nyuuzyou/wikis huggingface dataset:
Powers of 2 and 5 without Zeroes

== Theorem ==
The following $n \in \Z$ are such that both $2^n$ and $5^n$ have no zeroes in their decimal representation:
:$0, 1, 2, 3, 4, 5, 6, 7, 9, 18, 33$
== Proof ==
:
$n$ 	$2^n$ 	$5^n$
$0$ 	$1$ 	$1$
$1$ 	$2$ 	$5$
$2$ 	$4$ 	$25$
$3$ 	$8$ 	$125$
$4$ 	$16$ 	$625$
$5$ 	$32$ 	$3125$
$6$ 	$64$ 	$15 \, 625$
$7$ 	$128$ 	$78 \, 125$
$9$ 	$512$ 	$1 \, 953 \, 125$
$18$	$262 \, 144$ 	$3 \, 814 \, 697 \, 265 \, 625$
$33$	$8 \, 589 \, 934 \, 592$	$116 \, 415 \, 321 \, 826 \, 934 \, 814 \, 453 \, 125$
It is probable that $n = 33$ is the final instance.
== Also see ==
* Powers of 2 with no Zero in Decimal Representation
* Powers of 5 with no Zero in Decimal Representation
== Sources ==
* : $33$
Category:Powers of 2
Category:Powers of 5
Powers of 2 not containing Digit Power of 2

== Theorem ==
$2^{16} = 65 \, 536$ is the only known power of $2$, up to $2^{31 \, 000}$, whose digits do not contain $1$, $2$, $4$ or $8$.
== Proof ==
This has been demonstrated by an exhaustive search.
== Historical Note ==
== Sources ==
* 
* 
* : $65,536$
Category:Powers of 2
Category:65,536
Powers of 2 not containing Digit Power of 2/Historical Note

== Historical Note on Powers of 2 not containing Digit Power of 2 ==
This question appears to have first been raised by in $1989$.
While no formal attempt has been made to solve it, several searches were made in response, as follows:
: verified the result up to $2^{167}$.
: showed that if the powers of $2$ contain between $500$ and $1000$ digits, the digits $1, 2, 4, 8$ occur fairly normally.
: checked the powers of $2$ up to $2^{3320}$, finding no other solution.
: searched through $2^{20703}$, also finding no other solution.
:Finally, went to $2^{31000}$, which contains $9332$ digits, with the same result.
== Sources ==
* 
* 
* : $65,536$
Category:Powers of 2
Category:65,536
Powers of 2 whose Digits are Powers of 2

== Open Question ==
The only known powers of $2$ whose digits are also all powers of $2$ are:
:$1, 2, 4, 8, 128$
Are there any more?
== Progress ==
Demonstrated up to at least $2^{10 \, 000 \, 000}$.
This can be achieved by looking at the lowest $20$ digits only, by calculating those of successive powers of $2$ after applying the modulo operation.
== Sources ==
* : $128$
* 
* : $128$
Category:Powers of 2
Category:Open Questions
Powers of 2 with no Zero in Decimal Representation

== Unproven Hypotheses ==
The following powers of $2$ which contain no zero in their decimal representation are believed to be all that exist:
:$1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 8192, 16 \, 384,$
:$32 \, 768, 65 \, 536, 262 \, 144, 524 \, 288, 16 \, 777 \, 216, 33 \, 554 \, 432, 134 \, 217 \, 728,$
:$268 \, 435 \, 456, 2 \, 147 \, 483 \, 648, 4 \, 294 \, 967 \, 296, 8 \, 589 \, 934 \, 592, 17 \, 179 \, 869 \, 184,$
:$34 \, 359 \, 738 \, 368, 68 \, 719 \, 476 \, 736, 137 \, 438 \, 953 \, 472, 549 \, 755 \, 813 \, 888,$
:$562 \, 949 \, 953 \, 421 \, 312, 2 \, 251 \, 799 \, 813 \, 685 \, 248, 147 \, 573 \, 952 \, 589 \, 676 \, 412 \, 928,$
:$4 \, 722 \, 366 \, 482 \, 869 \, 645 \, 213 \, 696, 75 \, 557 \, 863 \, 725 \, 914 \, 323 \, 419 \, 136, 151 \, 115 \, 727 \, 451 \, 828 \, 646 \, 838 \, 272,$
:$2 \, 417 \, 851 \, 639 \, 229 \, 258 \, 349 \, 412 \, 352, 77 \, 371 \, 252 \, 455 \, 336 \, 267 \, 181 \, 195 \, 264$
but this has not been proven.
The corresponding indices are:
:$0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 14, 15, 16, 18, 19, 24, 25, 27, 28, 31, 32, 33, 34, 35, 36, 37, 39, 49, 51, 67, 72, 76, 77, 81, 86$
== Progress ==
== Sources ==
* : $86$
Category:Powers of 2
Category:86
Powers of 3 Modulo 8

