nyuuzyou/wikis huggingface dataset:
Sets of Operations on Set of 3 Elements/Automorphism Group of A/Operations with Identity

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\AA$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$, that is, $\map \Gamma S$.
Then:
:None of the operations of $\AA$ has an identity element.
== Proof ==
Recall from Automorphism Group of $\AA$ the elements of $\AA$, expressed in Cayley table form:
:$\begin{array}{c|ccc}
\to & a & b & c \\
\hline
a & a & b & c \\
b & a & b & c \\
c & a & b & c \\
\end{array}
\qquad
\begin{array}{c|ccc}
\gets & a & b & c \\
\hline
a & a & a & a \\
b & b & b & b \\
c & c & c & c \\
\end{array}
\qquad
\begin{array}{c|ccc}
\circ & a & b & c \\
\hline
a & a & c & b \\
b & c & b & a \\
c & b & a & c \\
\end{array}$
The result is apparent by inspection.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(c)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of B/Isomorphism Classes

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\BB$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ forms the set $\set {I_S, \tuple {a, b, c}, \tuple {a, c, b} }$, where $I_S$ is the identity mapping on $S$.
Let $\oplus \in \BB$.
Then the isomorphism class of $\oplus$ consists of $\oplus$ and exactly one other operation $\otimes$ on $S$ such that $\otimes \in \BB$.
That is, the elements of $\BB$ are in doubleton isomorphism classes each of which is a subset of $\BB$.
== Proof ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(b)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of C n

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\CC_1$, $\CC_2$ and $\CC_3$ be respectively the set of all operations $\circ$ on $S$ such that the groups of automorphisms of $\struct {S, \circ}$ are as follows:
where $I_S$ is the identity mapping on $S$.
Then:
:Each of $\CC_1$, $\CC_2$ and $\CC_3$ has $3^4 - 3$ elements.
=== Isomorphism Classes ===
=== Operations with Identity ===
=== Commutative Operations ===
== Proof ==
Recall the definition of (group) automorphism:
:$\phi$ is an automorphism on $\struct {S, \circ}$ :
::$\phi$ is a permutation of $S$
::$\phi$ is a homomorphism on $\struct {S, \circ}$: $\forall a, b \in S: \map \phi {a \circ b} = \map \phi a \circ \map \phi b$
From Identity Mapping is Group Automorphism, $I_S$ is always an automorphism on $\struct {S, \circ}$.
Hence it is not necessary to analyse the effect of $I_S$ on $S$.
, we will analyse the nature of $\CC_1$.
Let $n$ be the number of operations $\circ$ on $S$ such that $\tuple {a, b}$ is an automorphism of $\struct {S, \circ}$.
Let us denote the permutation $\tuple {a, b}$ as $r: S \to S$, defined as:
:$r = \map r a = b, \map r b = a, \map r c = c$
We select various product elements $x \circ y \in S$ and determine how $r$ constrains other product elements.
=== Lemma 1 ===
=== Lemma 2 ===
Hence, selecting each of the $3$ options for:
:$a \circ a$
:$a \circ b$
:$a \circ c$
:$c \circ a$
generates a unique operation on $S$.
From the Product Rule for Counting it follows that:
:$n = 3 \times 3 \times 3 \times 3 = 3^4$
From Automorphism Group of $\AA$, $3$ of those operations are elements of the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ are $\map \Gamma S$.
Hence those $3$ elements are not in $\CC_1$, and are removed from the overall count.
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(a)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of C n/Commutative Operations

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\CC_1$, $\CC_2$ and $\CC_3$ be respectively the set of all operations $\circ$ on $S$ such that the groups of automorphisms of $\struct {S, \circ}$ are as follows:
where $I_S$ is the identity mapping on $S$.
Then:
$8$ of the operations of each of $\CC_1$, $\CC_2$ and $\CC_3$ is commutative.
== Proof ==
, we will analyse the nature of $\CC_1$.
Recall this lemma:
=== Lemma ===
The operations brought about by $a \circ a$, $b \circ b$ and $c \circ c$ have no effect on whether those operations are commutative.
The operations in which $a \circ b = a$ and $a \circ b = b$ are definitely not commutative, as these are such that $a \circ b \ne b \circ a$.
Hence all commutative operations of $\CC_1$ have $a \circ b = c = b \circ a$.
Hence, of the above, we have:
:$3$ options of $a \circ a$ each give rise to a partial operation.
:Only $1$ option of $a \circ b$ gives rise to a partial operation, that is $a \circ b$.
:The $3$ options of $a \circ c$ each give rise to a partial operation, but they force the corresponding options of $c \circ a$, $b \circ c$ and $c \circ b$.
Hence by the Product Rule for Counting there are $3 \times 3 = 9$ commutative operations which can be constructed thus.
However, note that one of these:
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & c & b \\
b & c & b & a \\
c & b & a & c \\
\end {array}$
has already been accounted for in Automorphism Group of $\AA$: Commutative Operations.
This commutative operations is such that the group of automorphisms of $\struct {S, \circ}$ forms the complete symmetric group on $S$, not just the given permutations.
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(d)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of C n/Isomorphism Classes

