avewright/proofwiki-math huggingface dataset:
Theorem: Centralizer of Ring Subset is Subring

Let $S$ be a subset of a ring $\struct {R, +, \circ}$
Then $\map {C_R} S$, the centralizer of $S$ in $R$, is a subring of $R$.

Proof:
Certainly $0_R \in \map {C_R} S$ as $0_R$ commutes (trivially) with all elements of $R$.
Suppose $x, y \in \map {C_R} S$.
Then:
{{begin-eqn}}
{{eqn | q = \forall s \in S
      | l = s \circ \paren {x + \paren {-y} }
      | r = s \circ x + s \circ \paren {-y}
      | c = Distributivity of $\circ$ over $+$
}}
{{eqn | r = x \circ s + \paren {-y} \circ s
      | c = $x$ and $y$ are in $\map {C_R} S$
}}
{{eqn | r = \paren {x + \paren {-y} } \circ s
      | c = Distributivity of $\circ$ over $+$
}}
{{end-eqn}}
So:
:$x + \paren {-y} \in \map {C_R} s$
Suppose $x, y \in \map {C_R} S$ again.
Then from Element Commutes with Product of Commuting Elements:
:$x \circ y \in \map {C_R} S$
Thus all the conditions are fulfilled for Subring Test, and $\map {C_R} S$ is a subring of $R$.
{{qed}}
Theorem: Center of Ring is Commutative Subring

The center $\map Z R$ of a ring $R$ is a commutative subring of $R$.

Proof:
Follows directly from the definition of center and Centralizer of Ring Subset is Subring.
{{qed}}
Theorem: Ideal is Subring

Let $\struct {R, +, \circ}$ be a ring, and let $J$ be an ideal of $R$.
Then $J$ is a subring of $R$.

Proof:
This follows directly from the definition of an ideal and Subring Test.
{{qed}}
Theorem: Ring is Ideal of Itself

Let $\struct {R, +, \circ}$ be a ring.
Then $R$ is an ideal of $R$.

Proof:
From Ring is Subring of Itself, $\struct {R, +, \circ}$ is a subring of $\struct {R, +, \circ}$.
Also:
:$\forall x, y \in \struct {R, +, \circ}: x \circ y \in R$
thus fulfilling the condition for $R$ to be an ideal of $R$.
{{qed}}
Theorem: Ideal of Unit is Whole Ring

Let $\struct {R, +, \circ}$ be a ring with unity.
Let $J$ be an ideal of $R$.
If $J$ contains a unit of $R$, then $J = R$.

Proof:
Let $u \in J$, where $u \in U_R$.
Also by definition, we have $u^{-1} \in U_R$.
Let $x \in R$.
{{begin-eqn}}
{{eqn | o = 
      | r = x \in R
      | c = 
}}
{{eqn | o = \leadsto
      | r = x \circ u^{-1} \in R
      | c = as $R$ is closed
}}
{{eqn | o = \leadsto
      | r = \paren {x \circ u^{-1} } \circ u \in J
      | c = {{Defof|Ideal of Ring}}
}}
{{eqn | o = \leadsto
      | r = x \in J
      | c = Ring properties: $u \circ u^{-1} = 1_R$
}}
{{end-eqn}}
Thus $R \subseteq J$.
As $J \subseteq R$ by definition, it follows that $J = R$.
{{Qed}}
Theorem: Test for Ideal

Let $J$ be a subset of a ring $\struct {R, +, \circ}$.
Then $J$ is an ideal of $\struct {R, +, \circ}$ {{iff}} these all hold:
:$(1): \quad J \ne \O$
:$(2): \quad \forall x, y \in J: x + \paren {-y} \in J$
:$(3): \quad \forall j \in J, r \in R: r \circ j \in J, j \circ r \in J$

Proof:
=== Necessary Condition ===
Let $J$ be an ideal of $\struct {R, +, \circ}$.
Then conditions $(1)$ to $(3)$ hold by virtue of the ring axioms and $J$ being an ideal.
{{qed|lemma}}
Theorem: Epimorphism Preserves Rings

Let $\struct {R_1, +_1, \circ_1}$ be a ring, and $\struct {R_2, +_2, \circ_2}$ be a closed algebraic structure.
Let $\phi: R_1 \to R_2$ be an epimorphism.
Then $\struct {R_2, +_2, \circ_2}$ is a ring.