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a strictly positive integer.
Then:
:$3^n \equiv \begin {cases} 1 \pmod 8 & : \text {$n$ even} \\ 3 \pmod 8 & : \text {$n$ odd} \end {cases}$
== Proof 1 ==
== Proof 2 ==
Category:Powers of 3
Category:Modulo Arithmetic
Category:Powers of 3 Modulo 8
Powers of 3 Modulo 8/Proof 1

== Theorem ==
== Proof ==
Let the statement be rewritten as:
For all $r \in \Z_{\ge 0}$: 
:$3^r \equiv \begin {cases} 1 \pmod 8 & : r = 2 n \\ 3 \pmod 8 & : r = 2 n + 1 \end {cases}$
where $n \in \Z_{\ge 0}$.
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$3^{2 n} \equiv 1 \pmod 8$
$\map P 0$ is the case:
:$3^{2 \times 0} = 3^0 = 1 \equiv 1 \pmod 8$
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$3^{2 k} \equiv 1 \pmod 8$
from which it is to be shown that:
:$3^{2 \paren {k + 1} } \equiv 1 \pmod 8$
=== Induction Step ===
This is the induction step:
We have:
So $\map P k \implies \map P {k + 1}$ and then it follows by the Principle of Mathematical Induction that:
:$\forall n \in \Z_{\ge 0}: 3^{2 n} \equiv 1 \pmod 8$
Then we have:
and the result follows.
Category:Powers of 3 Modulo 8
Powers of 3 Modulo 8/Proof 2

== Theorem ==
== Proof ==
Let the statement be rewritten as:
For all $r \in \Z_{\ge 0}$: 
:$3^r \equiv \begin {cases} 1 \pmod 8 & : r = 2 n \\ 3 \pmod 8 & : r = 2 n + 1 \end {cases}$
where $n \in \Z_{\ge 0}$.
We have:
Then we have:
and the result follows.
Category:Powers of 3 Modulo 8
Powers of 5 with no Zero in Decimal Representation

== Unproven Hypotheses ==
The following powers of $5$ which contain no zero in their decimal representation are believed to be all that exist:
:$1, 5, 25, 125, 625, 3125, 15 \, 625, 78 \, 125, 1 \, 953 \, 125, 9 \, 765 \, 625,$
:$48 \, 828 \, 125, 762 \, 939 \, 453 \, 125, 3 \, 814 \, 697 \, 265 \, 625, 931 \, 322 \, 574 \, 615 \, 478 \, 515 \, 625,$
:$116 \, 415 \, 321 \, 826 \, 934 \, 814 \, 453 \, 125, 34 \, 694 \, 469 \, 519 \, 536 \, 141 \, 888 \, 238 \, 489 \, 627 \, 838 \, 134 \, 765 \, 625$
but this has not been proven.
The corresponding indices are:
:$0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 17, 18, 30, 33, 58$
== Progress ==
Category:Powers of 5
Category:58
Powers of Commutative Elements in Groups

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $a, b \in G$ such that $a$ and $b$ commute.
Then the following results hold:
=== Commutativity of Powers ===
=== Product of Commutative Elements ===
== Also see ==
* Powers of Commutative Elements in Semigroups
* Powers of Commutative Elements in Monoids
Category:Group Theory
Category:Commutativity
Powers of Commutative Elements in Monoids

== Theorem ==
These results are an extension of the results in Powers of Commutative Elements in Semigroups in which the domain of the indices is extended to include all integers.
Let $\left ({S, \circ}\right)$ be a monoid whose identity is $e_S$.
Let $a, b \in S$ be invertible elements for $\circ$ that also commute.
Then the following results hold.
=== Commutativity of Powers ===
=== Product of Commutative Elements ===
== Also see ==
* Powers of Commutative Elements in Semigroups
* Powers of Commutative Elements in Groups
Category:Monoids
Powers of Commutative Elements in Semigroups

