avewright/proofwiki-math huggingface dataset:
Theorem: Union Operation on Supersets of Subset is Closed

Let $S$ be a set.
Let $T \subseteq S$ be a given subset of $S$.
Let $\powerset S$ denote the power set of $S$
Let $\mathscr S$ be the subset of $\powerset S$ defined as:
:$\mathscr S = \set {Y \in \powerset S: T \subseteq Y}$
Then the algebraic structure $\struct {\mathscr S, \cup}$ is closed.

Proof:
Let $A, B \in \mathscr S$.
We have that:
{{begin-eqn}}
{{eqn | l = T
      | o = \subseteq
      | r = A
      | c = Definition of $\mathscr S$
}}
{{eqn | l = T
      | o = \subseteq
      | r = B
      | c = Definition of $\mathscr S$
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = T
      | o = \subseteq
      | r = A \cup B
      | c = Set is Subset of Union
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = A
      | o = \subseteq
      | r = S
      | c = {{Defof|Power Set}}
}}
{{eqn | l = B
      | o = \subseteq
      | r = S
      | c = {{Defof|Power Set}}
}}
{{eqn | ll= \leadsto
      | l = A \cup B
      | o = \subseteq
      | r = S
      | c = Union is Smallest Superset
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = A \cup B
      | o = \in
      | r = \powerset S
      | c = {{Defof|Power Set}}
}}
{{end-eqn}}
Thus we have:
{{begin-eqn}}
{{eqn | l = T
      | o = \subseteq
      | r = A \cup B
      | c = from $(1)$
}}
{{eqn | l = A \cup B
      | o = \in
      | r = \powerset S
      | c = from $(2)$
}}
{{eqn | ll= \leadsto
      | l = A \cup B
      | o = \in
      | r = \mathscr S
      | c = Definition of $\mathscr S$
}}
{{end-eqn}}
Hence the result by definition of closed algebraic structure.
{{qed}}
Theorem: Intersection Operation on Supersets of Subset is Closed

Let $S$ be a set.
Let $T \subseteq S$ be a given subset of $S$.
Let $\powerset S$ denote the power set of $S$
Let $\mathscr S$ be the subset of $\powerset S$ defined as:
:$\mathscr S = \set {Y \in \powerset S: T \subseteq Y}$
Then the algebraic structure $\struct {\mathscr S, \cap}$ is closed.

Proof:
Let $A, B \in \mathscr S$.
We have that:
{{begin-eqn}}
{{eqn | l = T
      | o = \subseteq
      | r = A
      | c = Definition of $\mathscr S$
}}
{{eqn | l = T
      | o = \subseteq
      | r = B
      | c = Definition of $\mathscr S$
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = T
      | o = \subseteq
      | r = A \cap B
      | c = Intersection is Largest Subset
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = A
      | o = \subseteq
      | r = S
      | c = {{Defof|Power Set}}
}}
{{eqn | l = B
      | o = \subseteq
      | r = S
      | c = {{Defof|Power Set}}
}}
{{eqn | ll= \leadsto
      | l = A \cap B
      | o = \subseteq
      | r = S
      | c = Intersection is Subset
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = A \cap B
      | o = \in
      | r = \powerset S
      | c = {{Defof|Power Set}}
}}
{{end-eqn}}
Thus we have:
{{begin-eqn}}
{{eqn | l = T
      | o = \subseteq
      | r = A \cap B
      | c = from $(1)$
}}
{{eqn | l = A \cap B
      | o = \in
      | r = \powerset S
      | c = from $(2)$
}}
{{eqn | ll= \leadsto
      | l = A \cap B
      | o = \in
      | r = \mathscr S
      | c = Definition of $\mathscr S$
}}
{{end-eqn}}
Hence the result by definition of closed algebraic structure.
{{qed}}
Theorem: Existence of Magma with no Proper Submagma

Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $S$ be a set of cardinality $n$:
:$\card S = n$
Then there exists an operation $\circ$ on $S$ such that:
:$\struct {S, \circ}$ is a magma
:$\struct {S, \circ}$ has no submagma $\struct {T, \circ}$ such that $T$ is a non-empty proper subset of $S$.