Proof:
From Epimorphism Preserves Groups, we have that if $\struct {R_1, +_1}$ is a group then so is $\struct {R_2, +_2}$.
From Epimorphism Preserves Semigroups, we have that if $\struct {R_1, \circ_1}$ is a semigroup then so is $\struct {R_2, \circ_2}$.
From Epimorphism Preserves Distributivity, we have that if $\circ_1$ distributes over $+_1$ then $\circ_2$ distributes over $+_2$.
So it follows from the definition of a ring that if $\struct {R_1, +_1, \circ_1}$ is a ring then so is $\struct {R_2, +_2, \circ_2}$.
{{qed}}
Theorem: Ring Homomorphism of Addition is Group Homomorphism

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Then $\phi: \struct {R_1, +_1} \to \struct {R_2, +_2}$ is a group homomorphism.

Proof:
From the definition of a ring, both $\struct {R_1, +_1}$ and $\struct {R_2, +_2}$ are abelian groups.
The result follows from the definitions of ring homomorphism and group homomorphism.
{{Qed}}
Category:Ring Homomorphisms
Category:Group Homomorphisms
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Theorem: Element of Integral Domain Divides Zero

Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.
Then every element of $D$ divides $0_D$:
:$\forall x \in D: x \divides 0_D$

Proof:
By definition, an integral domain is a ring.
So, from Ring Product with Zero:
:$\forall x \in D: 0_D = x \circ 0_D$
The result follows from the definition of divisor.
{{qed}}
Theorem: Unity Divides All Elements

Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Then unity is a divisor of every element of $D$:
:$\forall x \in D: 1_D \divides x$
Also:
:$\forall x \in D: -1_D \divides x$

Proof:
The element $1_D$ is the unity of $\struct {D, +, \circ}$, and so:
:$1_D \in D: x = 1_D \circ x$
Similarly, from Product of Ring Negatives:
:$-1_D \in D: x = \paren {-1_D} \circ \paren {-x}$
The result follows from the definition of divisor.
{{qed}}
Theorem: Unity Divides All Elements

Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Then unity is a divisor of every element of $D$:
:$\forall x \in D: 1_D \divides x$
Also:
:$\forall x \in D: -1_D \divides x$

Proof:
This is a special case of Unit of Integral Domain divides all Elements, as Unity is Unit.
Furthermore, from Unity and Negative form Subgroup of Units we also have that $-1_D$ is a unit of $D$.
Hence the result.
{{qed}}
Theorem: Element of Integral Domain is Divisor of Itself

Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Then every element of $D$ is a divisor of itself:
:$\forall x \in D: x \divides x$

Proof:
Follows directly from the definition of divisor:
:$\forall x \in D: \exists 1_D \in D: x = 1_D \circ x$
{{qed}}
Category:Integral Domains
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Theorem: Unit of Integral Domain divides all Elements

Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Let $\struct {U_D, \circ}$ be the group of units of $\struct {D, +, \circ}$.
Then:
:$\forall x \in D: \forall u \in U_D: u \divides x$
That is, every unit of $D$ is a divisor of every element of $D$.

Proof:
{{begin-eqn}}
{{eqn | q = \forall x \in D, u \in U_D
      | l = x
      | r = u \circ \paren {u^{-1} \circ x}
      | c = {{Defof|Unit of Ring}}
}}
{{eqn | ll= \leadsto
      | l = u
      | o = \divides
      | r = x
      | c = {{Defof|Divisor of Ring Element}}
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor of Unit is Unit

Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Let $\struct {U_D, \circ}$ be the group of units of $\struct {D, +, \circ}$.
Then:
:$x \in D, u \in U_D: x \divides u \implies x \in U_D$
That is, if $x$ is a divisor of a unit, $x$ must itself be a unit.