== Theorem ==
Let $\left ({S, \circ}\right)$ be a semigroup.
Let $a, b \in S$ both be cancellable elements of $S$.
Then the following results hold:
=== Commutativity of Powers ===
=== Product of Commutative Elements ===
== Also see ==
* Powers of Commutative Elements in Monoids
* Powers of Commutative Elements in Groups
Category:Semigroups
Category:Commutativity
Powers of Commuting Elements of Monoid Commute

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.
For $a \in S$, let $\circ^n a = a^n$ denote the $n$th power of $a$.
Let $a, b \in S$ such that $a$ commutes with $b$:
:$a \circ b = b \circ a$
Then:
:$\forall m, n \in \N: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$
That is:
:$\forall m, n \in \N: a^m \circ b^n = b^n \circ a^m$
== Proof ==
Because $\struct {S, \circ}$ is a monoid, it is a fortiori also a semigroup.
From Powers of Commuting Elements of Semigroup Commute:
:$\forall m, n \in \N_{>0}: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$
That is:
:$\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m$
It remains to be shown that the result holds for the cases where $m = 0$ and $n = 0$.
Let $n \in \N$:
Similarly, let $m \in \N$:
and:
Thus:
:$a^m \circ b^n = b^n \circ a^m$
holds for $n = 0$ and $m = 0$.
Thus:
:$\forall m, n \in \N: a^m \circ b^n = b^n \circ a^m$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.8$
Category:Monoids
Category:Commutativity
Category:Powers (Abstract Algebra)
Powers of Commuting Elements of Semigroup Commute

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
For $a \in S$, let $\circ^n a = a^n$ denote the $n$th power of $a$.
Let $a, b \in S$ such that $a$ commutes with $b$:
:$a \circ b = b \circ a$
Then:
:$\forall m, n \in \N_{>0}: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$
That is:
:$\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m$
== Proof ==
The proof proceeds by the Principle of Mathematical Induction:
Let $\map P n$ be the proposition:
:$\paren {\circ^n a} \circ b = b \circ \paren {\circ^n a}$
=== Basis of the Induction ===
demonstrating that $\map P 1$ holds.
This is the basis for the induction.
=== Induction Hypothesis ===
Suppose that $\map P k$ holds:
:$\paren {\circ^k a} \circ b = b \circ \paren {\circ^k a}$
This is the induction hypothesis.
It remains to be shown that:
:$\map P k \implies \map P {k + 1}$
That is, that:
:$\paren {\circ^{k + 1} a} \circ b = b \circ \paren {\circ^{k + 1} a}$
=== Induction Step ===
This is the induction step:
Thus:
So $\map P {k + 1}$ is true.
Thus:
:$\forall m \in \N_{>0}: \paren {\circ^m a} \circ b = b \circ \paren {\circ^m a}$
By repeating the argument above, replacing $a$ with $b$ and $b$ with $\circ^m a$, we have:
Hence the result:
:$\forall m, n \in \N_{>0}: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$
That is:
:$\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.8 \ (2)$
Category:Semigroups
Category:Commutativity
Category:Powers (Abstract Algebra)
Powers of Coprime Numbers are Coprime

== Theorem ==
Let $a, b$ be coprime integers:
:$a \perp b$
Then:
:$\forall n \in \N_{>0}: a^n \perp b^n$
== Proof ==
Proof by induction:
Let $a \perp b$.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$a^n \perp b^n$
$\map P 1$ is true, as this just says:
:$a \perp b$
=== Basis for the Induction ===
By :
:$a^2 \perp b$
Again, by :
:$a^2 \perp b^2$
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:
:$a^k \perp b^k$
Then we need to show:
:$a^{k + 1} \perp b^{k + 1}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N_{>0}: a^n \perp b^n$
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Coprime Integers
Powers of Disjoint Permutations

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\rho, \sigma$ be disjoint permutations.
Then:
: $\forall k \in \Z: \paren {\sigma \rho}^k = \sigma^k \rho^k$
== Proof ==
A direct application of Power of Product of Commutative Elements in Group, and the fact that Disjoint Permutations Commute.
== Sources ==
* : Chapter $9$: Permutations: Proposition $9.8$
Category:Symmetric Groups
Powers of Element form Subgroup