Proof:
For $n = 1$ the result follows trivially: there are no non-empty proper subsets of a singleton.
Let $S = \set {s_1, s_2, \ldots, s_n}$.
Let $\circ$ be defined on $S$ such that:
:$\forall s_a \in S: s_a \circ s_a = \begin{cases} s_{a + 1} & : a < n \\ s_1 & : a = n \end{cases}$
For $a \ne b$ the operation $s_a \circ s_b$ can be arbitrary as long as $s_a \circ s_b \in S$.
Let $T$ be such that $\O \subsetneq T \subsetneq S$.
Then either:
:$\exists s_k \in T: s_{k + 1} \notin T$
or:
:$s_n \in T$ but $s_1 \notin T$
otherwise $T = S$.
Thus either:
:$s_k \circ s_k \notin T$
or:
:$s_n \circ s_n \notin T$
In either case, $\circ$ is not closed in $T$.
Thus by definition, $\struct {T, \circ}$ is not a submagma of $\struct {S, \circ}$.
As $T$ is arbitrary, the result follows.
{{qed}}
Theorem: Subset of Abelian Group Generated by Product of Element with Inverse Element is Subgroup

Let $\struct {G, \circ}$ be an abelian group.
Let $S \subset G$ be a non-empty subset of $G$ such that $\struct {S, \circ}$ is closed.
Let $H$ be the set defined as:
:$H := \set {x \circ y^{-1}: x, y \in S}$
Then $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$.

Proof:
Let $x \in S$.
Then:
:$x \circ x^{-1} \in H$
and so $H \ne \O$.
Now let $a, b \in H$.
Then:
:$a = x_a \circ y_a^{-1}$
and:
:$b = x_b \circ y_b^{-1}$
for some $x_a, y_a, x_b, y_b \in S$.
Thus:
{{begin-eqn}}
{{eqn | l = a \circ b^{-1}
      | r = \paren {x_a \circ y_a^{-1} } \circ \paren {x_b \circ y_b^{-1} }^{-1}
      | c = 
}}
{{eqn | r = \paren {x_a \circ y_a^{-1} } \circ \paren {\paren {y_b^{-1} }^{-1} \circ x_b^{-1} }
      | c = Inverse of Group Product
}}
{{eqn | r = \paren {x_a \circ y_a^{-1} } \circ \paren {y_b \circ x_b^{-1} }
      | c = Inverse of Group Inverse
}}
{{eqn | r = x_a \circ \paren {y_a^{-1} \circ y_b} \circ x_b^{-1}
      | c = {{Group-axiom|1}}
}}
{{eqn | r = x_a \circ \paren {y_b\circ y_a^{-1} } \circ x_b^{-1}
      | c = {{Defof|Abelian Group}}
}}
{{eqn | r = \paren {x_a \circ y_b} \circ \paren {y_a^{-1} \circ x_b^{-1} }
      | c = {{Group-axiom|1}}
}}
{{eqn | r = \paren {x_a \circ y_b} \circ \paren {x_b \circ y_a}^{-1}
      | c = Inverse of Group Product
}}
{{end-eqn}}
As $\struct {S, \circ}$ is closed, both $x_a \circ y_b \in S$ and $x_b \circ y_a \in S$.
Thus $a \circ b$ is in the form $x \circ y^{-1}$ for $x, y \in S$.
Thus $a \circ b \in H$
Hence by the One-Step Subgroup Test, $H$ is a subgroup of $G$.
{{qed}}
Theorem: Local Basis Test

Let $\struct {S, \tau}$ be a topological space.
Let $x \in S$.
Let $\BB$ be a local basis for $x$ in $\struct {S, \tau}$.
Let $\CC$ be a set of open neighborhoods of $x$.
Then:
:$\CC$ is a local basis {{iff}}:
::$\forall B \in \BB \implies \exists C \in \CC: C \subseteq B$

Proof:
=== Necessary Condition ===
Let $\CC$ be a local basis.
Let $B \in \BB$.
Since $\BB$ is a local basis, by the definition of a local basis then $B$ is open.
Since $\CC$ is a local basis, by the definition of a local basis then:
:$\exists C \in \CC : C\subseteq B$.
{{qed|lemma}}
Theorem: Non-Zero Integers under Multiplication are not Subgroup of Reals

Let $\struct {\Z_{\ne 0}, \times}$ denote the algebraic structure formed by the set of non-zero integers under multiplication.
Let $\struct {\R_{\ne 0}, \times}$ denote the algebraic structure formed by the set of non-zero real numbers under multiplication.
Then, while $\struct {\Z_{\ne 0}, \times}$ is closed, it is not a subgroup of $\struct {\R_{\ne 0}, \times}$.

