hcju/proofwikips huggingface dataset:
Intersection of Congruence Classes
Let $\mathcal R_m$ denote congruence modulo $m$ on the set of integers $\Z$. Then: :$\mathcal R_m \cap \mathcal R_n = \mathcal R_{\lcm \set {m, n} }$ where $\lcm \set {m, n}$ is the lowest common multiple of $m$ and $n$. In the language of modulo arithmetic, this is equivalent to: :$a \equiv b \pmod m, a \equiv b \pmod n \implies a \equiv b \pmod {\lcm \set {m, n} }$
Mappings Between Residue Classes
Let $\eqclass a m$ be the residue class of $a$ (modulo $m$). Let $\phi: \Z_m \to \Z_n$ be a mapping given by: :$\map \phi {\eqclass x m} = \eqclass x n$ Then $\phi$ is well defined  $m$ is a divisor of $n$.
Prime Number has 4 Integral Divisors
Let $p$ be an integer. Then $p$ is a prime number  $p$ has exactly four integral divisors: $1, -1, p, -p$.
Prime not Divisor implies Coprime
Let $p, a \in \Z$. If $p$ is a prime number then: : $p \nmid a \implies p \perp a$ where: : $p \nmid a$ denotes that $p$ does not divide $a$ : $p \perp a$ denotes that $p$ and $a$ are coprime. It follows directly that if $p$ and $q$ are primes, then: : $p \divides q \implies p = q$ : $p \ne q \implies p \perp q$.
Composite Number has Two Divisors Less Than It
Let $n \in \Z_{> 1}$ such that $n \notin \mathbb P$. Then: :$\exists a, b \in \Z: 1 < a < n, 1 < b < n: n = a b$ That is, a non-prime number greater than $1$ can be expressed as the product of two positive integers strictly greater than $1$ and less than $n$. Note that these two numbers are not necessarily distinct.
Exponents of Primes in Prime Decomposition are Less iff Divisor
Let $a, b \in \Z_{>0}$. Then $a \divides b$ : :$(1): \quad$ every prime $p_i$ in the prime decomposition of $a$ appears in the prime decomposition of $b$ and: :$(2): \quad$ the exponent of each $p_i$ in $a$ is less than or equal to its exponent in $b$.
Not Coprime means Common Prime Factor
Let $a, b \in \Z$. If $d \divides a$ and $d \divides b$ such that $d > 1$, then $a$ and $b$ have a common divisor which is prime.
Set of Divisors of Integer
Let $n \in \Z_{>1}$. Let $n$ be expressed in its prime decomposition: :$n = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$ where $p_1 < p_2 < \dotsb < p_r$ are distinct primes and $k_1, k_2, \ldots, k_r$ are positive integers. The set of divisors of $n$ is: :$\set {p_1^{h_1} p_2^{h_2} \dotsm p_r^{h_r}: 0 \le h_i \le k_i, i = 1, 2, \ldots, r}$
Sum Less Minimum is Maximum
For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$: :$a + b - \min \left({a, b}\right) = \max \left({a, b}\right)$
Sum Less Maximum is Minimum
For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$: :$a + b - \max \left({a, b}\right) = \min \left({a, b}\right)$.
GCD and LCM from Prime Decomposition
Let $m, n \in \Z$. Let: :$m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$ :$n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}$ :$p_i \divides m \lor p_i \divides n, 1 \le i \le r$. That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers $m$ or $n$. Note that if one of the primes $p_i$ does not appear in the prime decompositions of either one of $m$ or $n$, then its corresponding index $k_i$ or $l_i$ will be zero. Then the following results apply: :$\gcd \set {m, n} = p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \ldots p_r^{\min \set {k_r, l_r} }$ :$\lcm \set {m, n} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \ldots p_r^{\max \set {k_r, l_r} }$
GCD and LCM Distribute Over Each Other
Let $a, b, c \in \Z$. Then: : $\lcm \set {a, \gcd \set {b, c} } = \gcd \set {\lcm \set {a, b}, \lcm \set {a, c} }$ : $\gcd \set {a, \lcm \set {b, c} } = \lcm \set {\gcd \set {a, b}, \gcd \set {a, c} }$ That is, greatest common divisor and lowest common multiple are distributive over each other.
