$ is a homomorphism.
From Lagrange's Theorem (Group Theory):
:$\card {G / K} = \card H$
Then we have:
But by definition of internal group direct product, a representation of $g \in G$ in the form $h \odot k$ where $h \in H$ and $k \in K$ is unique.
That is:
Hence the restriction of $q_K$ to $H$ is an injection.
Hence by Injection to Equivalent Finite Set is Bijection, it follows that:
:$q_K$ is a bijection.
A bijective homomorphism is an isomorphism by definition.
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.9 \ \text{(a)}$
Category:Quotient Epimorphism Condition for Normal Subgroup Product to be Internal Group Direct Product
Quotient Epimorphism from Integers by Principal Ideal

== Theorem ==
Let $m$ be a strictly positive integer.
Let $\ideal m$ be the principal ideal of $\Z$ generated by $m$.
The restriction to $\N_m$ of the quotient epimorphism $q_m$ from the ring $\struct {\Z, +, \times}$ onto $\struct {\Z, +, \times} / \ideal m$ is an isomorphism from the ring $\struct {\N_m, +_m, \times_m}$ of integers modulo $m$ onto the quotient ring $\struct {\Z, +, \times} / \ideal m$.
In particular, $\struct {\Z, +, \times} / \ideal m$ has $m$ elements.
== Proof ==
Let $x, y \in \N_m$.
By the Division Theorem:
Then $x +_m y = r$ and $x \times_m y = s$, so:
and similarly:
:$\map {q_m} {x \times_m y} = \map {q_m} {x y} = \map {q_m} x \map {q_m} y$
So the restriction of $q_m$ to $\N_m$ is a homomorphism from $\struct {\N_m, +_m, \times_m}$ into $\struct {\Z / \ideal m, +_{\ideal m}, \times_{\ideal m} }$.
Let $a \in \Z$.
Then:
:$\exists q, r \in \Z: a = q m + r: 0 \le r < m$
so:
:$\map {q_m} a = \map {q_m} r \in q_m \sqbrk {\N_m}$
Therefore:
:$\Z / \ideal m = q_m \sqbrk \Z = q_m \sqbrk {\N_m}$
Therefore the restriction of $q_m$ to $\N_m$ is surjective.
If $0 < r < m$, then $r \notin \ideal m$ and thus $\map {q_m} r \ne 0$.
Thus the kernel of the restriction of $q_m$ to $\N_m$ contains only zero.
Therefore by the Quotient Theorem for Group Epimorphisms, the restriction of $q_m$ to $\N_m$ is an isomorphism from $\N_m$ to $\Z / \ideal m$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.4$
Category:Modulo Arithmetic
Category:Ideal Theory
Category:Quotient Epimorphisms
Quotient Epimorphism is Epimorphism/Group

== Theorem ==
Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
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 $q_N$ is a group epimorphism whose kernel is $N$.
== Proof ==
The proof follows from Quotient Mapping on Structure is Epimorphism.
When $N \lhd G$, we have:
Therefore $q$ is a homomorphism.
We have that:
:$\forall x \in G: x N \in G / N = \map {q_N} x$
so $q$ is surjective.
Therefore $q$ is an epimorphism.
Let $x \in G$.
thus proving that $\map \ker {q_N} = N$ from definition of subset.
== Motivation ==
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $6$
* : $\S 1.10$: Theorem $24$
* : Chapter $\text{II}$: Groups: Quotient Groups: Theorem $3$
* : $\S 2.2$: Homomorphisms
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 60 \beta$
* : $\S 50.5$ Quotient groups
* : Chapter $8$: The Homomorphism Theorem: Remark
Category:Quotient Epimorphism is Epimorphism
Category:Quotient Groups
Category:Group Epimorphisms
Category:Quotient Epimorphisms
Quotient Epimorphism is Epimorphism/Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$.
Let $\phi: R \to R / J$ be the quotient (ring) epimorphism from $R$ to $R / J$:
:$x \in R: \map \phi x = x + J$
Then $\phi$ is a ring epimorphism whose kernel is $J$.
== Proof ==
Let $x, y \in R$.
Then:
and:
Thus $\phi$ is a homomorphism.
$\phi$ is surjective because:
:$\forall x + J \in R / J: x + J = \map \phi x$
Therefore $\phi$ is an epimorphism.
Let $x \in \map \ker \phi$.
Then:
Thus:
:$\map \ker \phi = J$
== Sources ==
* : Chapter $5$: Rings: $\S 24$. Homomorphisms: Example $46$
* : $\S 2.2$: Homomorphisms: Lemma $2.7$
* : $\S 60.2$ Factor rings: $\text{(ii)}$
Category:Quotient Epimorphism is Epimorphism
Category:Quotient Rings
Category:Quotient Epimorphisms
Category:Kernels of Ring Homomorphisms
Quotient Group by Intersection of Normal Subgroups not necessarily Cyclic if Quotient Groups are

