nyuuzyou/wikis huggingface dataset:
External Center of Similitude of Circles with respect to Radii

== Theorem ==
Let $A$ and $B$ be the centers of two circles $\bigcirc Ar$ and $\bigcirc BR$ whose radii are $r$ and $R$ respectively, $r \ne R$.
Let $\bigcirc Ar$ and $\bigcirc BR$ be such that neither is completely enclosed inside the other.
Let $T$ be the external center of similitude of $\bigcirc Ar$ and $\bigcirc BR$.
Let $P$ and $Q$ be points on the circumference of $\bigcirc Ar$ and $\bigcirc BR$ respectively.
Then:
:the radii $AP$ and $QR$ are parallel
:
:$T$, $P$ and $Q$ are collinear.
== Proof ==
:
=== Sufficient Condition ===
Let $AP$ and $QR$ are parallel.
We note that:
:$\angle TAP = \angle TBQ$
:$\angle PTA = \angle QTB$
and so by Triangles with Two Equal Angles are Similar:
:$\triangle TAP$ and $\triangle TBQ$ are similar.
Hence:
:$\dfrac {AT} {BT} = \dfrac r R$
As $P$ and $Q$ are arbitrary, it follows that the straight lines connecting the ends of all such parallel radii pass through $T$.
The same applies to the tangents, as they touch $\bigcirc Ar$ and $\bigcirc BR$ at parallel radii.
=== Necessary Condition ===
Let $T$, $P$ and $Q$ be collinear.
Let $TP$ and $TQ$ cut $\bigcirc Ar$ and $\bigcirc BR$ at $P'$ and $Q'$.
Then $AP$ and $BQ$ are parallel as are $AP'$ and $BQ'$.
== Sources ==
* : $\text {IV}$. Pure Geometry: Plane Geometry: The Centre of Similitude
Category:Centers of Similitude
External Direct Product Associativity

== Theorem ==
Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$\circ_1$ and $\circ_2$ are associative
:
:$\circ$ is associative.
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
== Also see ==
* External Direct Product Commutativity
* External Direct Product Identity
* External Direct Product Inverses
Category:External Direct Products
Category:Associativity
Category:External Direct Product Associativity
External Direct Product Associativity/General Result

== Theorem ==
Let $\ds \struct {S, \circ} = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
If $\circ_1, \ldots, \circ_n$ are all associative, then so is $\circ$.
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:If $\circ_1, \ldots, \circ_n$ are all associative, then so is the external direct product $\circ$ of $\circ_1, \ldots, \circ_n$.
=== Basis for the Induction ===
$\map P 1$ is true, as this just says:
:$\circ_1$ is associative.
$\map P 2$ is the case:
:If $\circ_1$ and $\circ_2$ are both associative, then so is the external direct product $\circ$ of $\circ_1$ and $\circ_2$.
This has been proved in External Direct Product Associativity.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:If $\circ_1, \ldots, \circ_k$ are all associative, then so is the external direct product $\circ$ of $\circ_1, \ldots, \circ_k$.
Then we need to show:
:If $\circ_1, \ldots, \circ_{k + 1}$ are all associative, then so is the external direct product $\circ$ of $\circ_1, \ldots, \circ_{k + 1}$.
=== Induction Step ===
This is our induction step:
Let $a, b, c \in S^{k + 1}$:
:$a = \tuple {a_1, a_2, \ldots, a_k, a_{k + 1} }$
:$b = \tuple {b_1, b_2, \ldots, b_k, b_{k + 1} }$
:$c = \tuple {c_1, c_2, \ldots, c_k, c_{k + 1} }$
Note that in the below, by abuse of notation, $\circ$ is to be used for two separate operations:
:$\tuple {a_1, a_2, \ldots, a_k, a_{k + 1} } \circ \tuple {b_1, b_2, \ldots, b_k, b_{k + 1} }$
and:
:$\tuple {a_1, a_2, \ldots, a_k} \circ \tuple {b_1, b_2, \ldots, b_k}$
Thus:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
For all $n \in \N_{> 0}$:
:If $\circ_1, \ldots, \circ_n$ are all associative, then so is the external direct product $\circ$ of $\circ_1, \ldots, \circ_n$.
== Also see ==
* External Direct Product Commutativity
* External Direct Product Identity
* External Direct Product Inverses
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.10: \ 1^\circ$
Category:External Direct Product Associativity
External Direct Product Associativity/Necessary Condition