Proof:
Let $x \in D, u \in U_D$ such that $x \divides u$.
By definition:
:$\exists t \in D: u = t \circ x$
Thus:
:$1_D = u^{-1} \circ u = u^{-1} \circ t \circ x$
Also, as $D$ is an integral domain and hence a commutative ring, we have:
:$u^{-1} \circ t \circ x = 1_D = x \circ u^{-1} \circ t$
It follows by definition that $x$ is a unit, as it has an inverse, namely $u^{-1} \circ t$.
{{qed}}
Theorem: Associatehood is Equivalence Relation

Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.
Let $\sim$ be the relation defined on $D$ as:
:$\forall x, y \in D: x \sim y$ {{iff}} $x$ is an associate of $y$
Then $\sim$ is an equivalence relation.

Proof:
Checking in turn each of the criteria for equivalence:
Theorem: Divisor Relation in Integral Domain is Transitive

Let $\struct {D, +, \circ}$ be an integral domain.
Let $x, y, z \in D$.
Then:
:$x \divides y \land y \divides z \implies x \divides z$

Proof:
Let $x \divides y \land y \divides z$.
Then from the definition of divisor, we have:
:$x \divides y \iff \exists s \in D: y = s \circ x$
:$y \divides z \iff \exists t \in D: z = t \circ y$
Then:
:$z = t \circ \paren {s \circ x} = \paren {t \circ s} \circ x$
Thus:
:$\exists \paren {t \circ s} \in D: z = \paren {t \circ s} \circ x$
and the result follows.
{{Qed}}
Theorem: Integers form Unique Factorization Domain

The integers $\struct {\Z, +, \times}$ form a unique factorization domain.

Proof:
Follows directly from:
:$(1) \quad$ the fundamental theorem of arithmetic
:$(2) \quad$ the fact that $\struct {\Z, +, \times}$ is an integral domain
:$(3) \quad$ the definitions complete factorization and equivalent factorizations.
{{qed}}
Category:Integral Domains
Category:Integers
Category:Factorization
Category:Unique Factorization Domains
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Theorem: Trivial Ordering Compatibility in Boolean Ring

Let $\struct {S, +, \circ}$ be a Boolean ring.
Then the trivial ordering is the only ordering on $S$ compatible with both its operations.

Proof:
That the trivial ordering is compatible with $\circ$ and $*$ follows from Trivial Ordering is Universally Compatible.
Conversely, suppose that $\preceq$ is an ordering compatible with $\circ$ and $*$.
We recall the definition of the trivial ordering:
{{:Definition:Trivial Ordering}}
Let $a, b \in S$ such that $a \preceq b$.
Since $\preceq$ is compatible with $\circ$ and $*$, we have:
{{begin-eqn}}
{{eqn | l = a
      | o = \preceq
      | r = b
}}
{{eqn | ll= \leadsto
      | l = 0
      | o = \preceq
      | r = b - a
}}
{{end-eqn}}
We have {{afortiori}} that a Boolean ring is an idempotent ring.
Hence we have:
{{begin-eqn}}
{{eqn | l = a
      | o = \preceq
      | r = b
}}
{{eqn | ll= \leadsto
      | l = a - b
      | o = \preceq
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = b - a
      | o = \preceq
      | r = 0
      | c = {{Corollary|Idempotent Ring has Characteristic Two}}
}}
{{end-eqn}}
We have shown that:
:$0 \preceq b - a$
and:
:$b - a \preceq 0$
By definition of ordering, $\preceq$ is antisymmetric.
This means:
:$b - a = 0$
and so:
:$a = b$
Hence $\preceq$ is the trivial ordering.
{{qed}}
Theorem: Smallest Field is Field

The ring $\struct {\set {0_R, 1_R}, +, \circ}$ is the smallest algebraic structure which is a field.