== Theorem ==
Let $G$ be a group.
Let $N$ and $K$ be normal subgroups of $G$.
Let the quotient groups $G / N$ and $G / K$ be cyclic.
Then the quotient group $G / \paren {N \cap K}$ is not necessarily cyclic.
== Proof ==
Proof by Counterexample:
Let $G = \set {e, a, b, c}$ be the Klein $4$-group whose identity element is $e$.
Let $N = \set {e, a}$ and $K = \set {e, b}$.
By Subgroups of Klein Four-Group, both $N$ and $K$ are subgroups of $G$.
By Prime Group is Cyclic, both $N$ and $K$ are cyclic.
By Subgroup of Abelian Group is Normal, both $N$ and $K$ are normal in $G$.
Then we have that:
:$N \cap K = e$
and so by Trivial Quotient Group is Quotient Group:
:$G / \paren {N \cap K} \cong G$
But $G$ is the Klein $4$-group, which is not cyclic.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $16$
Category:Quotient Groups
Quotient Group is Abelian iff All Commutators in Divisor

== Theorem ==
Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $G / N$ be the quotient group of $G$ by $N$.
Then the quotient group $G / N$ is abelian :
:$\forall x, y \in G: \sqbrk {x, y} \in N$
where $\sqbrk {x, y}$ denotes the commutator of $x$ and $y$.
That is, $x y x^{-1} y^{-1} \in N$ for all $x, y \in G$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.13$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47 \zeta$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $16$
Category:Group Commutators
Category:Quotient Groups
Category:Abelian Groups
Category:Quotient Group is Abelian iff All Commutators in Divisor
Quotient Group is Abelian iff All Commutators in Divisor/Proof 1

== Theorem ==
== Proof ==
Let $x, y \in G$.
First we establish the following:
=== Sufficient Condition ===
Let $G / N$ be abelian.
Then by definition:
:$\forall x, y \in G: \paren {x N} \paren {y N} = \paren {y N} \paren {x N}$
and it follows from $(1)$ that:
:$\forall x, y \in G: \sqbrk {x, y} \in N$
=== Necessary Condition ===
Conversely, let:
:$\forall x, y \in G: \sqbrk {x, y} \in N$
Again it follows from $(1)$ that:
:$\forall x, y \in G: \paren {x N} \paren {y N} = \paren {y N} \paren {x N}$
That is, that $G / N$ is abelian.
Category:Quotient Group is Abelian iff All Commutators in Divisor
Quotient Group is Abelian iff All Commutators in Divisor/Proof 2

== Theorem ==
== Proof ==
Let $G / N$ be abelian.
Then:
The argument reverses.
Suppose that $\forall a, b \in G: a b a^{-1} b^{-1} \in N$.
Hence the result.
Category:Quotient Group is Abelian iff All Commutators in Divisor
Quotient Group is Group