== Theorem ==
Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\circ$ be associative.
Then $\circ_1$ and $\circ_2$ are both also associative.
== Proof ==
Let $\circ$ be associative. 
 it is not the case that both $\circ_1$ and $\circ_2$ are associative.
, suppose $\circ_1$ is not associative.
Hence:
This contradicts our assumption that $\circ$ is associative.
The same argument can be applied, , to the case where $\circ_2$ is not associative.
Hence, by Proof by Contradiction, $\circ_1$ and $\circ_2$ are both associative.
== Also see ==
* External Direct Product Commutativity
* External Direct Product Identity
* External Direct Product Inverses
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.5$
Category:External Direct Product Associativity
External Direct Product Associativity/Sufficient Condition

== Theorem ==
Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\circ_1$ and $\circ_2$ be associative.
Then $\circ$ is also associative.
=== General Result ===
== Proof ==
Let $\circ_1$ and $\circ_2$ be associative. 
Thus $\circ$ is associative.
== Also see ==
* External Direct Product Commutativity
* External Direct Product Identity
* External Direct Product Inverses
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.1: \ 1^\circ$
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.1$
Category:External Direct Product Associativity
External Direct Product Closure

== Theorem ==
Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be algebraic structures.
Let $\struct {S \times T, \circ}$ be the external direct product of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be closed
:
:$\struct {S \times T, \circ}$ is also closed. 
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
Category:Closed Algebraic Structures
Category:External Direct Products
Category:External Direct Product Closure
External Direct Product Closure/General Result

== Theorem ==
Let $\ds \struct {S, \circ} = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
Let $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$ all be closed algebraic structures.
Then $\struct {S, \circ}$ is also a closed algebraic structure.
== Proof ==
Follows directly from External Direct Product Closure.
Category:External Direct Product Closure
External Direct Product Closure/Necessary Condition

== Theorem ==
Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be algebraic structures.
Let $\struct {S \times T, \circ}$ be the external direct product of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\struct {S \times T, \circ}$ be a closed algebraic structure. 
Then $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ are also closed.
== Proof ==
Let $\struct {S \times T, \circ}$ be closed.
Let $s_1, s_2 \in S$ and $t_1, t_2 \in T$.
Then:
demonstrating that $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ are closed.
Category:External Direct Product Closure
External Direct Product Closure/Sufficient Condition

== Theorem ==
Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be algebraic structures.
Let $\struct {S \times T, \circ}$ be the external direct product of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be closed.
Then $\struct {S \times T, \circ}$ is also closed. 
=== General Result ===
== Proof ==
Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be closed.
Let $\tuple {s_1, t_1} \in S \times T$ and $\tuple {s_2, t_2} \in S \times T$.
Then:
demonstrating that $\struct {S \times T, \circ}$ is closed.
Category:External Direct Product Closure
External Direct Product Commutativity

== Theorem ==
Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$\circ_1$ and $\circ_2$ are commutative
:
:$\circ$ is commutative.
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
== Also see ==
* External Direct Product Associativity
* External Direct Product Identity
* External Direct Product Inverses
Category:External Direct Products
Category:Commutativity
Category:External Direct Product Commutativity
External Direct Product Commutativity/General Result

== Theorem ==
Let $\ds \struct {S, \circ} = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
If $\circ_1, \ldots, \circ_n$ are all commutative, then so is $\circ$.
== Proof ==
Suppose that, for all $k \in \N^*_n$, $\circ_k$ is commutative.
Let $\tuple {s_1, s_2, \ldots, s_n}$ and $\tuple {t_1, t_2, \ldots, t_n}$ be elements of $\struct {S_1, \circ_1} \times \struct {S_2, \circ_2} \times \cdots \times \struct {S_n, \circ_n}$.
Hence the result.
== Also see ==
* External Direct Product Associativity
* External Direct Product Identity
* External Direct Product Inverses
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.10: \ 1^\circ$
Category:External Direct Product Commutativity
External Direct Product Commutativity/Necessary Condition

== Theorem ==
Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\circ$ be commutative.
Then $\circ_1$ and $\circ_2$ are both also commutative.
== Proof ==
Let $\circ$ be commutative. 
 it is not the case that both $\circ_1$ and $\circ_2$ are commutative.
, suppose $\circ_1$ is not commutative.
Hence:
This contradicts our assumption that $\circ$ is commutative.
The same argument can be applied, , to the case where $\circ_2$ is not commutative.
Hence, by Proof by Contradiction, $\circ_1$ and $\circ_2$ are both commutative.
== Also see ==
* External Direct Product Associativity
* External Direct Product Identity
* External Direct Product Inverses
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.5$
Category:External Direct Product Commutativity
External Direct Product Commutativity/Sufficient Condition

