nyuuzyou/wikis huggingface dataset:
Group Action on Subgroup by Right Regular Representation

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $*: H \times G \to G$ be the operation defined as:
:$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$
where $\map {\rho_{h^{-1} } } g$ is the right regular representation of $g$ by $h^{-1}$.
Then $*$ is a group action.
== Proof ==
The group action axioms are investigated in turn.
Let $h_1, h_2 \in H$ and $g \in G$.
Thus:
demonstrating that holds.
Then:
demonstrating that holds.
The group action axioms are thus seen to be fulfilled, and so $*$ is a group action.
== Also defined as ==
Some sources denote a mapping by placing the symbol defining that mapping on the left of its operand.
Thus under such a convention:
:$\map {\rho_h} a$ is written $a \rho_h$
and:
:$h * a$ is written $a * h$ (or even $a h$ in sources which do not place a high regard on clarity).
Thus the right regular representation itself can directly be defined as being a group action without the need to take the inverse:
Such a treatment can be found in .
As it runs contrary to the conventions used on , beyond its mention here it will not be used.
However, this example indicates how the arbitrary nature of notational conventions can cause the details of results to be equally arbitrarily dependent upon the convention used.
== Also see ==
* Group Action on Subgroup by Left Regular Representation
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions
Category:Group Actions
Group Action on Subgroup by Right Regular Representation is not Transitive

== Theorem ==
Let $G$ be a group.
Let $H$ be a proper subgroup of $G$.
Let $*: H \times G \to G$ be the group action defined as:
:$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$
where $\map {\rho_{h^{-1} } } g$ is the right regular representation of $g$ by $h^{-1}$.
Then $*$ is not transitive.
== Proof ==
From Group Action on Subgroup by Right Regular Representation it is established that $*$ is a group action.
From Orbit of Group Action on Subgroup by Right Regular Representation is Right Coset:
:$\forall x \in G: \Orb x = H x$
where $H x$ is the right coset of $H$ by $x$.
From Right Coset Space forms Partition it is apparent that $H x \ne G$ unless $H = G$.
Thus $\Orb x \ne G$ and the result follows by definition of transitive group action.
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions
Category:Transitive Group Actions
Group Action on Subgroup of Symmetric Group

== Theorem ==
Let $S_n$ be the symmetric group of $n$ elements.
Let $H$ be a subgroup of $S_n$.
Let $X$ be any set with $n$ elements.
Then $H$ acts on $X$ as a group of transformations on $X$.
== Proof ==
The identity permutation takes each element of $X$ to itself, thus fulfilling .
The group operation in $S_n$ ensures fulfilment of .
== Also see ==
* Stabilizer in Group of Transformations
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 76$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.2$
Category:Group Actions
Category:Symmetric Groups
Group Acts by Homeomorphisms Implies Projection on Quotient Space is Open

== Theorem ==
Let $G$ be a group acting by homeomorphisms on a topological space $X$.
Then the projection map $\pi: X \to X / G$ is open.
== Proof ==
Let $U \subset X$ be open.
We have to show that $\pi \sqbrk U$ is open.
By definition of quotient topology, this is the case $\pi^{-1} \sqbrk {\pi \sqbrk U}$ is open.
By definition of saturation under group action:
:$\ds \pi^{-1} \sqbrk {\pi \sqbrk U} = \bigcup_{g \mathop \in G} g U$
Because $G$ acts by homeomorphisms, $\pi^{-1} \sqbrk {\pi \sqbrk U}$ is open.
== Also see ==
* Projection of Subset is Open iff Saturation is Open
Category:Topology
Category:Group Actions
Group Acts on Itself

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then $\struct {G, \circ}$ acts on itself by the rule:
:$\forall g, h \in G: g * h = g \circ h$
== Proof ==
Follows directly from the group axioms and the definition of a group action.
== Also see ==
* Stabilizer of Element of Group Acting on Itself is Trivial
* Orbit of Element of Group Acting on Itself is Group
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 53$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.4$
Category:Group Actions
Group Automorphism/Examples/Constant Product on Real Numbers