Tau Function from Prime Decomposition
Let $n$ be an integer such that $n \ge 2$. Let the prime decomposition of $n$ be: :$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$ Let $\map \tau n$ be the tau function of $n$. Then: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$
N less than M to the N
:$\forall m, n \in \Z_{>0}: m > 1 \implies n < m^n$
Basis Representation Theorem
Let $b \in \Z: b > 1$. For every $n \in \Z_{> 0}$, there exists one and only one sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le t}$ such that: : $(1): \quad \displaystyle n = \sum_{k \mathop = 0}^t r_k b^k$ : $(2): \quad \displaystyle \forall k \in \closedint 0 t: r_k \in \N_b$ : $(3): \quad r_t \ne 0$ This unique sequence is called the '''representation of $n$ to the base $b$''', or, informally, we can say '''$n$ is (written) in base $b$'''.
Sum of Geometric Sequence
Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$. Let $n \in \N_{>0}$. Then: :$\displaystyle \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$
Congruence of Sum of Digits to Base Less 1
Let $x \in \Z$, and $b \in \N, b > 1$. Let $x$ be written in base $b$: :$x = \sqbrk {r_m r_{m - 1} \ldots r_2 r_1 r_0}_b$ Then: :$\displaystyle \map {s_b} x = \sum_{j \mathop = 0}^m r_j \equiv x \pmod {b - 1}$ where $\map {s_b} x$ is the digit sum of $x$ in base $b$ notation. That is, the digit sum of any integer $x$ in base $b$ notation is congruent to $x$ modulo $b - 1$.
Euler Phi Function of Integer
Let $n \in \Z_{>0}$, that is, a (strictly) positive integer. Let $\phi: \Z_{>0} \to \Z_{>0}$ be the Euler $\phi$-function. Then for any $n \in \Z_{>0}$, we have: :$\map \phi n = n \paren {1 - \dfrac 1 {p_1} } \paren {1 - \dfrac 1 {p_2} } \cdots \paren {1 - \dfrac 1 {p_r} }$ where $p_1, p_2, \ldots, p_r$ are the distinct primes dividing $n$. Or, more compactly: :$\displaystyle \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$ where $p \divides n$ denotes the primes which divide $n$.
Euler Phi Function is Multiplicative
The Euler $\phi$ function is a multiplicative function: :$m \perp n \implies \map \phi {m n} = \map \phi m \, \map \phi n$ where $m, n \in \Z_{>0}$.
Euler Phi Function of Prime Power
Let $p^n$ be a prime power for some prime number $p > 1$. Then: :$\map \phi {p^n} = p^n \paren {1 - \dfrac 1 p} = \paren {p - 1} p^{n - 1}$ where $\phi: \Z_{>0} \to \Z_{>0}$ is the Euler $\phi$ function.
Möbius Function is Multiplicative
The Möbius function $\mu$  is a multiplicative function: :$m \perp n \implies \map \mu {m n} = \map \mu m \map \mu n$ where $m, n \in \Z_{>0}$.
Sum of Möbius Function over Divisors
Let $n \in \Z_{>0}$ be a strictly positive integer. Then: :$\displaystyle \sum_{d \mathop \divides n} \map \mu d \frac n d = \map \phi n$ where: :$\displaystyle \sum_{d \mathop \divides n}$ denotes the sum over all of the divisors of $n$ :$\map \phi n$ is the Euler $\phi$ function, the number of integers less than $n$ that are prime to $n$ :$\map \mu d$ is the Möbius function. Equivalently, this says that: :$\phi = \mu * I_{\Z_{>0} }$ where: :$*$ denotes Dirichlet convolution :$I_{\Z_{>0} }$ denotes the identity mapping on $\Z_{>0}$, that is: ::$\forall n \in \Z_{>0}: I_{\Z_{>0} }: n \mapsto n$ {{wtd|Add a link to a page proving this equivalence.}}
Binomial Coefficient of Prime
Let $p$ be a prime number. Then: :$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$ where $\dbinom p k$ is defined as a binomial coefficient.
Prime Power of Sum Modulo Prime
Let $p$ be a prime number. Then: :$\forall n \in \N_{> 0}: \paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$
Binomial Coefficient involving Power of Prime
: $\dbinom {p^n k} {p^n} \equiv k \pmod p$ where $\dbinom {p^n k} {p^n}$ is a binomial coefficient.
Cassini's Identity
:$F_{n + 1} F_{n - 1} - F_n^2 = \paren {-1}^n$
Rational Numbers are Countably Infinite
The set $\Q$ of rational numbers is countably infinite.
Real Numbers form Ordered Field
The set of real numbers $\R$ forms an ordered field under addition and multiplication: $\struct {\R, +, \times, \le}$.
