nyuuzyou/wikis huggingface dataset:
Identity Matrix from Upper Triangular Matrix

== Theorem ==
Let $\mathbf A = \sqbrk a_{m n}$ be an upper triangular matrix of order $m \times n$ with no zero diagonal elements.
Let $k = \min \set {m, n}$.
Then $\mathbf A$ can be transformed into a matrix such that the first $k$ rows and columns form the unit matrix of order $k$.
== Proof ==
By definition of $k$:
:if $\mathbf A$ has more rows than columns, $k$ is the number of columns of $\mathbf A$.
:if $\mathbf A$ has more columns than rows, $k$ is the number of rows of $\mathbf A$.
Thus let $\mathbf A'$ be the square matrix of order $k$ consisting of the first $k$ rows and columns of $\mathbf A$:
:$\mathbf A' = \begin {bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} & \cdots & a_{1, k - 1} & a_{1 k} \\
 0 & a_{2 2} & a_{2 3} & \cdots & a_{2, k - 1} & a_{2 k} \\
 0 & 0 & a_{3 3} & \cdots & a_{3, k - 1} & a_{3 k} \\
 \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
 0 & 0 & 0 & \cdots & a_{k - 1, k - 1} & a_{k - 1, k} \\
 0 & 0 & 0 & \cdots & 0 & a_{k k} \\
\end {bmatrix}$
$\mathbf A$ can be transformed into echelon form $\mathbf B$ by using the elementary row operations:
:$\forall j \in \set {1, 2, \ldots, k}: e_j := r_j \to \dfrac 1 {a_{j j} } r_j$
Again, let $\mathbf B'$ be the square matrix of order $k$ consisting of the first $k$ rows and columns of $\mathbf B$:
:$\mathbf B' = \begin {bmatrix}
 1 & b_{1 2} & b_{1 3} & \cdots & b_{1, k - 1} & b_{1 k} \\
 0 & 1 & b_{2 3} & \cdots & b_{2, k - 1} & b_{2 k} \\
 0 & 0 & 1 & \cdots & b_{3, k - 1} & b_{3 k} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
 0 & 0 & 0 & \cdots & 1 & b_{k - 1, k} \\
 0 & 0 & 0 & \cdots & 0 & 1 \\
\end {bmatrix}$
$\mathbf B$ is then transformed into reduced echelon form $\mathbf C$ by means of the elementary row operations:
:$\forall j \in \set {1, 2, \ldots, k - 1}: e_{j k} := r_j \to r_j - b_{j k} r_k$
:$\forall j \in \set {1, 2, \ldots, k - 2}: e_{j, k - 1} := r_j \to r_j - b_{j, k - 1} r_{k - 1}$
and so on, until:
:$e_{1 2} := r_1 \to r_1 - b_{1 2} r_2$
Again, let $\mathbf C'$ be the square matrix of order $k$ consisting of the first $k$ rows and columns of $\mathbf C$:
:$\mathbf C' = \begin {bmatrix}
 1 & 0 & 0 & \cdots & 0 & 0 \\
 0 & 1 & 0 & \cdots & 0 & 0 \\
 0 & 0 & 1 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
 0 & 0 & 0 & \cdots & 1 & 0 \\
 0 & 0 & 0 & \cdots & 0 & 1 \\
\end {bmatrix}$
By inspection, $\mathbf C$ is seen to be the unit matrix of order $k$.
== Examples ==
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations
Category:Triangular Matrices
Category:Identity Matrix from Upper Triangular Matrix
Identity Matrix from Upper Triangular Matrix/Examples/Arbitrary Matrix 1