== Examples of Group Automorphisms ==
Let $\struct {\R, +}$ denote the real numbers under addition.
Let $\alpha \in \R$ be a real number.
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \alpha x$
Then $f$ is a (group) automorphism $\alpha \ne 0$.
== Proof ==
Let $\alpha \in \R$.
We have that:
Thus $f$ is a (group) homomorphism for all $\alpha \in \R$.
=== Sufficient Condition ===
Let $\alpha \in \R$ such that $\alpha \ne 0$.
Then $\dfrac 1 \alpha$ is the multiplicative inverse of $\alpha$ in the field of real numbers $\R$.
We have that:
and so $f$ is injective by definition.
Then we have:
This demonstrates that $f$ is surjective.
So by definition $f$ is a bijection.
Thus $f$ is an (group) isomorphism from $\R$ to $\R$ and so a (group) automorphism.
=== Necessary Condition ===
Let $f$ be a (group) automorphism.
 $\alpha = 0$.
Then we have:
:$\forall x, y \in \R: \map f x = \map f y = 0$
and so $f$ is not injective.
Hence $f$ is not a bijection.
Therefore $f$ is not an automorphism.
Hence by Proof by Contradiction $f$ is not an automorphism.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.4$
Category:Examples of Group Automorphisms
Group Commutators are Commuting Elements

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\sqbrk {g, h}$ denote the commutator of $g$ and $h$.
Then $\sqbrk {g, h}$ commutes with $\sqbrk {h, g}$, in the sense that:
:$\sqbrk {g, h} \circ \sqbrk {h, g} = \sqbrk {h, g} \circ \sqbrk {g, h}$
== Proof ==
From Inverse of Group Commutator:
:$\forall g, h \in G: \sqbrk {g, h} = \sqbrk {h, g}^{-1}$
The result follows from Group Element Commutes with Inverse.
== Sources ==
* 
Category:Group Commutators
Category:Commutativity
Group Direct Product/Examples/C2 x C2

== Example of Group Direct Product ==
The direct product of $C_2$, the cyclic group of order $2$, with itself is as follows.
Let us represent $C_2$ as the group $\struct {\set {1, -1}, \times}$:
:$\begin {array} {r|rr}
\struct {\set {1, -1} , \times} & 1 & -1 \\
\hline
1 & 1 & -1 \\
-1 & -1 & 1 \\
\end{array}$
Then the Cayley table for $C_2 \times C_2$ can be portrayed as:
:$\begin {array} {c|cccc}
C_2 \times C_2 & \tuple { 1, 1} & \tuple { 1, -1} & \tuple {-1, 1} & \tuple {-1, -1} \\
\hline
\tuple { 1, 1} & \tuple { 1, 1} & \tuple { 1, -1} & \tuple {-1, 1} & \tuple {-1, -1} \\
\tuple { 1, -1} & \tuple { 1, -1} & \tuple { 1, 1} & \tuple {-1, -1} & \tuple {-1, 1} \\
\tuple {-1, 1} & \tuple {-1, 1} & \tuple {-1, -1} & \tuple { 1, 1} & \tuple { 1, -1} \\
\tuple {-1, -1} & \tuple {-1, -1} & \tuple {-1, 1} & \tuple { 1, -1} & \tuple { 1, 1} \\
\end{array}$
This is seen by inspection to be an instance of the Klein $4$-group.
== Subgroups ==
== Sources ==
* : Chapter $1$: Definitions and Examples: Example $1.10$
Category:Examples of Group Direct Products
Category:Klein Four-Group
Group Direct Product/Examples/C2 x C2/Subgroups