Proof:
From Field Contains at least 2 Elements, a field must contain at least two elements.
Hence the null ring, which contains one element, is ''not'' a field.
For $\struct {\set {0_R, 1_R}, +, \circ}$ to be a field:
:$\struct {\set {0_R, 1_R}, +}$ must be an abelian group.
This is fulfilled as this is the parity group.
:$\struct {\set {0_R, 1_R}, \circ}$ must be a commutative division ring.
This is fulfilled, as $\struct {\set {0_R, 1_R}^*, \circ} = \struct {\set {1_R}, \circ}$ is the trivial group.
:$\circ$ needs to distribute over $+$. This follows directly from Ring Product with Zero and the behaviour of the identity element in a group.
{{qed}}
Theorem: Field is Integral Domain

Every field is an integral domain.

Proof:
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Suppose $\exists x, y \in F: x \circ y = 0_F$.
Suppose $x \ne 0_F$.
Then, by the definition of a field, $x^{-1}$ exists in $F$ and:
:$y = 1_F \circ y = x^{-1} \circ x \circ y = x^{-1} \circ 0_F = 0_F$.
Otherwise $x = 0_F$.
So if $x \circ y = 0_F$, either $x = 0_F$ or $y = 0_F$ as we were to show.
{{qed}}
Theorem: Field is Integral Domain

Every field is an integral domain.

Proof:
This result follows directly from:
:Field has no Proper Zero Divisors
:By definition, that a field is a commutative ring.
{{qed}}
Theorem: Finite Integral Domain is Galois Field

A finite integral domain is a Galois field.

Proof:
Let $\struct {R, +, \circ}$ be a finite integral domain whose unity is $1$ and whose zero is $0$.
Let $a \in R$ be arbitrary such that $a \ne 0$.
We wish to show that $a$ has a product inverse in $R$.
Consider the map $f: R \to R$ defined by $f: x \mapsto a \circ x$.
We first show that the kernel of $f$ is the null ring.
We have:
{{begin-eqn}}
{{eqn | l = \map \ker f
      | r =  \set {x \in R: \map f x = 0}
      | c = {{Defof|Kernel of Ring Homomorphism}}
}}
{{eqn | r = \set {x \in R: a \circ x = 0}
      | c = Definition of $f$
}}
{{end-eqn}}
By definition of integral domain, $R$ has no proper zero divisors.
Thus:
:$a \circ x = 0 \implies a = 0 \text { or } x = 0$
Because {{hypothesis}} $a \ne 0$, it follows that:
:$x = 0$
Therefore:
:$\map \ker f = \set 0$
Hence by Kernel is Trivial iff Monomorphism, $f$ is injective.
Next, the Pigeonhole Principle gives us that an injective mapping from a finite set onto itself is surjective.
Since $R$ is finite, the mapping $f$ is surjective.
Finally, since $f$ is surjective and $1 \in R$, we have:
:$\exists x \in R: \map f x = a \circ x = 1$
Thus $x$ is the right inverse of $a$ with respect to the ring product.
As $a$ is arbitrary, it follows that every element in $R \setminus \set 0$ has a right inverse for $\circ$.
By definition of integral domain, $\struct {R \setminus \set 0, \circ}$ is a monoid. 
Thus by Right Inverse for All is Left Inverse it follows that $x$ is also a left inverse of $a$, that is:
:$a \circ x = 1 \implies x \circ a = 1$
It follows that $x$ is the product inverse of $a$ and the proof is complete.
{{qed}}
Theorem: Finite Integral Domain is Galois Field

A finite integral domain is a Galois field.