== Example of Use of Identity Matrix from Upper Triangular Matrix ==
Let $\mathbf A$ be the matrix defined as:
:$\mathbf A = \begin {bmatrix}
1 & 1 & 1 & 1 \\ 
0 & 2 & 2 & 2 \\
0 & 0 & 3 & 4 \\
\end {bmatrix}$
is row equivalent to the reduced echelon matrix:
:$\mathbf E = \begin {bmatrix}
1 & 0 & 0 & -\dfrac 1 2 \\ 
0 & 1 & 0 & \dfrac 1 6 \\
0 & 0 & 1 & \dfrac 4 3 \\
\end {bmatrix}$
== Proof ==
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\mathbf A$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\mathbf A_k$.
$e_1 := r_2 \to \dfrac {r_2} 2$
$e_2 := r_3 \to \dfrac {r_3} 3$
Hence:
:$\mathbf A_3 = \begin {bmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & \dfrac 3 2 \\
0 & 0 & 1 & \dfrac 4 3 \\
\end {bmatrix}$
$e_3 := r_1 \to r_1 - r_3$
$e_4 := r_2 \to r_2 - r_3$
:$\mathbf A_4 = \begin {bmatrix}
1 & 1 & 0 & -\dfrac 1 3 \\
0 & 1 & 0 & \dfrac 1 6 \\
0 & 0 & 1 & \dfrac 4 3 \\
\end {bmatrix}$
$e_5 := r_1 \to r_1 - r_2$
:$\mathbf A_5 = \begin {bmatrix}
1 & 0 & 0 & -\dfrac 1 2 \\
0 & 1 & 0 & \dfrac 1 6 \\
0 & 0 & 1 & \dfrac 4 3 \\
\end {bmatrix}$
and it is seen that $\mathbf A_5$ is the required reduced echelon form.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.6$
Category:Identity Matrix from Upper Triangular Matrix
Identity Matrix is Permutation Matrix

== Theorem ==
An identity matrix is an example of a permutation matrix.
== Proof ==
An identity matrix, by definition, has instances of $1$ on the main diagonal and $0$ elsewhere.
Each diagonal element is by definition on one row and one column of the matrix.
Also by definition, each diagonal element is on a different row and 
column from each other diagonal element.
The result follows by definition of permutation matrix.
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.3$: $m \times n$ matrices
Category:Unit Matrices
Category:Permutation Matrices
Identity Minus Compact Linear Operator on Banach Space has Index Zero

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm {\,\cdot\,}_X}$ be a Banach space over $\Bbb F$.
Let $I : X \to X$ be the identity mapping.
Let $C : X \to X$ be a compact linear operator.
Then $I - C$ is a Fredholm operator with the index:
:$\map {\mathrm{ind} } {I - C} = 0$
== Proof ==
== Sources ==
* : $21.1$: Basic Properties of Compact Maps
Category:Banach Spaces
Identity Morphism is Initial Object in Coslice Category

== Theorem ==
Let $\mathbf C$ be a metacategory, and let $C \in \mathbf C_0$ be an object of $\mathbf C$.
Let $\operatorname{id}_C: C \to C$ be the identity morphism for $C$.
Then $\operatorname{id}_C$ is an initial object in the coslice category $C / \mathbf C$.
== Proof ==
Let $f: C \to D$ be an object of $C / \mathbf C$.
Then there is a morphism $a: \operatorname{id}_C \to f$ :
:$f = a \circ \operatorname{id}_C = a$
Thus, $f$ itself defines the unique morphism $\operatorname{id}_C \to f$ in $C \mathop / \mathbf C$.
We therefore have the following commutative diagram in $\mathbf C$:
$\quad\quad \begin{xy}
<-3em,0em>*+{C} = "X",
<3em,0em>*+{D} = "X2",
<0em,4em>*+{C} = "C",
"X";"X2" **@{--} ?>*@{>} ?*!/^1em/{f},
"C";"X" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_C},
"C";"X2" **@{-} ?>*@{>} ?<>(.6)*!/_1em/{f'},
\end{xy}$
Hence the result, by definition of initial object.
== Sources ==
* : $\S 2.2$: Example $2.11$: $6$
Category:Slice Categories
Identity Morphism is Terminal Object in Slice Category