== Subgroups of Group Direct Product of $C_2 \times C_2$ ==
Let us represent $C_2$ as the group $\struct {\set {1, -1}, \times}$:
:$\begin {array} {r|rr}
\struct {\set {1, -1} , \times} & 1 & -1 \\
\hline
1 & 1 & -1 \\
-1 & -1 & 1 \\
\end{array}$
The subsets of $C_2 \times C_2$ which form its subgroups are:
:$\set {\tuple {1, 1} }$
:$\set {\tuple {1, 1}, \tuple {1, -1} }$
:$\set {\tuple {1, 1}, \tuple {-1, 1} }$
:$\set {\tuple {1, 1}, \tuple {-1, -1} }$
:$\set {\tuple {1, 1}, \tuple {1, -1}, \tuple {-1, 1} , \tuple {-1, -1} }$ (that is, $C_2 \times C_2$ itself)
== Proof ==
From Trivial Subgroup is Subgroup and Group is Subgroup of Itself we have that $\set {\tuple {1, 1} }$ and $C_2 \times C_2$ are two of the subgroups
From Lagrange's Theorem (Group Theory), any subgroup of $C_2 \times C_2$ contains $1$, $2$ or $4$ elements.
There is only one subgroup for each of $1$ and $4$, both of which have been covered.
Thus all other subgroups can contain only $2$ elements.
Every subgroup of $C_2 \times C_2$ contains its identity $\set {\tuple {1, 1} }$.
So a subgroup of $C_2 \times C_2$ contains $\set {\tuple {1, 1} }$ and one other element.
Each of the other elements is of order $2$.
So, along with $\set {\tuple {1, 1} }$, each one forms a subgroup of $C_2 \times C_2$.
== Sources ==
* : Chapter $5$: Cosets and Lagrange's Theorem: Example $5.11$
Category:Examples of Group Direct Products
Category:Examples of Subgroups
Category:Klein Four-Group
Group Direct Product/Examples/C2 x C3

== Example of Group Direct Product ==
The direct product of $C_2$, the cyclic group of order $2$, with $C_3$, the cyclic group of order $3$, is as follows.
Let us represent $C_2$ as the group $\struct {\Z_2, +_2}$:
:$\begin {array} {r|rr}
+_2 & \eqclass 0 2 & \eqclass 1 2 \\
\hline
\eqclass 0 2 & \eqclass 0 2 & \eqclass 1 2 \\
\eqclass 1 2 & \eqclass 1 2 & \eqclass 0 2 \\
\end{array}$
and $C_3$ as the group $\struct {\Z_3, +_3}$:
:$\begin {array} {r|rrr}
+_3 & \eqclass 0 3 & \eqclass 1 3 & \eqclass 2 3 \\
\hline
\eqclass 0 3 & \eqclass 0 3 & \eqclass 1 3 & \eqclass 2 3 \\
\eqclass 1 3 & \eqclass 1 3 & \eqclass 2 3 & \eqclass 0 3 \\
\eqclass 2 3 & \eqclass 2 3 & \eqclass 0 3 & \eqclass 1 3 \\
\end{array}$
Then the Cayley table for $\struct{C_2 \times C_3, +_6}$ can be portrayed as:
:$\begin {array} {r|rrrrrr}
+_6 & \tuple {\eqclass 0 2, \eqclass 0 3} & \tuple {\eqclass 0 2, \eqclass 1 3} & \tuple {\eqclass 0 2, \eqclass 2 3} & \tuple {\eqclass 1 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 1 3} & \tuple {\eqclass 1 2, \eqclass 2 3} \\
\hline
\tuple {\eqclass 0 2, \eqclass 0 3} & \tuple {\eqclass 0 2, \eqclass 0 3} & \tuple {\eqclass 0 2, \eqclass 1 3} & \tuple {\eqclass 0 2, \eqclass 2 3} & \tuple {\eqclass 1 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 1 3} & \tuple {\eqclass 1 2, \eqclass 2 3} \\
\tuple {\eqclass 0 2, \eqclass 1 3} & \tuple {\eqclass 0 2, \eqclass 1 3} & \tuple {\eqclass 0 2, \eqclass 2 3} & \tuple {\eqclass 0 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 1 3} & \tuple {\eqclass 1 2, \eqclass 2 3} & \tuple {\eqclass 1 2, \eqclass 0 3} \\
\tuple {\eqclass 0 2, \eqclass 2 3} & \tuple {\eqclass 0 2, \eqclass 2 3} & \tuple {\eqclass 0 2, \eqclass 0 3} & \tuple {\eqclass 0 2, \eqclass 1 3} & \tuple {\eqclass 1 2, \eqclass 2 3} & \tuple {\eqclass 1 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 1 3} \\
\tuple {\eqclass 1 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 1 3} & \tuple {\eqclass 1 2, \eqclass 2 3} & \tuple {\eqclass 0 2, \eqclass 0 3} & \tuple {\eqclass 0 2, \eqclass 1 3} & \tuple {\eqclass 0 2, \eqclass 2 3} \\
\tuple {\eqclass 1 2, \eqclass 1 3} & \tuple {\eqclass 1 2, \eqclass 1 3} & \tuple {\eqclass 1 2, \eqclass 2 3} & \tuple {\eqclass 1 2, \eqclass 0 3} & \tuple {\eqclass 0 2, \eqclass 1 3} & \tuple {\eqclass 0 2, \eqclass 2 3} & \tuple {\eqclass 0 2, \eqclass 0 3} \\
\tuple {\eqclass 1 2, \eqclass 2 3} & \tuple {\eqclass 1 2, \eqclass 2 3} & \tuple {\eqclass 1 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 1 3} & \tuple {\eqclass 0 2, \eqclass 2 3} & \tuple {\eqclass 0 2, \eqclass 0 3} & \tuple {\eqclass 0 2, \eqclass 1 3} \\
\end{array}$
== Sources ==
* : Chapter $1$: Definitions and Examples: Exercise $5$
Category:Examples of Group Direct Products
Group Direct Product/Examples/R-0 x R