Proof:
Let $R$ be a finite integral domain with unity $1$ and zero $0$.
Let $R^*$ denote $R \setminus \set 0$, the set $R$ without the zero.
As $R$ is finite, we may enumerate the elements of $R$ as:
:$x_0 = 0, x_1 = 1, x_2, x_3, \ldots, x_n$
Let $R^*$.
Consider the elements:
:$x_k x_1, x_k x_2, \ldots, x_k x_n$ 
which are members of $R$ by closure of the ring product.
Since $R$ is an integral domain, by definition it has no proper zero divisors.
Therefore all of $x_k x_1, x_k x_2, \ldots, x_k x_n$ are elements of $R^*$.
Recall:
:$x_k x_p = x_k x_q \implies x_p = x_q$ 
by the Cancellation Law for Ring Product of Integral Domain.
So the list of products consists of $n$ distinct elements and is therefore equal to the whole $R^*$.
Since $1 \in R^*$, we have that $x_k x_{k^*} = 1$ for some $k^*$.
Therefore, there exists an inverse for every non-zero element, so $R$ is a field.
As the underlying set of $R$ is finite, $R$ is by definition a Galois field.
{{qed}}
Theorem: Finite Integral Domain is Galois Field

A finite integral domain is a Galois field.

Proof:
An integral domain is by definition a ring which has no proper zero divisors.
By definition, a Galois field is a field whose underlying set is finite.
The result follows from Finite Ring with No Proper Zero Divisors is Field.
{{qed}}
Theorem: Finite Integral Domain is Galois Field

A finite integral domain is a Galois field.

Proof:
{{AimForCont}} $\struct {D, +, \circ}$ is a finite integral domain which is not a field.
From Non-Field Integral Domain has Infinite Number of Ideals, $\struct {D, +, \circ}$ has an infinite number of distinct ideals.
But this contradicts the assertion that $\struct {D, +, \circ}$ is finite.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Field is Subfield of Itself

Let $\struct {F, +, \circ}$ be a field.
Then $\struct {F, +, \circ}$ is a subfield of $\struct {F, +, \circ}$.

Proof:
$F$ is a field and $F \subseteq F$ from Set is Subset of Itself.
{{Qed}}
Theorem: Center of Division Ring is Subfield

Let $\struct {K, +, \circ}$ be an division ring.
Let $\map Z K$ be the center of $K$.
Then $\map Z K$ is a subfield of $K$.

Proof:
For $\map Z K$ to be a subfield of $K$, it needs to be a division ring that is commutative.
Thus the result follows directly from Center of Ring is Commutative Subring.
{{qed}}
Theorem: Ideals of Field

Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Then $\struct {R, +, \circ}$ is a field {{iff}} the only ideals of $\struct {R, +, \circ}$ are $\struct {R, +, \circ}$ and $\set {0_R}$.

Proof:
=== Necessary Condition ===
Let $\struct {R, +, \circ}$ be a field.
The result follows from Field has 2 Ideals.
{{qed|lemma}}
Theorem: Epimorphism from Division Ring to Ring

Let $\struct {K, +, \circ}$ be a division ring whose zero is $0_K$.
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\phi: K \to R$ be a ring epimorphism.
Then one of the following applies:
:$(1): \quad R$ is a null ring
:$(2): \quad R$ is a division ring and $\phi$ is a ring isomorphism.

Proof:
We have that the kernel of $K$ is an ideal.
From Ideals of Division Ring, $\map \ker K$ must therefore either be $0_K$ or $K$.
Let $\map \ker K = 0_K$.
Then by Ring Epimorphism with Trivial Kernel is Isomorphism $\phi$ is an ring isomorphism.
Thus $R$ a division ring like $K$.
Let $\map \ker K = K$.
Then:
:$\forall x \in K: \map \phi x = 0_R$
As $\phi$ is an epimorphism, it is surjective.
Therefore:
:$R = \set {0_R}$
Hence the result.
{{qed}}
Theorem: Cosets are Equivalent

All left cosets of a group $G$ with respect to a subgroup $H$ are equivalent.
That is, any two left cosets are in one-to-one correspondence.
The same applies to right cosets.
As a special case of this:
:$\forall x \in G: \order {x H} = \order H = \order {H x}$
where $H$ is a subgroup of $G$.