Proof:
We have that Non-Zero Real Numbers under Multiplication form Group.
We also have that the set of non-zero integers $\Z_{\ne 0}$ form a subset of $\R_{\ne 0}$.
From Non-Zero Integers Closed under Multiplication:
:$\forall a, b \in \Z_{\ne 0}: a \times b \in \Z_{\ne 0}$
We have that:
:$\forall x \in \Z_{\ne 0}: 1 \times x= x = x \times 1$
and so $1$ is the identity of $\struct {\Z_{\ne 0}, \times}$.
But for $x \in \Z_{\ne 0}: x \ne 1$, there exists no $y \in \Z_{\ne 0}: x \times y = 1$.
Thus $\struct {\Z_{\ne 0}, \times}$ does not have inverses for all $x \in \Z_{\ne 0}$.
Thus, by definition, $\struct {\Z_{\ne 0}, \times}$ is not a group.
It follows that $\struct {\Z_{\ne 0}, \times}$ is not a subgroup of $\struct {\R_{\ne 0}, \times}$.
{{qed}}
Theorem: Condition for Elements of Group to be in Subgroup

Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$ be such that $2$ elements of $\set {x, y, x y}$ are elements of $h$.
Then ''all'' the elements of $\set {x, y, x y}$ are in $H$.

Proof:
As $H$ is a subgroup of $G$, it is a group in its own right.
Thus the group axioms all apply to $H$.
Let $x, y \in H$.
Then by {{Group-axiom|0}}, $x y \in H$.
Let $x, x y \in H$.
As $x \in H$, it follows that $x^{-1} \in H$ by {{Group-axiom|3}}.
Thus by {{Group-axiom|0}}, $x^{-1} \paren {x y} = y \in H$.
Let $y, x y \in H$.
As $y \in H$, it follows that $y^{-1} \in H$ by {{Group-axiom|3}}.
Thus by {{Group-axiom|0}}, $\paren {x y} y^{-1} = x \in H$.
{{qed}}
Theorem: Union of Subgroups/Corollary 1

Let $H \cup K$ be a subgroup of $G$.
Then either $H \subseteq K$ or $K \subseteq H$.

Proof:
{{AimForCont}} neither $H \subseteq K$ nor $K \subseteq H$.
Then from Union of Subgroups it follows that $H \cup K$ is not a subgroup of $G$.
The result follows by Proof by Contradiction.
{{qed}}
Theorem: Subgroup Generated by Commuting Elements is Abelian

Let $\struct {G, \circ}$ be a group.
Let $S \subseteq G$ such that:
:$\forall x, y \in S: x \circ y = y \circ x$
Then the subgroup generated by $S$ is abelian.

Proof:
Let $H = \gen S$ denote the subgroup generated by $S$.
Let $a, b \in H$.
Then:
:$a = s_1$
:$b = s_2$
for some words $s_1, s_2$ of the set of words $\map W S$ of $S$.
Then:
{{begin-eqn}}
{{eqn | l = a \circ b
      | r = s_1 \circ s_2
      | c = 
}}
{{eqn | r = s_2 \circ s_1
      | c = as all elements of $S$ commute with each other
}}
{{eqn | r = b \circ a
      | c = 
}}
{{end-eqn}}
Hence the result, by definition of abelian group.
{{qed}}
Theorem: Equivalence of Definitions of Generated Subgroup

{{TFAE|def = Generated Subgroup}}
Let $G$ be a group.
Let $S \subset G$ be a subset.