== Example of Group Direct Product ==
Let $G$ be the Cartesian product of $\R \setminus \set 0$ with $\R$:
:$G = \set {\tuple {a, b} \in \R^2: a \ne 0}$
Let $\circ$ be a group product on $G$ defined as:
:$\tuple {a_1, b_1} \circ \tuple {a_2, b_2} = \tuple {a_1 a_2, a_1 b_2 + b_1}$
Then the algebraic structure $\struct {G, \circ}$ is a group which is non-abelian.
=== Isomorphism to Set of Affine Mappings on Real Line under Composition ===
== Proof ==
Taking the group axioms in turn:
=== ===
By Definition of $\circ$:
:$\tuple {a_1, b_1} \circ \tuple {a_2, b_2} = \tuple {a_1 a_2, a_1 b_2 + b_1}$
where $a_1, a_2 \ne 0$.
Hence:
:$a_1 a_2 \ne 0$
from which:
:$\tuple {a_1 a_2, a_1 b_2 + b_1} \in \paren {\R \setminus \set 0} \times \R$
Thus $\circ$ is closed in $G$.
=== ===
Thus $\circ$ is associative.
=== ===
Thus $\tuple {1, 0}$ is the identity element of $\struct {G, \circ}$.
=== ===
We have that $\tuple {1, 0}$ is the identity element of $\struct {G, \circ}$.
We need to find $\tuple {a_2, b_2} \in G$ such that:
Hence:
We verify:
Thus every element $\tuple {a_1, b_1}$ of $\struct {G, \circ}$ has an inverse $\tuple {\dfrac 1 {a_1}, -\dfrac {b_1} {a_1} }$.
All the group axioms are thus seen to be fulfilled, and so $\struct {G, \circ}$ is a group.
It remains to show that $\struct {G, \circ}$ is not abelian.
Selecting arbitrary values, let:
:$\tuple {a_1, b_1} = \tuple {1, 2}$
:$\tuple {a_2, b_2} = \tuple {3, 4}$
Then:
:$a_1 b_2 + b_1 = 1 \times 4 + 2 = 6$
:$a_2 b_1 + b_2 = 3 \times 2 + 4 = 10$
As it is not generally true that $a_1 b_2 + b_1$ equals $a_2 b_1 + b_2$, it follows that $\circ$ is not commutative on $G$.
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{N}$
Category:Examples of Group Direct Products
Group Direct Product/Examples/R-0 x R/Isomorphism to Set of Affine Mappings on Real Line under Composition

== Theorem ==
Let $G$ be the Cartesian product of $\R \setminus \set 0$ with $\R$:
:$G = \set {\tuple {a, b} \in \R^2: a \ne 0}$
Let $\circ$ be the group product on $G$ defined as:
:$\tuple {a_1, b_1} \circ \tuple {a_2, b_2} = \tuple {a_1 a_2, a_1 b_2 + b_1}$
Let $S$ be the set of all real functions $f: \R \to \R$ of the form:
:$\forall x \in \R: \map f x = r x + s$
where $r \in \R_{\ne 0}$ and $s \in \R$
Let $\struct {S, \circ}$ be the algebraic structure formed from $S$ and the composition operation $\circ$.
Then $\struct {G, \circ}$ is (group) isomorphic to $\struct {S, \circ}$.
== Proof ==
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{N}$
Category:Examples of Group Direct Products
Group Direct Product is Product in Category of Groups