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Then:
:$\forall a \in G: H = \set {a^n: n \in \Z} \le G$
That is, the subset of $G$ comprising all elements possible as powers of $a \in G$ is a subgroup of $G$.
== Proof ==
Clearly $a \in H$, so $H \ne \O$.
Let $x, y \in H$.
Thus by the One-Step Subgroup Test:
: $H \le G$
== Sources ==
* : $\S 37.4$ Some important general examples of subgroups
Category:Subgroups
Powers of Elements in Group Direct Product

== Theorem ==
Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be group whose identities are $e_G$ and $e_H$.
Let $\struct {G \times H, \circ}$ be the group direct product of $G$ and $H$.
Then:
:$\forall n \in \Z: \forall g \in G, h \in H: \tuple {g, h}^n = \tuple {g^n, h^n}$
== Proof ==
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition $\forall g \in G, h \in H: \tuple {g, h}^n = \tuple {g^n, h^n}$.
=== Basis for the Induction ===
$\map P 0$ is true, as this says:
:$\tuple {g, h}^0 = \tuple {e_G, e_H}$
$\map P 1$ is true, as this says:
:$\tuple {g, h} = \tuple {g, h}$
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\tuple {g, h}^k = \tuple {g^k, h^k}$
Then we need to show:
:$\tuple {g, h}^{k + 1} = \tuple {g^{k + 1}, h^{k + 1} }$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: \forall g \in G, h \in H: \tuple {g, h}^n = \tuple {g^n, h^n}$
So we have shown the result holds true for all $n \ge 0$.
The result for $n < 0$ follows directly from Powers of Group Elements for Negative Indices.
Category:Powers (Abstract Algebra)
Category:Group Direct Products
Powers of Elements of Geometric Sequence are in Geometric Sequence

== Theorem ==
Let $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers.
Then the sequence $Q = \sequence {b_j}_{0 \mathop \le j \mathop \le n}$ defined as:
:$\forall j \in \set {0, 1, \ldots, n}: b_j = a_j^m$
where $m \in \Z_{>0}$, is a geometric sequence.
== Proof ==
From Form of Geometric Sequence of Integers, the $j$th term of $P$ is given by:
:$a_j = k q^j p^{n - j}$
Thus the $j$th term of $Q$ is given by:
:$b_j = k^m \paren {q^m}^j \paren {p^m}^{n - j}$
From Form of Geometric Sequence of Integers, this is a geometric sequence.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Geometric Sequences of Integers
Powers of Field Elements Commute

== Theorem ==
Let $\struct {F, +, \circ}$ be a field with zero $0_F$ and unity $1_F$.
Let $F^* = F \setminus {0_F}$ denote the set of elements of $F$ without the zero $0_F$.
Then:
:$(a):\quad \forall a \in \F^* : \forall n, m \in \Z : a^m \circ a^n = a^n \circ a^m$
:$(b):\quad \forall a \in F : \forall n, m \in \Z_{\ge 0} : a^m \circ a^n = a^n \circ a^m$
== Proof ==
=== Statement $(a)$ ===
By :
:$\struct{F^*, \circ}$ is an Abelian group
By :
:For all $a \in F^*$ and $n \in \Z$, $a^n$ is defined as the $n$th power of $a$ with respect to the Abelian group $\struct {F^*, \circ}$
From Powers of Group Element Commute:
:$\forall a \in \F^* : \forall n, m \in \Z : a^m \circ a^n = a^n \circ a^m$
=== Statement $(b)$ ===
Let $m,n \in \Z_{\ge 0}$ be arbitrary elements of $\Z_{\ge 0}$.
For $a \in F^*$, $(b)$ follows from $(a)$.
It remains to show that $(b)$ holds for $0_F$.
==== Case 1: $m = 0$ ====
Let $m = 0$.
We have:
==== Case 2: $m \ne 0$ ====
Let $m \ne 0$.
We have:
In both cases:
:$\paren{0_F}^m \circ \paren{0_F}^n = \paren{0_F}^n \circ \paren{0_F}^m$
Since $m,n$ were arbitrary elements of $\Z_{\ge 0}$:
:$\forall n, m \in \Z_{\ge 0} : \paren{0_F}^m \circ \paren{0_F}^n = \paren{0_F}^n \circ \paren{0_F}^m$
Category:Field Theory
Category:Powers (Abstract Algebra)
Category:Commutativity
Powers of Group Element Commute