== Theorem ==
Let $\mathbf C$ be a metacategory, and let $C \in \mathbf C_0$ be an object of $\mathbf C$.
Let $\operatorname{id}_C: C \to C$ be the identity morphism for $C$.
Then $\operatorname{id}_C$ is a terminal object in the slice category $\mathbf C \mathop / C$.
== Proof ==
Let $f: B \to C$ be an object of $\mathbf C \mathop / C$.
Then there is a morphism $a: f \to \operatorname{id}_C$ :
:$f = \operatorname{id}_C \circ a = a$
Thus, $f$ itself defines the unique morphism $f \to \operatorname{id}_C$ in $\mathbf C \mathop / C$.
We therefore have the following commutative diagram in $\mathbf C$:
$\quad\quad \begin{xy}
<-3em,0em>*+{B} = "X",
<3em,0em>*+{C} = "X2",
<0em,-4em>*+{C} = "C",
"X";"X2" **@{--} ?>*@{>} ?*!/_1em/{f},
"X";"C" **@{-} ?>*@{>} ?<>(.3)*!/^1em/{f},
"X2";"C" **@{-} ?>*@{>} ?<>(.3)*!/_1em/{\operatorname{id}_C},
\end{xy}$
Hence the result, by definition of terminal object.
== Sources ==
* : $\S 2.2$: Example $2.11$: $6$
Category:Slice Categories
Identity Morphism is Unique

== Theorem ==
Let $\mathbf C$ be a category.
Let $X$ be an object of $\mathbf C$.
Then the identity morphism $\operatorname{id}_X : X \to X$ is unique.
== Proof ==
Let $\operatorname{id}_X^1$, $\operatorname{id}_X^2$ be two identity morphisms for $X$.
By definition, for any morphism $f : Y \to X$, we have:
:$\operatorname{id}_X^1 \circ f = f$
In particular, taking $Y = X$ and $f = \operatorname{id}_X^2$, we have:
:$\operatorname{id}_X^1 \circ \operatorname{id}_X^2 = \operatorname{id}_X^2$
Similarly, for any morphism $g : X \to Y$, we have:
:$g \circ \operatorname{id}_X^2 = g$
So taking $Y = X$ and $g = \operatorname{id}_X^1$ we have:
:$\operatorname{id}_X^1 \circ \operatorname{id}_X^2 = \operatorname{id}_X^1$
Putting this together we have:
:$\operatorname{id}_X^2 = \operatorname{id}_X^1 \circ \operatorname{id}_X^2 = \operatorname{id}_X^1$
as required.
Category:Category Theory
Identity Morphism of Product

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $C$ and $D$ be objects of $\mathbf C$, and let $C \times D$ be a binary product for $C$ and $D$.
Then:
:$\operatorname{id}_{\paren {C \mathop \times D} } = \operatorname{id}_C \times \operatorname{id}_D$
where $\operatorname{id}$ denotes an identity morphism, and $\times$ signifies a product of morphisms.
== Proof ==
By definition of the product morphism $\operatorname{id}_C \times \operatorname{id}_D$, it is the unique morphism making:
$\quad\quad \begin{xy}\xymatrix@+1em@L+5px{
C
\ar[d]_*+{\operatorname{id}_C}
&
C \times D
\ar[l]_*+{\pr_1}
\ar[r]^*+{\pr_2}
\ar@{-->}[d]^*+{\operatorname{id}_C \times \operatorname{id}_D}
&
D
\ar[d]^*+{\operatorname{id}_D}
\\
C
&
C \times D
\ar[l]^*+{\pr_1}
\ar[r]_*+{\pr_2}
&
D
}\end{xy}$
a commutative diagram.
It is immediate by the behaviour of identity morphisms that $\operatorname{id}_{\paren {C \mathop \times D} }$ is the unique morphism sought.
Category:Category Theory
Identity Operator is Compact iff Finite-Dimensional Normed Vector Space