== Theorem ==
Let $\mathbf{Grp}$ be the category of groups.
Let $G$ and $H$ be groups, and let $G \times H$ be their direct product.
Then $G \times H$ is a binary product of $G$ and $H$ in $\mathbf{Grp}$.
== Proof ==
Let $F$ be a group.
By Direct Product of Group Homomorphisms is Homomorphism, given group homomorphisms:
:$g: F \to G, h: F \to H$
their direct product $g \times h: F \to G \times H$ is a group homomorphism.
From Projections on Direct Product of Group Homomorphisms, the following diagram is commutative:
:$\begin{xy}\xymatrix@L+3mu@R=3em{
&
 F
 \ar@/_/[dl]_*{g}
 \ar[d]^*{\quad g \times h}
 \ar@/^/[dr]^*{h}
\\
 G
&
 G \times H
 \ar[l]^*{\pr_1}
 \ar[r]_*{\pr_2}
&
 H
}\end{xy}$
By Cartesian Product is Set Product, $g \times h$ is the only mapping $F \to G \times H$ that could fit into the diagram.
Since it is also a group homomorphism, we see that the universal mapping property for the binary product is satisfied.
Thus the direct product $G \times H$ indeed is a categorical product for $G$ and $H$ in $\mathbf{Grp}$.
== Sources ==
* : $\S 2.5.2$
Category:Category of Groups
Category:Group Theory
Group Direct Product of Cyclic Groups

== Theorem ==
Let $G$ and $H$ both be finite cyclic groups with orders $n = \order G$ and $m = \order H$ respectively.
Then:
:The group direct product $G \times H$ is cyclic
:
:$n$ and $m$ are coprime, that is:
::$n \perp m$
=== Corollary ===
== Proof ==
Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$ respectively.
=== Necessary condition ===
Suppose:
:$(1): \quad \order G = n, G = \gen x$
:$(2): \quad \order H = m, H = \gen y$
:$(3): \quad m \perp n$
Then:
But then:
:$\tuple {x, y}^{n m} = e_{G \times H} = \tuple {x^{n m}, y^{n m} }$
Thus:
:$k \divides n m$
So:
:$\order {\tuple {x, y} } = n m \implies \gen {\tuple {x, y} } = G \times H$
=== Sufficient condition ===
Suppose that $G \times H$ is cyclic.
Let $\tuple {x, y}$ be a generator of $G \times H$.
By Cardinality of Cartesian Product of Finite Sets the order of $G \times H$ is:
:$\order G \cdot \order H = n m$
Therefore by Order of Cyclic Group equals Order of Generator:
:$\order {\tuple {x, y} } = n m$
On the other hand, by Order of Group Element in Group Direct Product we have:
:$\order {\tuple {x, y} } = \lcm \set {\tuple {\order x, \order y} }$
Next we claim that $x$ generates $G$.
Let $x' \in G$.
Then:
:$\tuple {x', e_H} \in G \times H$
so there exists $k \in \N$ such that:
:$\tuple {x, y}^k = \tuple {x^k, y^k} = \tuple {x', e_H}$
and therefore $x^k = x'$.
Thus the powers of $x$ generate the whole group $G$.
In the same way, it is seen that $y$ generates $H$.
Therefore by Order of Cyclic Group equals Order of Generator:
:$\order x = n$
:$\order y = m$
Thus we have that:
:$n m = \order {\tuple {x, y} } = \lcm \set {n, m}$
Moreover by Product of GCD and LCM we have that:
:$\lcm \set {n, m} = \dfrac {n m} {\gcd \set {n, m} }$
These two equalities imply that:
:$\gcd \set {n, m} = 1$
That is, $n$ and $m$ are coprime.
== Examples ==
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 43 \gamma$
* : Chapter $6$: An Introduction to Groups: Exercise $18$
* : Chapter $13$: Direct products: Proposition $13.1 \ (3)$
Category:Group Direct Products
Category:Cyclic Groups
Category:Group Direct Product of Cyclic Groups
Group Direct Product of Cyclic Groups/Corollary