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $g \in G$.
Let $m, n \in \N_{>0}$.
Then:
:$\forall m, n \in \N_{>0}: g^n \circ g^m = g^m \circ g^n$
== Proof ==
By definition, a group is also a semigroup.
The result follows as a special case of Powers of Semigroup Element Commute
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.12)$
Category:Group Theory
Category:Powers (Abstract Algebra)
Category:Commutativity
Powers of Group Elements

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $g \in G$.
Then the following results hold:
=== Negative Index ===
=== Sum of Indices ===
=== Product of Indices ===
== Also see ==
* Exponent Combination Laws
Category:Group Theory
Category:Powers (Abstract Algebra)
Category:Index Laws
Powers of Group Elements/Negative Index

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $g \in G$.
Then:
:$\forall n \in \Z: \paren {g^n}^{-1} = g^{-n} = \paren {g^{-1} }^n$
=== Additive Notation ===
This can also be written in additive notation as:
== Proof ==
All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Sum of Indices:
:$\forall n \in \Z: \paren {g^n}^{-1} = g^{-n} = \paren {g^{-1} }^n$
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.10)$
* : $\S 4.6$. Elementary theorems on groups: Theorem $\text{(v)}$
* : Chapter $2$: The Definition of Group Structure: $\S 27$
* : $\S 35$: Elementary consequences of the group axioms
* : Chapter $3$: Elementary consequences of the definitions: Proposition $3.8 \ (3)$
Category:Group Theory
Category:Index Laws
Powers of Group Elements/Negative Index/Additive Notation

== Theorem ==
Let $\struct {G, +}$ be a group whose identity is $e$.
Let $g \in G$.
Then:
:$\forall n \in \Z: -\paren {n g} = \paren {-n} g = n \paren {-g}$
== Proof ==
All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Sum of Indices:
:$\forall n \in \Z: \paren {g^n}^{-1} = g^{-n} = \paren {g^{-1} }^n$
where in this context:
:the group operation is $+$
:the $n$th power of $g$ is denoted $n g$
:the inverse of $g$ is $-g$.
== Sources ==
* : $\S 2.1$: Subrings: Notation $2$
Category:Group Theory
Category:Index Laws
Powers of Group Elements/Product of Indices

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $g \in G$.
Then:
:$\forall m, n \in \Z: \paren {g^m}^n = g^{m n} = \paren {g^n}^m$
=== Additive Notation ===
This can also be written in additive notation as:
== Proof ==
All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Product of Indices:
:$\forall m, n \in \Z: g^{m n} = \paren {g^m}^n = \paren {g^n}^m$
== Also see ==
* Powers of Group Elements/Sum of Indices
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.6)$
* : $\S 4.6$. Elementary theorems on groups: Example $87 \ \text{(vii)}$
* : $\S 35.10 \ \text {(ii)}$: Elementary consequences of the group axioms
* : Chapter $3$: Elementary consequences of the definitions: Proposition $3.8 \ (2)$
Category:Group Theory
Category:Powers (Abstract Algebra)
Category:Index Laws
Powers of Group Elements/Product of Indices/Additive Notation

== Theorem ==
Let $\struct {G, +}$ be a group whose identity is $e$.
Let $g \in G$.
Then:
:$\forall m, n \in \Z: n \cdot \paren {m \cdot g} = \paren {m \times n} \cdot g = m \cdot \paren {n \cdot g}$
== Proof ==
All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Product of Indices:
:$\forall m, n \in \Z: g^{m n} = \paren {g^m}^n = \paren {g^n}^m$
where in this context:
:the group operation is $+$
:the $n$th power of $g$ is denoted $n \cdot g$
== Sources ==
* : $\S 2.1$: Subrings: Notation $2$
Category:Group Theory
Category:Index Laws
Powers of Group Elements/Sum of Indices

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $g \in G$.
Then:
:$\forall m, n \in \Z: g^m \circ g^n = g^{m + n}$
=== Additive Notation ===
This can also be written in additive notation as:
== Proof ==
All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Sum of Indices:
:$\forall m, n \in \Z: g^m \circ g^n = g^{m + n}$
== Also see ==
* Powers of Group Elements/Product of Indices
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.5)$
* : $\S 4.6$. Elementary theorems on groups: Example $87 \ \text{(vi)}$
* : $\S 2.1$: Subrings: Notation $2$
* : $\S 35.10 \ \text {(i)}$: Elementary consequences of the group axioms
* : Chapter $3$: Elementary consequences of the definitions: Proposition $3.8 \ (1)$
Category:Group Theory
Category:Powers (Abstract Algebra)
Category:Index Laws
Powers of Group Elements/Sum of Indices/Additive Notation