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space. 
Let $I : X \to X$ be the identity operator. 
Then $I$ is compact $X$ is finite-dimensional.
== Proof ==
Let $\operatorname {ball} X$ be the closed unit ball in $\struct {X, \norm {\, \cdot \,} }$.
From the definition of a compact operator, we have that $I$ is compact :
:$\overline {\operatorname {ball} X}$ is compact in $\struct {X, \norm \cdot_X}$.
From Closed Ball is Closed and Set is Closed iff Equals Topological Closure, this is equivalent to: 
:$\operatorname {ball} X$ is compact in $\struct {X, \norm {\, \cdot \,} }$.
From Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact, this is equivalent to:
:$X$ is finite-dimensional.
Category:Finite Dimensional Vector Spaces
Category:Compact Linear Transformations
Identity Permutation is Disjoint from All

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $e \in S_n$ be the identity permutation on $S_n$.
Then $e$ is disjoint from every permutation $\pi$ on $S_n$ (including itself).
== Proof ==
By definition of the identity permutation:
:$\forall i \in \N_{>0}: \map e i = i$
Thus $e$ fixes all elements of $S_n$.
Thus each element moved by a permutation $\pi$ is fixed by $e$.
The set of elements moved by $e$ is $\O$, so the converse is true vacuously.
Category:Symmetric Groups
Category:Identity Mappings
Identity Property in Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $s \in S$ be such that:
:$\forall a \in S: \exists x, y \in S: s \circ x = a = y \circ s$
Then $\struct {S, \circ}$ has an identity.
== Proof ==
Suppose that:
:$\forall a \in S: \exists x, y \in S: s \circ x = a = y \circ s$.
Since $s \in S$, it follows that:
:$\exists v, w \in S: s \circ v = s = w \circ s$.
Let $a \in S$.
Then:
:$\exists x, y \in S: s \circ x = a = y \circ s$.
Thus:
Hence:
:$w \circ a = a$ and $a \circ v = a$
for any $a \in S$.
In particular:
:Letting $a = v$ in the first of these gives $w \circ v = v$
:Letting $a = w$ in the second gives $w \circ v = w$.
Thus $v = w \circ v = w$ is the identity element in $S$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Exercise $4.10$
Category:Semigroups
Category:Identity Elements
Identity Theorem

== Theorem ==
Let $U$ be an open connected subset of the complex plane $\C$.
Let $f$ and $g$ be complex functions whose domain is $U$.
Let $S = \left\{{z \in U: f \left({z}\right) = g \left({z}\right)}\right\}$.
Let $f$ and $g$ be analytic on $U$.
Let $S$ have a limit point in $U$.
Then:
:$\forall z \in U : f \left({z}\right) = g \left({z}\right)$
== Proof ==
Category:Complex Analysis
Identity is Only Group Element of Order 1

== Theorem ==
In every group, the identity, and only the identity, has order $1$.
== Proof ==
Let $G$ be a group with identity $e$.
Then:
:$e^1 = e$
and:
:$\forall a \in G: a \ne e: a^1 = a \ne e$.
Hence the result.
== Sources ==
* : $\S 8$: Chapter $\text {I}$: The Group Concept: The Order (Period) of an Element: $\text{(i)}$
* : $\S 5.4$. Cyclic groups: Example $101$
* : $\S 38.1$ Period of an element
Category:Identity Elements
Category:Order of Group Elements
Identity is Unique

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure that has an identity element $e \in S$.
Then $e$ is unique.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Identity of Monoid is Unique
* Identity of Group is Unique
Category:Identity Elements
Category:Identity is Unique
Category:Algebraic Structures
Identity is Unique/Proof 1