== Theorem ==
Let $n_1, n_2, \ldots, n_s$ be a finite sequence of integers, all greater than $1$, such that for any pair of them $n_i$ and $n_j$, $n_i \perp n_j$.
Let $G_i$ be a cyclic group of order $n_i$ for each $i: 1 \le i \le s$.
Then $G_1 \times G_2 \times \cdots \times G_s$ is cyclic of order $n_1 n_2 \ldots n_s$.
== Proof ==
When $s = 1$ the result is trivial.
Assume the result holds for $s = k$.
Then $H = G_1 \times G_2 \times \ldots \times G_k$ is cyclic of order $n_1 n_2 \ldots n_k$.
Applying the main result to $H \times G_{k + 1}$ gives us the result for $s = k + 1$.
The result follows by induction.
== Sources ==
* : Chapter $13$: Direct products: Corollary $13.2$
Category:Group Direct Products
Category:Cyclic Groups
Group Direct Product of Cyclic Groups/Examples/C2 x C2

== Examples of Use of Group Direct Product of Cyclic Groups ==
Let $C_2$ denote the cyclic group of order $2$.
The group direct product $C_2 \times C_2$ is not a cyclic group.
== Proof ==
For the group direct product of the cyclic groups $C_m$ and $C_n$ to be cyclic, it is necessary and sufficient for $m$ and $n$ to be coprime.
That is not the case here.
It is seen from Group Direct Product of $C_2 \times C_2$ that this is an instance of the Klein $4$-group.
This is not a cyclic group.
== Sources ==
* : Chapter $13$: Direct products: Remark $1$
Category:Group Direct Product of Cyclic Groups
Group Direct Product of Cyclic Groups/Examples/C2 x C3

== Examples of Use of Group Direct Product of Cyclic Groups ==
Let $C_2$ denote the cyclic group of order $2$.
The group direct product $C_2 \times C_3$ is a cyclic group.
== Proof ==
Let us represent $C_2$ as the group $\struct {\Z_2, +_2}$:
:$\begin {array} {r|rr}
+_2 & \eqclass 0 2 & \eqclass 1 2 \\
\hline
\eqclass 0 2 & \eqclass 0 2 & \eqclass 1 2 \\
\eqclass 1 2 & \eqclass 1 2 & \eqclass 0 2 \\
\end{array}$
and $C_3$ as the group $\struct {\Z_3, +_3}$:
:$\begin {array} {r|rrr}
+_3 & \eqclass 0 3 & \eqclass 1 3 & \eqclass 2 3 \\
\hline
\eqclass 0 3 & \eqclass 0 3 & \eqclass 1 3 & \eqclass 2 3 \\
\eqclass 1 3 & \eqclass 1 3 & \eqclass 2 3 & \eqclass 0 3 \\
\eqclass 2 3 & \eqclass 2 3 & \eqclass 0 3 & \eqclass 1 3 \\
\end{array}$
So:
:$C_2 \times C_3 = \gen {\tuple {\eqclass 1 2, \eqclass 1 3} }$
as can be seen from its Cayley table:
:$\begin {array} {r|rrrrrr}
+_6 & \tuple {\eqclass 0 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 1 3} & \tuple {\eqclass 0 2, \eqclass 2 3} & \tuple {\eqclass 0 2, \eqclass 1 3} & \tuple {\eqclass 1 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 2 3} \\
\hline
\tuple {\eqclass 0 2, \eqclass 0 3} & \tuple {\eqclass 0 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 1 3} & \tuple {\eqclass 0 2, \eqclass 2 3} & \tuple {\eqclass 1 2, \eqclass 0 3} & \tuple {\eqclass 0 2, \eqclass 1 3} & \tuple {\eqclass 1 2, \eqclass 2 3} \\
\tuple {\eqclass 1 2, \eqclass 1 3} & \tuple {\eqclass 1 2, \eqclass 1 3} & \tuple {\eqclass 0 2, \eqclass 2 3} & \tuple {\eqclass 1 2, \eqclass 0 3} & \tuple {\eqclass 0 2, \eqclass 1 3} & \tuple {\eqclass 1 2, \eqclass 2 3} & \tuple {\eqclass 0 2, \eqclass 0 3} \\
\tuple {\eqclass 0 2, \eqclass 2 3} & \tuple {\eqclass 0 2, \eqclass 2 3} & \tuple {\eqclass 1 2, \eqclass 0 3} & \tuple {\eqclass 0 2, \eqclass 1 3} & \tuple {\eqclass 1 2, \eqclass 2 3} & \tuple {\eqclass 0 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 1 3} \\
\tuple {\eqclass 1 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 0 3} & \tuple {\eqclass 0 2, \eqclass 1 3} & \tuple {\eqclass 1 2, \eqclass 2 3} & \tuple {\eqclass 0 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 1 3} & \tuple {\eqclass 0 2, \eqclass 2 3} \\
\tuple {\eqclass 0 2, \eqclass 1 3} & \tuple {\eqclass 0 2, \eqclass 1 3} & \tuple {\eqclass 1 2, \eqclass 2 3} & \tuple {\eqclass 0 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 1 3} & \tuple {\eqclass 0 2, \eqclass 2 3} & \tuple {\eqclass 1 2, \eqclass 0 3} \\
\tuple {\eqclass 1 2, \eqclass 2 3} & \tuple {\eqclass 1 2, \eqclass 2 3} & \tuple {\eqclass 0 2, \eqclass 0 3} & \tuple {\eqclass 1 2, \eqclass 1 3} & \tuple {\eqclass 0 2, \eqclass 2 3} & \tuple {\eqclass 1 2, \eqclass 0 3} & \tuple {\eqclass 0 2, \eqclass 1 3} \\
\end{array}$
== Sources ==
* : Chapter $13$: Direct products: Direct products: Example $13.4$
Category:Group Direct Product of Cyclic Groups
Group Direct Product of Infinite Cyclic Groups