Rational Numbers form Subfield of Real Numbers
The (ordered) field $\struct {\Q, +, \times, \le}$ of rational numbers forms a subfield of the field of real numbers $\struct {\R, +, \times, \le}$. That is, the field of real numbers $\struct {\R, +, \times, \le}$ is an extension of the rational numbers $\struct {\Q, +, \times, \le}$.
Definition:Rational Number/Canonical Form
Let $r \in \Q$ be a rational number. The '''canonical form of $r$''' is the expression $\dfrac p q$, where: :$r = \dfrac p q: p \in \Z, q \in \Z_{>0}, p \perp q$ where $p \perp q$ denotes that $p$ and $q$ have no common divisor except $1$.
Ordering Properties of Real Numbers
=== Trichotomy Law === {{:Trichotomy Law for Real Numbers}} === Ordering is Transitive === {{:Real Number Ordering is Transitive}} === Ordering is Compatible with Addition === {{:Real Number Ordering is Compatible with Addition}} === Ordering is Compatible with Multiplication === {{:Real Number Ordering is Compatible with Multiplication}}
Order is Preserved on Positive Reals by Squaring
:$x < y \iff x^2 < y^2$
Real Plus Epsilon
Let $a, b \in \R$, such that: :$\forall \epsilon \in \R_{>0}: a < b + \epsilon$ where $\R_{>0}$ is the set of strictly positive real numbers. That is: :$\epsilon > 0$ Then: :$a \le b$
Difference of Two Squares
Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$. Let $x, y \in R$. Then: : $x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$ When $R$ is one of the standard sets of numbers, that is $\Z, \Q, \R$, and so on, then this translates into: :$x^2 - y^2 = \paren {x + y} \paren {x - y}$
Mediant is Between
Let $a, b, c, d$ be ''any'' real numbers such that $b > 0, d > 0$. Let $r = \dfrac a b < \dfrac c d = s$. Then: :$r < \dfrac {a + c} {b + d} < s$
Real Numbers Between Epsilons
Let $a, b \in \R$ such that $\forall \epsilon \in \R_{>0}: a - \epsilon < b < a + \epsilon$. Then $a = b$.
Exists Integer Below Any Real Number
Let $x$ be a real number. Then there exists an integer less than $x$: :$\forall x \in \R: \exists n \in \Z: n < x$
Archimedean Principle
Let $x$ be a real number. Then there exists a natural number greater than $x$. :$\forall x \in \R: \exists n \in \N: n > x$ That is, the set of natural numbers is unbounded above.
Real Number is between Floor Functions
:$\forall x \in \R: \floor x \le x < \floor {x + 1}$
Real Number is between Ceiling Functions
:$\forall x \in \R: \left \lceil {x - 1} \right \rceil \le x < \left \lceil {x} \right \rceil$
Real Number minus Floor
:$x - \floor x \in \hointr 0 1$
Ceiling minus Real Number
:$\forall x \in \R: \left \lceil {x} \right \rceil - x \in \left[{0 \,.\,.\, 1}\right)$
Real Number is Floor plus Difference
:There exists an integer $n \in \Z$ such that for some $t \in \hointr 0 1$: ::$x = n + t$ : :$n = \floor x$
Floor plus One
Let $x \in \R$. Then: :$\left \lfloor {x + 1} \right \rfloor = \left \lfloor {x} \right \rfloor + 1$ where $\left \lfloor {x} \right \rfloor$ is the floor function of $x$.
Real Number is Integer iff equals Floor
:$x = \floor x \iff x \in \Z$
Sum of Floor and Floor of Negative
Let $x \in \R$. Then: :$\floor x + \floor {-x} = \begin{cases} 0 & : x \in \Z \\ -1 & : x \notin \Z \end{cases}$ where $\floor x$ denotes the floor of $x$.
Ceiling defines Equivalence Relation
Let $\mathcal R$ be the relation defined on $\R$ such that: :$\forall x, y, \in \R: \left({x, y}\right) \in \mathcal R \iff \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil$ where $\left \lceil {x}\right \rceil$ is the ceiling of $x$. Then $\mathcal R$ is an equivalence, and $\forall n \in \Z$, the $\mathcal R$-class of $n$ is the half-open interval $\left({n - 1 \,.\,.\, n}\right]$.
Real Number is Ceiling minus Difference
Let $n$ be a integer.  :$(1): \quad$ There exists $t \in \hointr 0 1$ such that $x = n - t$ :$(2): \quad n = \ceiling x$
Cauchy's Inequality
:$\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$
Cancellable Finite Semigroup is Group
Let $\struct {S, \circ}$ be a non-empty finite semigroup in which all elements are cancellable. Then $\struct {S, \circ}$ is a group.