== Theorem ==
Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Then the quotient group $G / N$ is indeed a group.
=== Corollary ===
== Proof ==
By Subgroup is Normal iff Left Cosets are Right Cosets, the set of left cosets for $N$ equals the set of right cosets.
It follows that $G / N$ does not depend on whether left cosets are used to define it or right cosets.
, we will work with the left cosets.
By definition of quotient group, the elements of $G / N$ are the cosets of $N$ in $G$, where the group operation is defined as:
:$\paren {a N} \paren {b N} = \paren {a b} N$
The operation has been shown in Coset Product is Well-Defined to be a well-defined operation.
Now we need to demonstrate that $G / N$ is a group.
=== ===
This follows from Coset Product is Well-Defined.
As $a b \in G$, it follows that $\paren {a b} N$ is a left coset.
Thus $G / N$ is closed.
=== ===
The associativity of coset product follows directly from Subset Product within Semigroup is Associative:
Thus $G / N$ is associative.
=== ===
The left coset $e N = N$ serves as the identity:
Similarly $\paren {x N} N = x N$.
=== ===
We have $\paren {x N}^{-1} = x^{-1} N$:
Similarly $\paren {x^{-1} N} \paren {x N} = N$.
Thus $x^{-1} N$ is the inverse of $x N$.
All the group axioms are seen to be fulfilled, and $G / N$ has been shown to be a group.
== Also see ==
From Subgroup is Normal iff Left Cosets are Right Cosets, the left coset space of a normal subgroup is equal to its right coset space.
It follows that $G / N$ does not depend on whether left cosets are used to define it or right cosets.
Thus we do not need to distinguish between the left quotient group and the right quotient group - the two are one and the same.
== Sources ==
* : $\S 6.7$. Quotient groups
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.4$
* : $\S 1.10$: Theorem $24$
* : Chapter $\text{II}$: Groups: Quotient Groups: Theorem $3$
* : $\S 2.2$: Homomorphisms
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47$
* : $\S 50.3$ Quotient groups
* : Chapter $7$: Normal subgroups and quotient groups: Proposition $7.11$
Category:Quotient Groups
Quotient Group is Group/Corollary

== Corollary to Quotient Group is Group ==
Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
If $G$ is finite, then:
:$\index G N = \order {G / N}$
== Proof ==
From Quotient Group is Group, $G / N$ is a group.
From Lagrange's Theorem, we have:
:$\index G N = \dfrac {\order G} {\order N}$
From the definition of quotient group:
:$\order {G / N} = \dfrac {\order G} {\order N}$
Hence the result.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47$
* : $\S 50.4 \ \text{(iii)}$ Quotient groups
Category:Quotient Groups
Quotient Group is Subgroup of Power Structure of Group

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\struct {N, \circ}$ be a normal subgroup of $\struct {G, \circ}$.
Then $\struct {G / N, \circ_N}$ is a subgroup of $\struct {\powerset G, \circ_\PP}$, where:
:$\struct {G / N, \circ_N}$ is the quotient group of $G$ by $N$
:$\struct {\powerset G, \circ_\PP}$ is the power structure of $\struct {G, \circ}$.
== Proof ==
Follows directly from:
* Quotient Group is Group
* Cosets of $G$ by $N$ are subsets of $G$ and therefore elements of $\powerset G$
* The operation $\circ_N$ is defined as the subset product of cosets.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.4$
Category:Quotient Groups
Category:Power Structures
Quotient Group of Abelian Group is Abelian

== Theorem ==
Let $G$ be an abelian group.
Let $N \le G$.
Then the quotient group $G / N$ is abelian.
== Proof ==
First we note that because $G$ is abelian, from Subgroup of Abelian Group is Normal we have $N \lhd G$.
Thus $G / N$ exists for all subgroups of $G$.
Let $X = x N, Y = y N$ where $x, y \in G$.
From the definition of coset product:
Thus $G / N$ is abelian.
Category:Quotient Groups
Quotient Group of Cyclic Group

== Theorem ==
Let $G$ be a cyclic group which is generated by $g$.
Let $H$ be a subgroup of $G$.
Then $g H$ generates $G / H$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47 \gamma$
Category:Cyclic Groups
Category:Quotient Groups
Category:Quotient Group of Cyclic Group
Quotient Group of Cyclic Group/Proof 1