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\CC_1$, $\CC_2$ and $\CC_3$ be respectively the set of all operations $\circ$ on $S$ such that the groups of automorphisms of $\struct {S, \circ}$ are as follows:
where $I_S$ is the identity mapping on $S$.
Let $\CC = \CC_1 \cup \CC_2 \cup \CC_3$.
Let $\oplus \in \CC$.
Then the isomorphism class of $\oplus$ consists of $3$ elements: exactly one operation from each of $\CC_1$, $\CC_2$ and $\CC_3$.
== Proof ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(b)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of C n/Lemma 1

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\CC_1$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ forms the set $\set {I_S, \tuple {a, b} }$, where $I_S$ is the identity mapping on $S$.
Then:
$c$ is an idempotent element under $\circ$, that is:
:$c \circ c = c$
== Proof ==
Recall the definition of (group) automorphism:
:$\phi$ is an automorphism on $\struct {S, \circ}$ :
::$\phi$ is a permutation of $S$
::$\phi$ is a homomorphism on $\struct {S, \circ}$: $\forall a, b \in S: \map \phi {a \circ b} = \map \phi a \circ \map \phi b$
Let us denote $\tuple {a, b}$ as the mapping $r: S \to S$:
:$r := \map r a = b, \map r b = a, \map r c = c$
 $c$ is not idempotent.
Then $c \circ c = x$, where $x = a$ or $x = b$.
So it cannot be the case that $c \circ c = a$ or $c \circ c = b$.
Hence $c \circ c = c$, that is, $c$ is idempotent.
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of C n/Lemma 2

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\CC_1$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ forms the set $\set {I_S, \tuple {a, b} }$, where $I_S$ is the identity mapping on $S$.
Then:
== Proof ==
Recall the definition of (group) automorphism:
:$\phi$ is an automorphism on $\struct {S, \circ}$ :
::$\phi$ is a permutation of $S$
::$\phi$ is a homomorphism on $\struct {S, \circ}$: $\forall a, b \in S: \map \phi {a \circ b} = \map \phi a \circ \map \phi b$
Let us denote $\tuple {a, b}$ as the mapping $r: S \to S$:
:$r := \map r a = b, \map r b = a, \map r c = c$
In Lemma 1 it has been established that:
We select values for $x$ in the expression $a \circ b = x$ and determine how $r$ constrains other product elements.
Thus:
Hence:
Similarly:
Hence:
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of C n/Operations with Identity

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\CC_1$, $\CC_2$ and $\CC_3$ be respectively the set of all operations $\circ$ on $S$ such that the groups of automorphisms of $\struct {S, \circ}$ are as follows:
where $I_S$ is the identity mapping on $S$.
Then:
:$9$ of the operations of each of $\CC_1$, $\CC_2$ and $\CC_3$ has an identity element.
== Proof ==
, we will analyse the nature of $\CC_1$.
Recall this lemma:
=== Lemma ===
We observe that neither $a$ nor $b$ can be an identity, because:
However, we note that:
and so from here it may still be the case that $c$ may be an identity for at least one element of $\CC_1$.
There are still $3$ options for $a \circ a$ and $a \circ b$ which do not affect the behaviour of $c \circ x$ or $x \circ c$.
From the Product Rule for Counting it follows that there are $3 \times 3 = 9$ operations of $\CC_1$ which have an identity element.
As $\CC_2$ and $\CC_3$ can be obtained from $\CC_1$ just by renaming elements of $S$, the same applies to both $\CC_2$ and $\CC_3$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(c)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of D