Finite Semigroup Equal Elements for Different Powers
Let $\left({S, \circ}\right)$ be a finite semigroup. Then: : $\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$
Element has Idempotent Power in Finite Semigroup
Let $\struct {S, \circ}$ be a finite semigroup. For every element in $\struct {S, \circ}$, there is a power of that element which is idempotent. That is: :$\forall x \in S: \exists i \in \N: x^i = x^i \circ x^i$
Powers of Ring Elements
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$. Let $n \cdot x$ be an integral multiple of $x$: :$n \cdot x = \begin {cases} 0_R & : n = 0 \\ x & : n = 1 \\ \paren {n - 1} \cdot x + x & : n > 1 \end {cases}$ that is $n \cdot x = x + x + \cdots \paren n \cdots x$. For $n < 0$ we use: :$-n \cdot x = n \cdot \paren {-x}$ Then: :$\forall n \in \Z: \forall x \in R: \paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$
Power of Conjugate equals Conjugate of Power
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $x, y \in G$ such that $\exists a \in G: x \circ a = a \circ y$. That is, let $x$ and $y$ be conjugate. Then: : $\forall n \in \Z: y^n = \paren {a^{-1} \circ x \circ a}^n = a^{-1} \circ x^n \circ a$ It follows directly that: : $\exists b \in G: \forall n \in \Z: y^n = b \circ x^n \circ b^{-1}$ In particular: : $y^{-1} = \paren {a^{-1} \circ x \circ a}^{-1} = a^{-1} \circ x^{-1} \circ a$
Product of Conjugates equals Conjugate of Products
Let $\struct {G, \circ}$ be a group. Then: :$\forall a, x, y \in G: \paren {a \circ x \circ a^{-1} } \circ \paren {a \circ y \circ a^{-1} } = a \circ \paren {x \circ y} \circ a^{-1}$ That is, the product of conjugates is equal to the conjugate of the product.
Power of Product with Inverse
Let $G$ be a group whose identity is $e$. Let $a, b \in G: a b = b a^{-1}$. Then: : $\forall n \in \Z: a^n b = b a^{-n}$
Powers of Elements in Group Direct Product
Let $\left({G, \circ_1}\right)$ and $\left({H, \circ_2}\right)$ be group whose identities are $e_G$ and $e_H$. Let $\left({G \times H, \circ}\right)$ be the group direct product (either external or internal) of $G$ and $H$. Then: : $\forall n \in \Z: \forall g \in G, h \in H: \left({g, h}\right)^n = \left({g^n, h^n}\right)$
Powers of Commutative Elements in Semigroups
Let $\left ({S, \circ}\right)$ be a semigroup. Let $a, b \in S$ both be cancellable elements of $S$. Then the following results hold:
Powers of Commutative Elements in Monoids
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.
Powers of Commutative Elements in Groups
Let $\struct {G, \circ}$ be a group. Let $a, b \in G$ such that $a$ and $b$ commute. Then the following results hold:
General Morphism Property for Semigroups
Let $\struct {S, \circ}$ and $\struct {T, *}$ be semigroups. Let $\phi: S \to T$ be a homomorphism. Then: :$\forall s_k \in S: \map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$ Hence it follows that: :$\forall n \in \N_{>0}: \forall s \in S: \map \phi {s^n} = \paren {\map \phi s}^n$
Homomorphism of Power of Group Element
Let $\struct {G, \circ}$ and $\struct {H, \ast}$ be groups. Let $\phi: S \to T$ be a group homomorphism. Then: : $\forall n \in \Z: \forall g \in G: \map \phi {g^n} = \paren {\map \phi g}^n$
Finite Subgroup Test
Let $\struct {G, \circ}$ be a group. Let $H$ be a non-empty finite subset of $G$. Then: :$H$ is a subgroup of $G$ : :$\forall a, b \in H: a \circ b \in H$ That is, a non-empty finite subset of $G$ is a subgroup  it is closed.
Powers of Element form Subgroup
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 $G$.
Homomorphism of Generated Group
Let $\left({G, \circ}\right)$ and $\left({H, \circ}\right)$ be groups. Let $\phi: G \to H$ and $\psi: G \to H$ be homomorphisms.  Let $\left \langle {S} \right \rangle = G$ be the group generated by $S$. Let $\forall x \in S: \phi \left({x}\right) = \psi \left({x}\right)$ Then $\phi = \psi$.