== Theorem ==
Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\circ_1$ and $\circ_2$ be commutative operations.
Then $\circ$ is also a commutative operation.
=== General Result ===
== Proof ==
Let $\circ_1$ and $\circ_2$ be commutative operations.
Thus $\circ$ is commutative.
== Also see ==
* External Direct Product Associativity
* External Direct Product Identity
* External Direct Product Inverses
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.1: \ 1^\circ$
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.1$
Category:External Direct Product Commutativity
External Direct Product Identity

== Theorem ==
Let $\struct {S \times T, \circ}$ be the external direct product of two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$\struct {S, \circ_1}$ has identity element $e_S$ and $\struct {T, \circ_2}$ has identity element $e_T$
:
:$\tuple {e_S, e_T}$ is the identity element for $\struct {S \times T, \circ}$.
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
== Also see ==
* External Direct Product Associativity
* External Direct Product Commutativity
* External Direct Product Inverses
Category:External Direct Products
Category:Identity Elements
Category:External Direct Product Identity
External Direct Product Identity/General Result

== Theorem ==
Let $\ds \struct {\SS, \circ} = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
Let $e_1, e_2, \ldots, e_n$ be the identity elements of $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$ respectively.
Then $\tuple {e_1, e_2, \ldots, e_n}$ is the identity element of $\struct {\SS, \circ}$.
== Proof ==
Let $s := \tuple {s_1, s_2, \ldots, s_n}$ be an arbitrary element of $\struct {S_1, \circ_1} \times \struct {S_2, \circ_2} \times \cdots \times \struct {S_n, \circ_n}$.
Let $e := \tuple {e_1, e_2, \ldots, e_n}$.
Then:
and:
== Also see ==
* External Direct Product Associativity
* External Direct Product Commutativity
* External Direct Product Inverses
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.10: \ 2^\circ$
Category:External Direct Product Identity
External Direct Product Identity/Necessary Condition

== Theorem ==
Let $\struct {S \times T, \circ}$ be the external direct product of two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\struct {S \times T, \circ}$ have an identity element $\tuple {e_S, e_T}$.
Then:
:$\struct {S, \circ_1}$ has an identity element $e_S$
and:
:$\struct {T, \circ_2}$ has an identity element $e_T$.
== Proof ==
Let $\tuple {e_S, e_T}$ be an identity of $\struct {S \times T, \circ}$.
Then we have:
and:
Thus $e_S$ and $e_T$ are identity elements of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ respectively.
== Also see ==
* External Direct Product Associativity
* External Direct Product Commutativity
* External Direct Product Inverses
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.5$
Category:External Direct Product Identity
External Direct Product Identity/Sufficient Condition

== Theorem ==
Let $\struct {S \times T, \circ}$ be the external direct product of two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let:
:$\struct {S, \circ_1}$ have identity element $e_S$
and:
:$\struct {T, \circ_2}$ have identity element $e_T$.
Then $\tuple {e_S, e_T}$ is the identity element for $\struct {S \times T, \circ}$.
=== General Result ===
== Proof ==
Let $e_S$ and $e_T$ be the identity elements of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ respectively.
Thus $\tuple {e_S, e_T}$ is the identity of $\struct {S \times T, \circ}$.
== Also see ==
* External Direct Product Associativity
* External Direct Product Commutativity
* External Direct Product Inverses
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.1: \ 2^\circ$
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.1$
Category:External Direct Product Identity
External Direct Product Inverses

== Theorem ==
Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$s^{-1}$ and $t^{-1}$ are inverse elements of $s \in \struct {S, \circ_1}$ and $t \in \struct {T, \circ_2}$ respectively
:
:$\tuple {s^{-1}, t^{-1} }$ is an inverse element of $\tuple {s, t} \in \struct {S \times T, \circ}$.
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
== Also see ==
* External Direct Product Associativity
* External Direct Product Commutativity
* External Direct Product Identity
Category:External Direct Products
Category:Inverse Elements
Category:External Direct Product Inverses
External Direct Product Inverses/General Result

== Theorem ==
Let $\ds \struct {\SS, \circ} = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
Let $\tuple {x_1, x_2, \ldots, x_n} \in S$.
Let $y_k$ be an inverse of $x_k$ in $\struct {S_k, \circ_k}$ for each of $k \in \N^*_n$.
Then $\tuple {y_1, y_2, \ldots, y_n}$ is the inverse of $\tuple {x_1, x_2, \ldots, x_n} \in \SS$ in $\struct {\SS, \circ}$.
== Proof ==
Let $e_1, e_2, \ldots, e_n$ be the identity elements of $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$ respectively.
Let $x := \tuple {x_1, x_2, \ldots, x_n}$.
Let $y := \tuple {x_1, x_2, \ldots, x_n}$.
From External Direct Product Identity, $e := \tuple {e_1, e_2, \ldots, e_n}$ is the identity element of $\SS$.
Then:
and:
== Also see ==
* External Direct Product Associativity
* External Direct Product Commutativity
* External Direct Product Identity
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.10: \ 3^\circ$
Category:External Direct Product Inverses
External Direct Product Inverses/Necessary Condition