== Theorem ==
The group direct product of two infinite cyclic groups is not cyclic.
== Proof ==
Let $G_1 = \struct {G_1, \circ_1}$ and $G_2 = \struct {G_2, \circ_2}$ be infinite cyclic groups.
Let $G = \struct {G, \circ} = G_1 \times G_2$.
Let $G_1 = \gen {g_1}, G_2 = \gen {g_2}$.
From Generators of Infinite Cyclic Group:
:$g_1$ and $g_1^{-1}$ are the only generators of $G_1$
:$g_2$ and $g_2^{-1}$ are the only generators of $G_2$.
So a generator of $G$ must be of the form $\tuple { g_1^{\pm 1}, g_2^{\pm 1} }$ to have any hope to generate all of $G$.
, suppose that $G = \gen {\tuple {g_1, g_2} }$.
Let $e_1$ be the identity element of $G_1$.
Let $x_2 \in G_2$.
From the definition of an infinite cyclic group, both $g_1$ and $g_2$ are of infinite order.
Suppose now $\tuple {e_1, x_2} \in \gen {\tuple {g_1, g_2} }$.
Then there is an $i \in \Z$ such that we have:
:$\tuple {g_1, g_2}^i = \tuple {g_1^i, g_2^i} = \tuple {e_1, x_2}$
However, as $g_1$ is of infinite order, it must be that $i = 0$. 
Hence $x_2 = g_2^0 = e_2$, where $e_2$ is the identity element of $G_2$.
It follows that $\tuple {e_1, g_2} \notin \gen {\tuple {g_1, g_2} }$.
Therefore:
:$G \ne \gen {\tuple {g_1, g_2} }$
It follows that $G$ cannot be generated by one element.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 43 \delta$
Category:Group Direct Products
Category:Infinite Cyclic Group
Group Element Commutes with Inverse

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $x \in G$.
Then:
:$x \circ x^{-1} = x^{-1} \circ x$
That is, $x$ commutes with its inverse $x^{-1}$.
== Proof ==
By definition of inverse element:
:$x \circ x^{-1} = e = x^{-1} \circ x$
Hence the result by definition.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory
* : $\S 4.6$. Elementary theorems on groups: Example $86$
Category:Group Theory
Category:Commutativity
Group Element is Self-Inverse iff Order 2

== Theorem ==
Let $\struct {S, \circ}$ be a group whose identity is $e$.
An element $x \in \struct {S, \circ}$ is self-inverse :
:$\order x = 2$
== Proof ==
Let $x \in G: x \ne e$.
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $9$
Category:Order of Group Elements