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\DD$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ forms the set $\set {I_S}$, where $I_S$ is the identity mapping on $S$.
Then:
:$\DD$ has $19 \, 422$ elements.
=== Isomorphism Classes ===
=== Operations with Identity ===
=== Commutative Operations ===
== Proof ==
Let $n$ denote the cardinality of $\DD$.
Equivalently, $n$ equals the number of operations $\circ$ on $S$ on which the only automorphism is $I_S$.
Recall these definitions:
Let $\AA$, $\BB$, $\CC_1$, $\CC_2$ and $\CC_3$ be respectively the set of all operations $\circ$ on $S$ such that the groups of automorphisms of $\struct {S, \circ}$ are as follows:
=== Lemma ===
Let $N$ be the total number of operations on $S$.
From the lemma, and from the Fundamental Principle of Counting:
:$N = \card \AA + \card \BB + \card {\CC_1} + \card {\CC_2} + \card {\CC_3} + \card {\DD}$
From Count of Binary Operations on Set:
:$N = 3^{3^2} = 3^9 = 19 \, 683$
Then we have:
:From Automorphism Group of $\AA$: $\card \AA = 3$
:From Automorphism Group of $\BB$: $\card \BB = 3^3 - 3 = 24$
:From Automorphism Group of $\CC_n$: for $n = 1, 2, 3: \card {\CC_n} = 3^4 - 3 = 78$
Hence we have:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(a)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of D/Commutative Operations

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\DD$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ forms the set $\set {I_S}$, where $I_S$ is the identity mapping on $S$.
Then:
:$696$ of the operations of $\DD$ is commutative.
== Proof ==
Let $n$ denote the number of commutative operations of $\DD$.
Recall these definitions:
Let $\AA$, $\BB$, $\CC_1$, $\CC_2$ and $\CC_3$ be respectively the set of all operations $\circ$ on $S$ such that the groups of automorphisms of $\struct {S, \circ}$ are as follows:
Let $N$ be the total number of commutative operations on $S$.
Let:
:$A$ denote the number of commutative operations in $\AA$
:$B$ denote the number of commutative operations in $\BB$
:$C$ denote the total number of commutative operations in $\CC_1$, $\CC_2$ and $\CC_3$.
From the lemma, and from the Fundamental Principle of Counting:
:$N = A + B + C + D$
From Count of Commutative Binary Operations on Set:
:$N = 3^6 = 729$
Then we have:
:From Automorphism Group of $\AA$: Operations with Identity: $A = 1$
:From Automorphism Group of $\BB$: Operations with Identity: $B = 8$
:From Automorphism Group of $\CC_n$: Operations with Identity: $C = 3 \times 8$
Hence we have:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(d)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of D/Isomorphism Classes

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\DD$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ forms the set $\set {I_S}$, where $I_S$ is the identity mapping on $S$.
Let $\oplus \in \DD$.
Then the isomorphism class of $\oplus$ consists of $6$ elements, all of which are in $\DD$.
== Proof ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(b)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of D/Lemma

== Lemma for Sets of Operations on Set of 3 Elements ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\AA$, $\BB$, $\CC_1$, $\CC_2$, $\CC_3$ and $\DD$ be respectively the set of all operations $\circ$ on $S$ such that the groups of automorphisms of $\struct {S, \circ}$ are as follows:
Then:
:$\set {\AA, \BB, \CC_1, \CC_2, \CC_3, \DD}$ forms a partition of the set of all operations on $S$.
== Proof ==
Let $\OO$ denote the set of all operations on $S$.
Let $\SS := \set {\AA, \BB, \CC_1, \CC_2, \CC_3, \DD}$.
First we note that from Identity Mapping is Group Automorphism, $I_S$ is always an automorphism on $\struct {S, \circ}$.
Thus every operation on $S$ belongs to at least one of $\set {\AA, \BB, \CC_1, \CC_2, \CC_3, \DD}$.
That is:
:$\OO = \ds \bigcup \SS$
Then we note that from:
:Automorphism Group of $\AA$
:Automorphism Group of $\BB$
:Automorphism Group of $\CC_n$
each of $\AA$, $\BB$, $\CC_1$, $\CC_2$ and $\CC_3$ is non-empty.
Consider for example the constant operation on $S$:
:$\forall x, y \in S: x \circ y = a$
From the above cited results, $\circ$ as described does not belong to any of $\AA$, $\BB$, $\CC_1$, $\CC_2$ and $\CC_3$.
Hence it must belong to $\DD$.
Thus $\DD$ is also non-empty.
It remains to be shown that $\SS$ is a pairwise disjoint set of sets.
By definition, $\DD$ is disjoint with all other elements of $\SS$.
By construction, $\AA$ is also disjoint with all other elements of $\SS$.
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of D/Operations with Identity