== Theorem ==
== Proof ==
Suppose $e_1$ and $e_2$ are both identity elements of $\struct {S, \circ}$.
Then by the definition of identity element:
:$\forall s \in S: s \circ e_1 = s = e_2 \circ s$
Then:
:$e_1 = e_2 \circ e_1 = e_2$
So:
:$e_1 = e_2$
and there is only one identity element after all.
== Sources ==
* : $\S 4.3$. Units and zeros
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Theorem $4.1$
* : $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups: Theorem $1.2$
Category:Identity is Unique
Identity is Unique/Proof 2

== Theorem ==
== Proof ==
Let $e_S$ be an identity of $\struct {S, \circ}$.
Then by definition, $e_S$ is both a left identity and a right identity.
By More than one Left Identity then no Right Identity, if there is more than one of either, there cannot be one of the other.
So there can be only one of each.
By Left and Right Identity are the Same, they are one and the same thing.
Category:Identity is Unique
Identity is in Kernel of Group Homomorphism

== Theorem ==
Let $G$ and $H$ be groups.
Let $e_G$ and $e_H$ be the identity elements of $G$ and $H$ respectively.
Let $\phi: G \to H$ be a (group) homomorphism from $G$ to $H$.
Then:
:$e_G \in \map \ker \phi$
where $\map \ker \phi$ is the kernel of $\phi$.
== Proof ==
From the definition of kernel:
:$\map \ker \phi = \set {x \in G: \map \phi x = e_H}$
From Group Homomorphism Preserves Identity we have that:
:$\map \phi {e_G} = e_H$
Hence the result.
== Proof ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 65$
Category:Kernels of Group Homomorphisms
Identity is only Idempotent Cancellable Element

== Theorem ==
Let $e_S$ be the identity of an algebraic structure $\struct {S, \circ}$.
Then $e_S$ is the only cancellable element of $\struct {S, \circ}$ that is idempotent.
== Proof ==
By Identity Element is Idempotent, $e_S$ is idempotent.
Let $x$ be a cancellable idempotent element of $\struct {S, \circ}$.
So $x \circ x = x \circ e_S$.
But because $x$ is also by hypothesis cancellable, it follows that $x = e_S$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Theorem $8.2$
Category:Cancellability
Category:Identities are Idempotent
Identity is only Idempotent Element in Group/Proof 2

== Theorem ==
== Proof ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $x \in G$ such that $x \circ x = x$.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.14)$
* : $\S 1.4$: Lemma $1$
* : $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups: Theorem $1.2 \text{(i)}$
Category:Identity is only Idempotent Element in Group
Identity of Affine Group of One Dimension

== Theorem ==
Let $\map {\mathrm {Af}_1} \R$ denote the $1$-dimensional affine group on $\R$.
Then $\map {\mathrm {Af}_1} \R$ has $f_{1, 0}$ as an identity element.
== Proof ==
Let $f_{a b} \in \map {\mathrm {Af}_1} \R$.
Then:
Thus $f_{1, 0}$ is the identity element of $\map {\mathrm {Af}_1} \R$.
Category:Affine Groups
Identity of Algebraic Structure is Preserved in Substructure

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure with identity $e$.
Let $\struct {T, \circ}$ be a algebraic substructure of $\struct {S, \circ}$.
That is, let $T \subseteq S$.
Let $e \in T$.
Then $e$ is an identity of $\struct {T, \circ}$.
== Proof ==
Let $x \in T$.
By the definition of subset, $x \in S$.
Since $e$ is an identity of $\struct {S, \circ}$:
:$e \circ x = x \circ e = x$
Since this holds for all $x \in T$, $e$ is an identity of $\struct {T, \circ}$.
Category:Identity Elements
Identity of Cancellable Monoid is Identity of Submonoid