== Theorem ==
== Proof ==
Let $G$ be a cyclic group generated by $g$.
Let $H \le G$.
We need to show that every element of $G / H$ is of the form $\left({g H}\right)^k$ for some $k \in \Z$.
Suppose $x H \in G / H$.
Then, since $G$ is generated by $g$, $x = g^k$ for some $k \in \Z$.
But $\left({g H}\right)^k = \left({g^k}\right) H = x H$.
So $g H$ generates $G / H$.
Category:Quotient Group of Cyclic Group
Quotient Group of Cyclic Group/Proof 2

== Theorem ==
== Proof ==
Let $H$ be a subgroup of the cyclic group $G = \gen g$.
Then by Homomorphism of Powers for Integers:
:$\forall n \in \Z: \map {q_H} {g^n} = \paren {\map {q_H} g}^n = \paren {g H}^n$
As $G = \set {g^n: n \in \Z}$, we conclude that:
:$G / H = q_H \sqbrk G = \set {\paren {g H}^n: n \in \Z}$
Thus, by Epimorphism from Integers to Cyclic Group, $g H$ generates $G / H$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.3$
Category:Quotient Group of Cyclic Group
Quotient Group of Direct Products

== Theorem ==
Let $G$ and $G'$ be groups.
Let:
:$H \lhd G$
:$H' \lhd G'$
where $\lhd$ denotes the relation of being a normal subgroup.
Then:
:$\paren {G \times G'} / \paren {H \times H'}$ is isomorphic to $\paren {G / H} \times \paren {G' / H'}$
where:
:$G \times G'$ denotes the group direct product of $H$ and $H'$
:$G / H$ denotes the quotient group of $G$ by $H$.
== Proof ==
Let $\phi_1: G \to G / H$ and $\phi_2: G' \to G' / H'$ be the quotient epimorphisms with $H$ and $H'$ as their kernels, respectively.
Now define a homomorphism $\phi: G \times G' \to \paren {G / H} \times \paren {G' / H'}$ by:
:$\phi = \phi_1 \times \phi_2$
so:
:$\map \phi {x, x'} = \tuple {\map {\phi_1} x, \map {\phi_2} {x'} }$
The kernel of $\phi$ is clearly $H \times H'$, and $\phi$ is surjective.
So by the First Isomorphism Theorem for Groups:
:$\paren {G / H} \times \paren {G' / H'} \cong \paren {G \times G'} / \paren {H \times H'}$
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.14 \ \text{(a)}$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47 \theta$
Category:Quotient Groups
Category:Group Direct Products
Quotient Group of General Linear Group by Special Linear Group

== Theorem ==
Let $\GL {n, \R}$ denote the general linear group of degree $n$ over $\R$.
Let $\SL {n, \R}$ denote the special linear group of degree $n$ over $\R$.
Then the quotient group $\GL {n, \R} / \SL {n, \R}$ is the multiplicative group of real numbers $\struct {\R_{\ne 0}, \times}$.
== Proof ==
Let $\det: \GL {n, \R} \to \struct {\R_{\ne 0}, \times}$ be the group homomorphism:
:$\mathbf A \mapsto \map \det {\mathbf A}$
where $\map \det {\mathbf A}$ is the determinant of $\mathbf A$.
This is demonstrated to be a homomorphism in General Linear Group to Determinant is Homomorphism
From the the corollary to General Linear Group to Determinant is Homomorphism, the kernel of $\phi$ is $\SL {n, \R}$.
Thus from Kernel is Normal Subgroup of Domain, $\SL {n, \R}$ is normal in $\GL {n, \R}$.
From the First Isomorphism Theorem for Groups:
:$\Img \phi \cong \GL {n, \R} / \SL {n, \R}$
By definition, the image of $\phi$ is the multiplicative group of real numbers.
Hence the result.
== Sources ==
* : $\S 52$. The first isomorphism theorem
Category:General Linear Group
Category:Special Linear Group
Category:Examples of Quotient Groups
Quotient Group of Ideal is Coset Space