== Theorem ==
Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\tuple {s^{-1}, t^{-1} }$ be an inverse of $\tuple {s, t} \in \struct {S \times T, \circ}$.
Then:
:$s^{-1}$ is an inverse of $s \in \struct {S, \circ_1}$
and:
:$t^{-1}$ is an inverse of $t \in \struct {T, \circ_2}$.
== Proof ==
Let $\tuple {e_S, e_T}$ be the identity element of $\struct {S \times T, \circ}$.
Let $\tuple {s^{-1}, t^{-1} }$ be an inverse element of $\tuple {s, t} \in \struct {S \times T, \circ}$.
Then we have:
and:
Thus $s^{-1}$ and $t^{-1}$ are inverse elements of $s \in \struct {S, \circ_1}$ and $t \in \struct {T, \circ_2}$ respectively.
== Also see ==
* External Direct Product Associativity
* External Direct Product Commutativity
* External Direct Product Inverses
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.5$
Category:External Direct Product Identity
External Direct Product Inverses/Sufficient Condition

== Theorem ==
Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let:
:$s^{-1}$ be an inverse of $s \in \struct {S, \circ_1}$
and:
:$t^{-1}$ be an inverse of $t \in \struct {T, \circ_2}$.
Then $\tuple {s^{-1}, t^{-1} }$ is an inverse of $\tuple {s, t} \in \struct {S \times T, \circ}$.
=== General Result ===
== Proof ==
Let:
:$e_S$ be the identity for $\struct {S, \circ_1}$
and:
:$e_T$ be the identity for $\struct {T, \circ_2}$.
Also let:
:$s^{-1}$ be the inverse of $s \in \struct {S, \circ_1}$
and
:$t^{-1}$ be the inverse of $t \in \struct {T, \circ_2}$.
Then:
Thus the inverse of $\tuple {s, t}$ is $\tuple {s^{-1}, t^{-1} }$.
== Also see ==
* External Direct Product Associativity
* External Direct Product Commutativity
* External Direct Product Identity
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.1: \ 3^\circ$
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.1$
Category:External Direct Product Inverses
External Direct Product of Abelian Groups is Abelian Group

== Theorem ==
Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be groups.
Then the group direct product $\struct {G \times H, \circ}$ is abelian both $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ are abelian.
=== General Result ===
== Proof ==
Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be groups whose identities are $e_G$ and $e_H$ respectively.
From External Direct Product of Groups is Group, $\struct {G \times H, \circ}$ is indeed a group whose identity is $\tuple {e_G, e_H}$.
Suppose $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ are both abelian.
Then from External Direct Product Commutativity, $\struct {G \times H, \circ}$ is also abelian.
Now suppose that $\struct {G \times H, \circ}$ is abelian.
Then:
Thus:
: $g_1 \circ_1 g_2 = g_2 \circ_1 g_1$
and $\struct {G, \circ_1}$ is seen to be abelian.
The same argument holds for $\struct {H, \circ_2}$.
== Also see ==
* External Direct Product of Groups is Group
== Sources ==
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
* : Chapter $6$: An Introduction to Groups: Exercise $1$
* : Chapter $1$: Definitions and Examples: Exercise $6$
* : Chapter $13$: Direct products: Proposition $13.1 \ (1)$
Category:Group Direct Products
Category:Abelian Groups
External Direct Product of Abelian Groups is Abelian Group/General Result

== Theorem ==
The external direct product of a finite sequence of abelian groups is itself an abelian group.
== Proof ==
Let $\struct {G_1, \circ_1}, \struct {G_2, \circ_2}, \ldots, \struct {G_n, \circ_n}$ be abelian groups.
Let $\ds \struct {G, \circ} = \prod_{k \mathop = 1}^n G_k$ be the external direct product of $\struct {G_1, \circ_1}, \struct {G_2, \circ_2}, \ldots, \struct {G_n, \circ_n}$.
From External Direct Product of Groups is Group: Finite Product it follows that $\struct {G, \circ}$ is a group.
By definition, each of $\circ_1, \circ_2, \ldots, \circ_n$ are commutative operations.
From External Direct Product Commutativity: General Result it follows that $\circ$ is commutative.
Hence, by definition, $\struct {G, \circ}$ is an abelian group.
Category:Abelian Groups
Category:Group Direct Products