== Theorem ==
Let $\struct {G, +}$ be a group whose identity is $e$.
Let $g \in G$.
Then:
:$\forall m, n \in \Z: \paren {m \cdot g} + \paren {n \cdot g} = \paren {m + n} \cdot g$
== Proof ==
All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Sum of Indices:
:$\forall m, n \in \Z: g^m \circ g^n = g^{m + n}$
where in this context the group operation is $+$ and $n$th power of $g$ is denoted $n g$.
== Sources ==
* : $\S 2.1$: Subrings: Notation $2$
Category:Group Theory
Category:Index Laws
Powers of Infinite Order Element

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $a \in G$ have infinite order in $G$.
Then:
:$\forall m, n \in \Z: m \ne n \implies a^m \ne a^n$
== Proof ==
Let $m, n \in \Z$.
Then:
The result follows from transposition.
== Sources ==
* : $\S 38.2$ Period of an element
Category:Order of Group Elements
Powers of Permutation Element

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi \in S_n$, and let $i \in \N^*_n$.
Let $k \in \Z: k > 0$ be the smallest such that:
:$\map {\pi^k} i \in \set {i, \map \pi i, \map {\pi^2} i, \ldots, \map {\pi^{k - 1} } i}$
Then $\map {\pi^k} i = i$.
== Proof ==
 $\map {\pi^k} i = \map {\pi^r} i$ for some $r > 0$.
Then, since $\pi$ has an inverse, $\map {\pi^{k - r} } i = i$.
This contradicts the definition of $k$, so $r = 0$.
Category:Symmetric Groups
Powers of Primitive Complex Root of Unity form Complete Set

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $U_n$ denote the complex $n$th roots of unity:
:$U_n = \set {z \in \C: z^n = 1}$
Let $\alpha_k = \exp \paren {\dfrac {2 k \pi i} n}$ denote the $k$th complex root of unity.
Let $\alpha_k$ be a primitive complex root of unity.
Let $V_k = \set { {\alpha_k}^r: r \in \set {0, 1, \ldots, n - 1} }$.
Then:
: $V_k = U_n$
That is, $V_k = \set { {\alpha_k}^r: r \in \set {0, 1, \ldots, n - 1} }$ forms the complete set of complex $n$th roots of unity.
== Proof ==
From Roots of Unity under Multiplication form Cyclic Group, $\struct {U_n, \times}$ is a group.
The result follows from Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order.
== Sources ==
* : $\S 3$. Roots of Unity
Category:Complex Roots of Unity
Powers of Semigroup Element Commute

== Theorem ==
Let $\struct {S, \odot}$ be a semigroup.
Let $a \in S$.
Let $m, n \in \Z_{>0}$.
Then:
:$\forall m, n \in \Z_{>0}: a^n \odot a^m = a^m \odot a^n$
== Proof ==
== Sources ==
* : $\S 4.2$. Commutative and associative operations: $(4)$
Category:Semigroups
Category:Powers (Abstract Algebra)
Category:Commutativity
Powerset is not Subset of its Set

== Theorem ==
Let $A$ be a set.
Then:
:$\powerset A \not \subseteq A$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* No Injection from Power Set to Set
Category:Powerset is not Subset of its Set
Category:Power Set
Category:Subsets
Powerset is not Subset of its Set/Proof 1

== Theorem ==
== Proof ==
 that $\powerset A \subseteq A$, and define:
:$C = \set {x \in \powerset A : x \notin x}$
We have that $C \subseteq \powerset A$, as it contains only the $x \in \powerset A$ meeting the condition $x \notin x$.
Since $\powerset A \subseteq A$, we have:
:$C \subseteq A$
and thus
:$C \in \powerset A$
We can derive a similar contradiction to Russell's Paradox.
If $C \in C$, then it must meet $C$'s condition that $C \notin C$.
If $C \notin C$, then it meets $C$'s condition for $C \in C$.
== Sources ==
* : Chapter $1$: Exercise $1.2$
Category:Powerset is not Subset of its Set