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +}$ be the quotient group of $\struct {R, +}$ by $\struct {J, +}$.
Then each element of $\struct {R / J, +}$ is a coset of $J$ in $R$, that is, is of the form $x + J = \set {x + j: j \in J}$ for some $x \in R$.
The rule of addition of these cosets is: $\paren {x + J} + \paren {y + J} = \paren {x + y} + J$.
The identity of $\struct {R / J, +}$ is $J$ and for each $x \in R$, the inverse of $x + J$ is $\paren {-x} + J$.
== Proof ==
From Ideal is Additive Normal Subgroup that $J$ is a normal subgroup of $R$ under $+$.
From Quotient Ring Addition is Well-Defined, $+$ is a well-defined operation.
The rest follows directly from the definition of quotient group.
== Sources ==
* : $\S 60$. Factor rings
Category:Ideal Theory
Category:Quotient Groups
Category:Cosets
Quotient Group of Infinite Cyclic Group by Subgroup

== Theorem ==
Let $C_n$ be the cyclic group of order $n$.
Then:
:$C_n \cong \dfrac {\struct {\Z, +} } {\struct {n \Z, +} } = \dfrac \Z {n \Z}$
where:
:$\Z$ is the additive group of integers
:$n \Z$ is the additive group of integer multiples
:$\Z / n \Z$ is the quotient group of $\Z$ by $n \Z$.
Thus, every cyclic group is isomorphic to one of:
:$\Z, \dfrac \Z \Z, \dfrac \Z {2 \Z}, \dfrac \Z {3 \Z}, \dfrac \Z {4 \Z}, \ldots$
== Proof ==
Let $C_n = \gen {a: a^n = e_{C_n} }$, that is, let $a$ be a generator of $C_n$.
Let us define $\phi: \struct {\Z, +} \to C_n$ such that:
:$\forall k \in \Z: \map \phi k = a^k$
Then from the First Isomorphism Theorem:
:$\Img \phi = C_n = \struct {\Z, +} / \map \ker \phi$
We now need to show that $\map \ker \phi = n \Z$.
We have:
:$\map \ker \phi = \set {k \in \Z: a^k = e_{C_n} }$
Let $x \in \map \ker \phi$.
Then $a^x = e_{C_n}$ and thus:
:$n \divides x$
The result follows.
== Sources ==
* : $\S 7.4$. Kernel and image: Example $141$
Category:Infinite Cyclic Group
Category:Quotient Groups
Category:Additive Groups of Integer Multiples
Category:Additive Group of Integers
Quotient Group of Integers by Multiples

== Theorem ==
Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {m \Z, +}$ be the additive group of integer multiples of $m$.
Let $\struct {\Z_m, +_m}$ be the additive group of integers modulo $m$.
Then the quotient group of $\struct {\Z, +}$ by $\struct {m \Z, +}$ is $\struct {\Z_m, +_m}$.
Thus:
:$\index \Z {m \Z} = m$
== Proof ==
From Subgroups of Additive Group of Integers, $\struct {m \Z, +}$ is a subgroup of $\struct {\Z, +}$.
From Subgroup of Abelian Group is Normal, $\struct {m \Z, +}$ is normal in $\struct {\Z, +}$.
Therefore the quotient group $\dfrac {\struct {\Z, +} } {\struct {m \Z, +} }$ is defined.
Now $\Z$ modulo $m \Z$ is Congruence Modulo a Subgroup.
This is merely congruence modulo an integer.
Thus the quotient set $\Z / m \Z$ is $\Z_m$.
The left coset of $k \in \Z$ is denoted $k + m \Z$, which is the same thing as $\eqclass k m$ from the definition of residue class.
So $\index \Z {m \Z} = m$ follows from the definition of Subgroup Index.
== Sources ==
* : $\S 6.1$. The quotient sets of a subgroup: Example $113$
* : $\S 6.3$. Index. Transversals: Example $115$
* : $\S 6.7$. Quotient groups: Example $126$
* : Chapter $\text{II}$: Groups: A Little Number Theory
* : Chapter $2$: Subgroups and Cosets: $\S 37$
* : Chapter $2$: Subgroups and Cosets: $\S 39$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47 \alpha$
* : Chapter $7$: Normal subgroups and quotient groups: Example $7.12$
Category:Examples of Quotient Groups
Category:Modulo Arithmetic
Category:Additive Group of Integers
Category:Additive Groups of Integer Multiples
Category:Additive Groups of Integers Modulo m
Quotient Group of Quadratic Residues Modulo 2 of 2-adic Units