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\DD$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ forms the set $\set {I_S}$, where $I_S$ is the identity mapping on $S$.
Then:
:$216$ of the operations of $\DD$ has an identity element.
== Proof ==
Let $n$ denote the number of operations of $\DD$ which have an identity element.
Recall these definitions:
Let $\AA$, $\BB$, $\CC_1$, $\CC_2$ and $\CC_3$ be respectively the set of all operations $\circ$ on $S$ such that the groups of automorphisms of $\struct {S, \circ}$ are as follows:
Let $N$ be the total number of operations on $S$ which have an identity element.
Let:
:$A$ denote the number of operations in $\AA$ which have an identity element
:$B$ denote the number of operations in $\BB$ which have an identity element
:$C$ denote the total number of operations in $\CC_1$, $\CC_2$ and $\CC_3$ which have an identity element.
From the lemma, and from the Fundamental Principle of Counting:
:$N = A + B + C + D$
From Count of Binary Operations with Identity:
:$N = 3^5 = 243$
Then we have:
:From Automorphism Group of $\AA$: Operations with Identity: $A = 0$
:From Automorphism Group of $\BB$: Operations with Identity: $B = 0$
:From Automorphism Group of $\CC_n$: Operations with Identity: $C = 3 \times 9$
Hence we have:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(c)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Commutative Operations

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\PP$ be the set of all commutative operations $\circ$ on $S$.
Then the elements of $\PP$ are divided in $129$ isomorphism classes.
That is, up to isomorphism, there are $129$ commutative operations on $S$ which have an identity element.
== Proof ==
From Automorphism Group of $\AA$: Commutative Operations:
:there is exactly $1$ commutative operation in $\AA$.
From Automorphism Group of $\BB$: Commutative Operations:
:there are $8$ commutative operations in $\BB$.
From Automorphism Group of $\CC_n$: Commutative Operations:
:there are $3 \times 8$ commutative operations in $\CC$.
From Automorphism Group of $\DD$: Commutative Operations:
:there are $696$ commutative operations in $\DD$.
From Automorphism Group of $\CC_n$: Isomorphism Classes:
:the elements of $\BB$ form isomorphism classes in pairs.
From Automorphism Group of $\CC_n$: Isomorphism Classes:
:the elements of $\CC$ form isomorphism classes in threes.
From Automorphism Group of $\DD$: Isomorphism Classes:
:the elements of $\DD$ form isomorphism classes in sixes.
Hence there are:
:$\dfrac 8 2 = 4$ isomorphism classes of commutative operations in $\BB$.
:$\dfrac {3 \times 8} 3 = 8$ isomorphism classes of commutative operations in $\CC$.
:$\dfrac {696} 6 = 116$ isomorphism classes of commutative operations in $\DD$.
Thus there are $1 + 4 + 8 + 116 = 129$ isomorphism classes of operations $\circ$ on $S$ which have an identity element..
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(d)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Isomorphism Classes

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\MM$ be the set of all operations $\circ$ on $S$.
Then the elements of $\MM$ are divided in $3330$ isomorphism classes.
That is, up to isomorphism, there are $3330$ operations on $S$.
== Proof ==
From Automorphism Group of $\AA$: Isomorphism Classes:
:each element of $\AA$ is in its own isomorphism class.
Hence $\AA$ contributes $3$ isomorphism classes.
From Automorphism Group of $\BB$: Isomorphism Classes:
:the $24$ elements of $\BB$ form $12$ isomorphism classes in pairs.
From Automorphism Group of $\CC_n$: Isomorphism Classes:
:the $3 \times 78$ elements of $\CC$ form $78$ isomorphism classes in threes.
From Automorphism Group of $\DD$: Isomorphism Classes:
:the $19 \, 422$ elements of $\DD$ form $3237$ isomorphism classes in sixes.
Thus there are $3 + 12 + 78 + 3237 = 3330$ isomorphism classes.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(b)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Operations with Identity