== Theorem ==
Let $\struct {S, \circ}$ be a monoid, all of whose elements are cancellable.
Let $\struct {T, \circ}$ be a submonoid of $\struct {S, \circ}$.
Then the identity of $\struct {T, \circ}$ is the same element as the identity of $\struct {S, \circ}$.
== Proof ==
By Identity of Monoid is Unique:
: there is only one identity element of $\struct {S, \circ}$
and:
: there is only one identity element of $\struct {T, \circ}$.
Let $e_S$ be the identity of $\struct {S, \circ}$, and $e_T$ the identity of $\struct {T, \circ}$.
From Identity is only Idempotent Cancellable Element, $e_S$ is the only cancellable element of $\struct {S, \circ}$ which is idempotent.
But all elements of $\struct {S, \circ}$ are cancellable.
Thus $e_S$ is the only idempotent element of $\struct {S, \circ}$.
Again, all elements of $\struct {S, \circ}$ are cancellable.
Thus from Cancellable Element is Cancellable in Subset, all elements of $\struct {T, \circ}$ are cancellable.
Thus, $e_T$ is the only element of $\struct {T, \circ}$ which is idempotent.
Thus, as $e_T \in S$, we have $e_S \circ e_T = e_T = e_T \circ e_T$ and thus $e_S = e_T$.
== Also see ==
* Identity of Submonoid is not necessarily Identity of Monoid
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets
Category:Cancellable Monoids
Category:Identity Elements
Identity of Cardinal Product is One

== Theorem ==
Let $\mathbf a$ be a cardinal.
Then:
: $\bsone \mathbf a = \mathbf a$
where $\bsone \mathbf a$ denotes the product of the (cardinal) one and $\mathbf a$.
That is, $\bsone$ is the identity element of the product operation on cardinals.
== Proof ==
Let $\mathbf a = \card A$ for some set $A$.
From the definition of (cardinal) one, $\bsone$ is the cardinal associated with a singleton set, say, $\set \O$.
We have by definition of product of cardinals that $\bsone \mathbf a$ is the cardinal associated with $\set \O \times A$.
Consider the mapping $f: \set \O \times A \to A$ defined as:
:$\forall a \in A: \map f {\O, a} = a$
Let $a_1, a_2 \in A$ such that:
:$\map f {\O, a_1} = \map f {\O, a_2}$
Then:
Thus $f$ is an injection.
Let $a \in A$.
By definition of $f$:
:$a = \map f {\O, a}$
Thus $f$ is a surjection.
So $f$ is both an injection and a surjection, and so by definition a bijection.
Thus a bijection has been established between $\set \O \times A$ and $A$.
It follows by definition that $\set \O \times A$ and $A$ are equivalent.
The result follows by definition of cardinal.
== Sources ==
* : $\S 8$: Theorem $8.4: \ (3)$
Category:Cardinals
Identity of Cardinal Sum is Zero

== Theorem ==
Let $\mathbf a$ be a cardinal.
Then:
: $\mathbf a + \mathbf 0 = \mathbf a$
where $\mathbf a + \mathbf 0$ denotes the sum of the zero cardinal and $\mathbf a$.
That is, $\mathbf 0$ is the identity element of the sum operation on cardinals.
== Proof ==
Let $\mathbf a = \card A$ for some set $A$.
From Union with Empty Set we have $A \cup \O = A$.
From Intersection with Empty Set we have $A \cap \O = \O$.
So $A$ and $\O$ are disjoint and so:
== Sources ==
* : $\S 8$: Theorem $8.5: \ (3)$
Category:Cardinals
Identity of Group Direct Product

== Theorem ==
Let $\struct {G \times H, \circ}$ be the group direct product of the two groups $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Let $e_G$ be the identity for $\struct {G, \circ_1}$.
Let $e_H$ be the identity for $\struct {H, \circ_2}$.
Then $\tuple {e_G, e_H}$ is the identity for $\struct {G \times H, \circ}$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{II}$: Groups: Direct Products
Category:Group Direct Products
Category:Identity of Group Direct Product
Identity of Group is Unique