== Theorem ==
Let $\Q_2$ be the $2$-adic numbers.
Let $\Q_2^\times$ denote the set of invertible elements of $\Q_2$.
Let $\paren{\Q_2^\times}^2 = \set{a^2 : a \in \Q_2^\times}$
Then the multiplicative quotient group $\Q_2^\times \mathop/ \paren{\Q_2^\times}^2$ has order $8$ with:
:$\set{1, -1, 5, -5, 2, -2, 10, -10}$ is a transversal
=== Corollary ===
== Proof ==
== Sources ==
* : $\S 1.7$ Hensel's Lemma and Congruences: Exercise $44$
Category:P-adic Units
Quotient Group of Quadratic Residues Modulo 2 of 2-adic Units/Corollary

== Theorem ==
Let $\Q_2$ be the $2$-adic numbers.
Let $\Q_2^\times$ denote the set of invertible elements of $\Q_2$.
Let $\paren{\Q_2^\times}^2 = \set{a^2 : a \in \Q_2^\times}$
Then:
:$\Q_2^\times \mathop/ \paren{\Q_2^\times}^2$ is isomorphic to $\Z \mathop/ 2\Z \oplus \Z \mathop/ 2\Z \oplus \Z \mathop/ 2\Z$ 
== Proof ==
== Sources ==
* : $\S 1.7$ Hensel's Lemma and Congruences: Exercise $44$
Category:P-adic Units
Quotient Group of Quadratic Residues Modulo p of P-adic Units

== Theorem ==
Let $\Q_p$ be the $p$-adic numbers for some prime $p \ne 2$.
Let $\Q_p^\times$ denote the set of invertible elements of $\Q_p$.
Let $\paren{\Q_p^\times}^2 = \set{a^2 : a \in \Q_p^\times}$
Then the multiplicative quotient group $\Q_p^\times \mathop/ \paren{\Q_p^\times}^2$ has order $4$:
:$\exists c \in \Q_p^\times \setminus \paren{\Q_p^\times}^2 : \set{1, p, c, cp}$ is a transversal
=== Corollary ===
== Proof ==
By definition of field:
:$\Q_p^\times = \Q_p \setminus \set{0}$ is an abelian group
From Group of Units is Group:
:$\struct{\Q_p^\times, \times}$ is a subgroup of $\struct{\Q_p^*, \times}$
From Power of Elements is Subgroup:
:$\struct{\paren{\Q_p^\times}^2, \times}$ is a subgroup of $\struct{\Q_p^\times, \times}$
By definition of quotient group, the quotient group $\Q_p^\times \mathop/ \paren{\Q_p^\times}^2$ can be formed.
== Sources ==
* : $\S 1.7$ Hensel's Lemma and Congruences: Exercise $37$
Category:P-adic Units
Quotient Group of Quadratic Residues Modulo p of P-adic Units/Corollary

== Theorem ==
Let $\Q_p$ be the $p$-adic numbers for some prime $p \ne 2$.
Let $\Q_p^\times$ denote the set of invertible elements of $\Q_p$.
Let $\paren{\Q_p^\times}^2 = \set{a^2 : a \in \Q_p^\times}$
Then:
:$\Q_p^\times \mathop/ \paren{\Q_p^\times}^2$ is isomorphic to $\Z \mathop/ 2\Z \oplus \Z \mathop/ 2\Z$ 
== Proof ==
== Sources ==
* : $\S 1.7$ Hensel's Lemma and Congruences: Exercise $37$
Category:P-adic Units