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\NN$ be the set of all operations $\circ$ on $S$ which have an identity element.
Then the elements of $\NN$ are divided in $45$ isomorphism classes.
That is, up to isomorphism, there are $45$ operations on $S$ which have an identity element.
== Proof ==
From Automorphism Group of $\AA$: Operations with Identity:
:there are no elements of $\AA$ which have an identity element.
From Automorphism Group of $\BB$: Operations with Identity:
:there are no elements of $\BB$ which have an identity element.
From Automorphism Group of $\CC_n$: Operations with Identity:
:there are $3 \times 9$ elements of $\CC$ which have an identity element.
From Automorphism Group of $\DD$: Operations with Identity:
:there are $216$ elements of $\DD$ which have an identity element.
From Automorphism Group of $\CC_n$: Isomorphism Classes:
:the elements of $\CC$ form isomorphism classes in threes.
From Automorphism Group of $\DD$: Isomorphism Classes:
:the elements of $\DD$ form isomorphism classes in sixes.
Hence there are:
:$\dfrac {3 \times 9} 3 = 9$ isomorphism classes of elements of $\CC$ which have an identity element.
:$\dfrac {216} 6 = 36$ isomorphism classes of elements of $\DD$ which have an identity element.
Thus there are $3 + 36 = 45$ isomorphism classes of operations $\circ$ on $S$ which have an identity element..
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(c)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Permutations of Equivalent Sets are Equivalent

== Theorem ==
Let $A$ and $B$ be sets such that:
:$A \sim B$
where $\sim$ denotes set equivalence.
Let $\map \Gamma A$ denote the set of permutations on $A$.
Then:
:$\map \Gamma A \sim \map \Gamma B$
== Proof ==
By definition of set equivalence, let $f: A \to B$ be a bijection.
Define $\Phi : \map \Gamma A \to \map \Gamma B$ by:
:$\map \Phi \gamma := f \circ \gamma \circ f^{-1}$
By definition of permutation, each $\gamma \in \map \Gamma A$ is a bijection.
By Composite of Bijections is Bijection, each $f \circ \gamma$ is a bijection.
By Inverse of Bijection is Bijection, the inverse mapping $f^{-1}$ is a bijection.
Then by Composite of Bijections is Bijection, each $\map \Phi \gamma = f \circ \gamma \circ f^{-1}$ is a bijection.
Thus each $\map \Phi \gamma$ is a permutation of $B$, so $\Phi$ is well-defined.
Define $\Psi : \map \Gamma B \to \map \Gamma A$ by:
:$\map \Psi \delta := f^{-1} \circ \delta \circ f$
$\Psi$ is a well-defined mapping by the same reasoning as for $\Phi$.
Then for each $\gamma \in \map \Gamma A$:
Thus $\Psi \circ \Phi = I_{\map \Gamma A}$.
Similarly, for each $\delta \in \map \Gamma B$:
Thus $\Phi \circ \Psi = I_{\map \Gamma B}$.
By Bijection iff Left and Right Inverse, both $\Phi$ and $\Psi$ are bijections.
In particular:
:$\map \Gamma A \sim \map \Gamma B$
== Sources ==
* : Chapter $3$. Mappings: Exercise $12$
Category:Permutations
Category:Set Equivalence
Seven Touching Cylinders

== Theorem ==
It is possible to arrange $7$ identical cylinders so that each one touches each of the others.
The cylinders must be such that their heights must be at least $\dfrac {7 \sqrt 3} 2$ of the diameters of their bases.
== Proof ==
:
It remains to be proved that the heights of the cylinders must be at least $\dfrac {7 \sqrt 3} 2$ of the diameters of their bases.
== Historical Note ==
== Sources ==
* : $12$: Nine More Problems: Answers: $1$.
* : Answers: $312$. The Six Submarines
Category:Right Circular Cylinders
Category:Recreational Mathematics
Seven Touching Cylinders/Historical Note

== Historical Note on Seven Touching Cylinders ==
 originally raised the similar Six Touching Cylinders, which was an old puzzle, in his Mathematical Games column in .
It was originally written in the context of cigarettes.
On its publication, he received letters from about $15$ readers who had discovered this $7$-cylinder solution.
He credited and for the diagram.
== Sources ==
* : $12$: Nine More Problems: Answers: $1$.
* : $12$: Nine More Problems: Answers: $1$.
Category:Historical Notes