== Theorem ==
Let $\struct {G, \circ}$ be a group which has an identity element $e \in G$.
Then $e$ is unique.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\S 4.6$. Elementary theorems on groups
* : Chapter $1$: Rings - Definitions and Examples: $1$: The definition of a ring: Definitions $1.1 \ \text{(b)}$
Category:Group Theory
Category:Identity Elements
Category:Identity of Group is Unique
Identity of Group is Unique/Proof 2

== Theorem ==
== Proof ==
Let $e$ and $f$ both be identity elements of a group $\struct {G, \circ}$.
Then:
So $e = f$ and there is only one identity after all.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory
* : $\S 1.4$: Lemma $2$
* : Chapter $\text{II}$: Groups: The Group Property
* : Chapter $3$: Elementary consequences of the definitions: Proposition $3.1$
Category:Identity of Group is Unique
Identity of Group is Unique/Proof 3

== Theorem ==
== Proof ==
From Group has Latin Square Property, there exists a unique $x \in G$ such that:
:$a x = b$
and there exists a unique $y \in G$ such that:
:$y a = b$
Setting $b = a$, this becomes:
There exists a unique $x \in G$ such that:
:$a x = a$
and there exists a unique $y \in G$ such that:
:$y a = a$
These $x$ and $y$ are both $e$, by definition of identity element.
== Sources ==
* : Chapter $3$: Elementary consequences of the definitions: Proposition $3.3$: Remark $3$
Category:Identity of Group is Unique
Identity of Group is in Center

== Theorem ==
Let $G$ be a group.
Let $e$ be the identity of $G$.
Then $e$ is in the center of $G$:
:$e \in \map Z G$
== Proof ==
From Center is Intersection of Centralizers:
:$\ds \map Z G = \bigcap_{g \mathop \in G} \map {C_G} g$
where $\map {C_G} g$ denotes the centralizer of $g$.
From Centralizer of Group Element is Subgroup, each of $\map {C_G} g$ is a subgroup of $G$.
From Identity of Subgroup:
:$\forall g \in G: e \in \map {C_G} g$
Hence by definition of set intersection:
:$e \in \ds \bigcap_{g \mathop \in G} \map {C_G} g$
whence the result.
Category:Centers of Groups
Category:Identity Elements
Identity of Group is in Singleton Conjugacy Class

== Theorem ==
Let $G$ be a group.
Let $e$ be the identity of $G$.
Then $e$ is in its own singleton conjugacy class:
:$\conjclass e = \set e$
== Proof ==
From Identity of Group is in Center:
:$e \in \map Z G$
where $\map Z G$ is the center of $G$.
From Conjugacy Class of Element of Center is Singleton:
:$\conjclass e = \set e$
== Sources ==
* 
Category:Identity Elements
Category:Conjugacy Classes
Identity of Inverse Completion of Commutative Monoid

== Theorem ==
Let $\struct {S, \circ}$ be a commutative monoid whose identity is $e$.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then $e \in T$ is the identity for $\circ'$.
== Proof ==
Let $e$ be the identity for $\circ$.
Let $e = x \circ' y^{-1}$, where $x \in S, y \in C$.
Then:
Thus $e = y^{-1} \circ' y$, and $y^{-1} \circ' y$ is the identity for $\circ'$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers: Theorem $20.1: \ 4^\circ$
Category:Inverse Completions
Identity of Monoid is Cancellable

== Theorem ==
Let $\struct {S, \circ}$ be a monoid.
Let $e$ be an identity element of $\struct {S, \circ}$.
Then $e$ is a cancellable element of $\struct {S, \circ}$.
== Proof ==
Let $x, y \in S$ such that $x \circ e = y \circ e$.
Then, by definition of identity element:
:$x = x \circ e = y \circ e = y$
Thus $x = y$.
== Also see ==
* Invertible Element of Monoid is Cancellable
Category:Cancellability
Category:Monoids