Set of Words Generates Group
Let $S \subseteq G$ where $G$ is a group. Let $\hat S$ be defined as $S \cup S^{-1}$, where $S^{-1}$ is the set of all the inverses of all the elements of $S$. Then $\gen S = \map W {\hat S}$, where $\map W {\hat S}$ is the set of words of $\hat S$.
Subset Product is Subset of Generator
Let $\struct {G, \circ}$ be a group. Let $X, Y \subseteq \struct {G, \circ}$. Then $X \circ Y \subseteq \gen {X, Y}$ where: :$X \circ Y$ is the Subset Product of $X$ and $Y$ in $G$. :$\gen {X, Y}$ is the subgroup of $G$ generated by $X$ and $Y$.
Order of Subset Product with Singleton
Let $\struct {G, \circ}$ be a group. Let $X, Y \subseteq \struct {G, \circ}$ such that $X$ is a singleton: :$X = \set x$ Then: :$\order {X \circ Y} = \order Y = \order {Y \circ X}$ where $\order S$ is defined as the order of $S$.
Product of Subset with Intersection
Let $\struct {G, \circ}$ be an algebraic structure. Let $X, Y, Z \subseteq G$. Then: :$X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$ :$\paren {Y \cap Z} \circ X \subseteq \paren {Y \circ X} \cap \paren {Z \circ X}$ where $X \circ Y$ denotes the subset product of $X$ and $Y$.
Order of Subgroup Product
Let $G$ be a group. Let $H$ and $K$ be subgroups of $G$. Then: :$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$ where: :$H K$ denotes subset product :$\order H$ denotes the order of $H$.
Index of Intersection of Subgroups
Let $G$ be a group. Let $H, K$ be subgroups of finite index of $G$. Then: :$\index G {H \cap K} \le \index G H \index G K$ where $\index G H$ denotes the index of $H$ in $G$. Note that here the symbol $\le$ is being used with its meaning '''less than or equal to'''. Equality holds  $H K = \set {h k: h \in H, k \in K} = G$.
Intersection of Subgroups of Prime Order
Let $G$ be a group whose identity is $e$. Let $H$ and $K$ be subsets of $G$ such that: :$\order H = \order K = p$ :$H \ne K$ :$p$ is prime. Then: : $H \cap K = \set e$ That is, the intersection of two unequal subgroups of a group, both of whose order is the same prime, consists solely of the identity.
Tower Law for Subgroups
Let $\struct {G, \circ}$ be a group. Let $H$ be a subgroup of $G$ with finite index. Let $K$ be a subgroup of $H$. Then: :$\index G K = \index G H \index H K$ where $\index G H$ denotes the index of $H$ in $G$.
Morphism from Integers to Group
Let $G$ be a group whose identity is $e$. Let $g \in G$. Let $\phi: \Z \to G$ be the mapping defined as: :$\forall n \in \Z: \map \phi n = g^n$. Then: : If $g$ has infinite order, then $\phi$ is a group isomorphism from $\struct {\Z, +}$ to $\gen g$. : If $g$ has finite order such that $\order g = m$, then $\phi$ is a group epimorphism from $\struct {\Z, +}$ to $\gen g$ whose kernel is the principal ideal $\paren m$. :Thus $\gen g$ is isomorphic to $\struct {\Z, +}$, and $m$ is the smallest (strictly) positive integer such that $g^m = e$.
Identity is Only Group Element of Order 1
In every group, the identity, and only the identity, has order $1$.
Group Element is Self-Inverse iff Order 2
Let $\struct {S, \circ}$ be a group whose identity is $e$. An element $x \in \struct {S, \circ}$ is self-inverse : :$\order x = 2$
Powers of Infinite Order Element
Let $G$ be a group whose identity is $e$. Let $a \in G$ have infinite order in $G$. Then: :$\forall m, n \in \Z: m \ne n \implies a^m \ne a^n$
Element of Finite Group is of Finite Order
In any finite group, each element has finite order.
Inverse Element is Power of Order Less 1
Let $G$ be a group whose identity is $e$. Let $g \in G$ be of finite order. Then: : $\order g = n \implies g^{n - 1} = g^{-1}$
Equal Powers of Finite Order Element
:$g^r = g^s \iff k \divides \paren {r - s}$
Order of Element Divides Order of Finite Group
In a finite group, the order of a group element divides the order of its group: :$\forall x \in G: \order x \divides \order G$
Element to Power of Group Order is Identity
Let $G$ be a group whose identity is $e$ and whose order is $n$. Then: :$\forall g \in G: g^n = e$
Boolean Group is Abelian
Let $G$ be a Boolean group. Then $G$ is abelian.