Proof:
Let us set up mappings $\theta: H \to H x$ and $\phi: H x \to H$ as follows:
:$\forall u \in H: \map \theta u = u x$
:$\forall v \in H x: \map \phi v = v x^{-1}$
From Element in Right Coset iff Product with Inverse in Subgroup:
:$v \in H x \implies v x^{-1} \in H$
Now:
:$\forall v \in H x: \theta \circ \map \phi v = v x^{-1} x = v$
:$\forall u \in H: \phi \circ \map \theta u = u x x^{-1} = u$
Thus $\theta \circ \phi = I_{H x}$ and $\phi \circ \theta = I_H$ are identity mappings.
So $\theta = \phi^{-1}$: both are bijections and one is the inverse of the other.
This establishes, for each $x \in G$, the set equivalence between $H$ and $H x$:
:$H \simeq H x$
In particular, for any $x, y \in G$, it follows from $H x \simeq H$ and $H y \simeq H$ that:
:$H x \simeq H y$
by Set Equivalence behaves like Equivalence Relation.
Similarly, we can set up mappings $\alpha: H \to x H$ and $\beta: x H \to H$ as follows:
:$\forall u \in H: \map \alpha u = x u$
:$\forall v \in x H: \map \beta v = x^{-1} v$
Analogous to above reasoning gives $\alpha = \beta^{-1}$ which establishes $H \simeq x H$. 
Also similarly, $x H \simeq y H$ for all $x, y \in G$.
Hence the result.
{{qed}}
Theorem: Cosets are Equivalent

All left cosets of a group $G$ with respect to a subgroup $H$ are equivalent.
That is, any two left cosets are in one-to-one correspondence.
The same applies to right cosets.
As a special case of this:
:$\forall x \in G: \order {x H} = \order H = \order {H x}$
where $H$ is a subgroup of $G$.

Proof:
Follows directly from Set Equivalence of Regular Representations.
{{qed}}
Theorem: Left and Right Coset Spaces are Equivalent

Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$.
Let:
:$x H$ denote the left coset of $H$ by $x$
:$H y$ denote the right coset of $H$ by $y$.
Then:
:$\order {\set {x H: x \in G} } = \order {\set {H y: y \in G} }$

Proof:
Let there be exactly $r$ different left cosets of $H$ in $G$.
Let a complete repetition-free list of these left cosets be:
:$a_1 H, a_2 H, a_3 H \ldots, a_r H: a_1, a_2, \ldots, a_r \in G$
From Left Coset Space forms Partition, every element of $G$ is contained in exactly one of the left cosets.
Let $x \in G$.
Then, for $1 \le i \le r$:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = H a_i^{-1}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = x \paren {a_i^{-1} }^{-1}
      | o = \in
      | r = H
      | c = Element in Right Coset iff Product with Inverse in Subgroup
}}
{{eqn | ll= \leadstoandfrom
      | l = x a_i
      | o = \in
      | r = H
      | c = Inverse of Group Inverse
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {x^{-1} }^{-1} a_i
      | o = \in
      | r = H
      | c = Inverse of Group Inverse
}}
{{eqn | ll= \leadstoandfrom
      | l = x^{-1}
      | o = \in
      | r = a_i H
      | c = Element in Left Coset iff Product with Inverse in Subgroup
}}
{{end-eqn}}
Since $x^{-1} \in a_i H$ is true for precisely one value of $i$, it follows that $x \in H a_i^{-1}$ is also true for precisely that value of $i$.
So there are exactly $r$ different right cosets of $H$ in $G$, and a complete repetition-free list of these is:
:$H a_1^{-1}, H a_2^{-1}, H a_3^{-1}, \ldots, H a_r^{-1}$
The result follows.
{{qed}}
Theorem: Left and Right Coset Spaces are Equivalent

Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$.
Let:
:$x H$ denote the left coset of $H$ by $x$
:$H y$ denote the right coset of $H$ by $y$.
Then:
:$\order {\set {x H: x \in G} } = \order {\set {H y: y \in G} }$

Proof:
Let $G$ be a group and let $H \le G$.
Consider the mapping $\phi$ from the left coset space to the right coset space defined as:
:$\forall g \in G: \map \phi {g H} = H g^{-1}$
We need to show that $\phi$ is a bijection.
First we need to show that $\phi$ is well-defined.
That is, that $a H = b H \implies \map \phi {a H} = \map \phi {b H}$.
Suppose $a H = b H$.
{{begin-eqn}}
{{eqn | l = a H = b H
      | o = \iff
      | r = a^{-1} b \in H
      | c = Left Cosets are Equal iff Product with Inverse in Subgroup
}}
{{eqn | l = H a^{-1} = H b^{-1}
      | o = \iff
      | r = a^{-1} \paren {b^{-1} }^{-1} \in H
      | c = Right Cosets are Equal iff Product with Inverse in Subgroup
}}
{{end-eqn}}
But $a^{-1} \paren {b^{-1} }^{-1} = a^{-1} b \in H$ as $a H = b H$.
So $H a^{-1} = H b^{-1}$ and $\phi$ is well-defined.
Next we show that $\phi$ is injective:
Suppose $\exists x, y \in G: \map \phi {x H} = \map \phi {y H}$.
Then $H x^{-1} = H y^{-1}$, so $x^{-1} = e_G x^{-1} = h y^{-1}$ for some $h \in H$.
Thus $h = x^{-1} y \implies h^{-1} = y^{-1} x$.
As $H$ is a subgroup, $h^{-1} \in H$.
Thus:
: $y^{-1} x \in H$
So by Left Cosets are Equal iff Product with Inverse in Subgroup:
:$x H = y H$
Thus $\phi$ is injective.
Next we show that $\phi$ is surjective:
Let $H x$ be a right coset of $H$ in $G$.
Since $x = \paren {x^{-1} }^{-1}$, $H x = \map \phi {x^{-1} H}$ and so $\phi$ is surjective.
Thus $\phi$ constitutes a bijection from the left coset space to the right coset space, and the result follows.
{{qed}}
Theorem: Cosets in Abelian Group

Let $G$ be an abelian group.
Then every right coset modulo $H$ is a left coset modulo $H$.
That is:
:$\forall x \in G: x H = H x$
In an abelian group, therefore, we can talk about '''congruence modulo $H$''' and not worry about whether it is left or right.

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = \forall x, y \in G: x^{-1} y = y x^{-1}
      | c = 
}}
{{eqn | o = \leadsto
      | r = \paren {x \equiv^l y \pmod H \iff y \equiv^r x \pmod H}
      | c = {{Defof|Congruence Modulo Subgroup}}
}}
{{eqn | o = \leadsto
      | r = \paren {x \equiv^l y \pmod H \iff x \equiv^r y \pmod H}
      | c = Congruence Modulo Subgroup is Equivalence Relation, therefore Symmetric
}}
{{end-eqn}}
{{qed}}
Theorem: Conjugate of Set by Identity

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $S \subseteq G$.
Then the conjugate of $S$ by $e$ is $S$:
:$S^e = S$

Proof:
{{begin-eqn}}
{{eqn | l = S^e
      | r = \set {y \in G: \exists x \in S: y = e \circ x \circ e^{-1} }
      | c = {{Defof|Conjugate of Group Subset}}
}}
{{eqn | r = \set {y \in G: \exists x \in S: y = x}
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = S
      | c=
}}
{{end-eqn}}
{{Qed}}