Order of Group Element equals Order of Inverse
Let $G$ be a group whose identity is $e$. Then: : $\forall x \in G: \order x = \order {x^{-1} }$ where $\order x$ denotes the order of $x$.
Order of Conjugate Element equals Order of Element
Let $\struct {G, \circ}$ be a group whose identity is $e$. Then :$\forall a, x \in \struct {G, \circ}: \order {x \circ a \circ x^{-1} } = \order a$ where $\order a$ denotes the order of $a$ in $G$.
Order of Homomorphic Image of Group Element
Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$ respectively. Let $\phi: G \to H$ be a homomorphism. Let $g \in G$ be of finite order. Then: :$\forall g \in G: \order {\map \phi g} \divides \order g$ where $\divides$ denotes divisibility.
Non-Trivial Group has Non-Trivial Cyclic Subgroup
Let $G$ be a group whose identity element is $e$. Let $g \in G$. If $g$ has infinite order, then $\gen g$ is an infinite cyclic group. If $\order g = n$, then $\gen g$ is a cyclic group with $n$ elements. Thus, every group which is non-trivial has at least one cyclic subgroup which is also non-trivial. In the case that $G$ is itself cyclic, that cyclic subgroup may of course be itself.
Epimorphism from Integers to Cyclic Group
Let $\gen a = \struct {G, \circ}$ be a cyclic group. Let $f: \Z \to G$ be a mapping defined as: $\forall n \in \Z: \map f n = a^n$. Then $f$ is a (group) epimorphism from $\struct {\Z, +}$ onto $\gen a$.
Cyclic Group is Abelian
Let $G$ be a cyclic group. Then $G$ is abelian.
Cyclic Groups of Same Order are Isomorphic
Two cyclic groups of the same order are isomorphic to each other.
Order of Subgroup of Cyclic Group
Let $C_n = \gen g$ be the cyclic group of order $n$ which is generated by $g$ whose identity is $e$. Let $a \in C_n: a = g^i$. Let $H = \gen a$. Then: :$\order H = \dfrac n {\gcd \set {n, i} }$ where: :$\order H$ denotes the order of $H$ :$\gcd \set {n, i}$ denotes the greatest common divisor of $n$ and $i$.
Number of Powers of Cyclic Group Element
Let $G$ be a cyclic group of order $n$, generated by $g$. Let $d \divides n$. Then the element $g^{n/d}$ has $d$ distinct powers.
Subgroup of Finite Cyclic Group is Determined by Order
Let $G = \gen g$ be a cyclic group whose order is $n$ and whose identity is $e$. Let $d \divides n$, where $\divides$ denotes divisibility. Then there exists exactly one subgroup $G_d = \gen {g^{n / d} }$ of $G$ with $d$ elements.
Prime Group is Cyclic
Let $p$ be a prime number. Let $G$ be a group whose order is $p$. Then $G$ is cyclic.
Group of Order less than 6 is Abelian
All groups with less than $6$ elements are abelian.
Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order
Let $C_n$ be the cyclic group of order $n$. Let $C_n = \gen a$, that is, that $C_n$ is generated by $a$. Then: :$C_n = \gen {a^k} \iff k \perp n$ That is, $C_n$ is also generated by $a^k$  $k$ is coprime to $n$.
Order of Conjugate of Subgroup
Let $G$ be a group. Let $H$ be a subgroup of $G$ such that $H$ is of finite order. Then $\order {H^a} = \order H$.
Subgroup of Index 2 is Normal
A subgroup of index $2$ is always normal.
Intersection of Normal Subgroups is Normal
Let $G$ be a group. Let $I$ be an indexing set. Let $\family {N_i}_{i \mathop \in I}$ be a non-empty indexed family of normal subgroups of $G$. Then $\displaystyle \bigcap_{i \mathop \in I} N_i$ is a normal subgroup of $G$.
Union of Conjugacy Classes is Normal
Let $G$ be a group. Let $H \le G$. Then $H$ is normal in $G$  $H$ is a union of conjugacy classes of $G$.
Unique Subgroup of a Given Order is Normal
Let a group $G$ have only one subgroup of a given order. Then that subgroup is normal.
Smallest Normal Subgroup containing Set
Let $S \subseteq G$ where $G$ is a group. Then there exists a unique smallest normal subgroup of $G$ which contains $S$.