Proof:
=== $(1)$ is equivalent to $(2)$ ===
Let $H$ be the smallest subgroup containing $S$.
Let $\mathbb S$ be the set of subgroups containing $S$.
To show the equivalence of the two definitions, we need to show that $H = \bigcap \mathbb S$.
Since $H$ is a subgroup containing $S$:
:$H \in \mathbb S$
By Intersection is Subset:
:$\bigcap \mathbb S \subseteq H$
On the other hand, by Intersection of Subgroups is Subgroup:
:$\bigcap \mathbb S$ is a subgroup containing $S$.
Since $H$ be the smallest subgroup containing $S$:
:$H \subseteq \bigcap \mathbb S$
By definition of set equality:
:$H = \bigcap \mathbb S$
Hence the result.
{{qed|lemma}}
Theorem: Stabilizer of Element after Group Action

Let $\struct {G, \circ}$ be a group.
Let $S$ be a set.
Let $*_S: G \times S \to S$ be a group action.
Let $x \in S, a \in G$.
Then:
:$\Stab {a * x} = a^{-1} \circ \Stab x \circ a$

Proof:
{{begin-eqn}}
{{eqn | l = \Stab {a * x}
      | r = \set {g \in G: g * \paren {a * x} = a * x}
      | c = {{Defof|Stabilizer}}
}}
{{eqn | r = \set {g \in G: \paren {g \circ a} * x = a * x}
      | c = {{GroupActionAxiom|2}}
}}
{{eqn | r = \set {g \in G: a^{-1} * \paren {g \circ a} * x = a^{-1} * \paren {a * x} }
      | c = 
}}
{{eqn | r = \set {g \in G: \paren {a^{-1} \circ \paren {g \circ a} } * x = \paren {a^{-1} \circ a} * x}
      | c = {{GroupActionAxiom|2}}
}}
{{eqn | r = \set {g \in G: \paren {a^{-1} \circ g \circ a} * x = x}
      | c = {{Group-axiom|1}} and {{Group-axiom|3}}
}}
{{eqn | r = a^{-1} \circ \set {g \in G: g * x = x} \circ a
      | c = {{Defof|Subset Product}}
}}
{{eqn | r = a^{-1} \circ \Stab x \circ a
      | c = {{Defof|Stabilizer}}
}}
{{end-eqn}}
{{qed}}
Theorem: Group Action of Symmetric Group/Subset

Let $r \in \N: 0 < r \le n$.
Let $B_r$ denote the set of all subsets of $\N_n$ of cardinality $r$:
:$B_r  := \set {S \subseteq \N_n: \card S = r}$
Let $*$ be the mapping $*: S_n \times B_r \to B_r$ defined as:
:$\forall \pi \in S_n, \forall S \in B_r: \pi * S = \pi \sqbrk S$
where $\pi \sqbrk S$ denotes the image of $S$ under $\pi$.
Then $*$ is a group action.

Proof:
The group action axioms are investigated in turn.
Let $\pi, \rho \in S_n$.
Let $S \in B_r$.
Thus:
{{begin-eqn}}
{{eqn | l = \pi * \paren {\rho * S}
      | r = \pi * \rho \sqbrk S
      | c = Definition of $*$
}}
{{eqn | r = \pi \sqbrk {\rho \sqbrk S}
      | c = Definition of $*$
}}
{{eqn | r = \paren {\pi \circ \rho} \sqbrk S
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \paren {\pi \circ \rho} * n
      | c = Definition of $*$
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|1}} holds.
Then:
{{begin-eqn}}
{{eqn | l = I_{\N_n} * S
      | r = I_{\N_n} \sqbrk S
      | c = where $I_{\N_n}$ is the identity mapping on $\N_n$
}}
{{eqn | r = S
      | c = {{Defof|Identity Mapping}}
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|2}} holds.
{{qed}}
Theorem: Group Action of Symmetric Group on Subset is Transitive

Let $r \in \N: 0 < r \le n$.
Let $B_r$ denote the set of all subsets of $\N_n$ of cardinality $r$:
:$B_r  := \set {S \subseteq \N_n: \card S = r}$
Let $*$ be the mapping $*: S_n \times B_r \to B_r$ defined as:
:$\forall \pi \in S_n, \forall S \in B_r: \pi * B_r = \pi \sqbrk S$
where $\pi \sqbrk S$ denotes the image of $S$ under $\pi$.
Then $*$ is a transitive group action.

Proof:
From Group Action of Symmetric Group on Subset it is established that $*$ is a group action.
Let $U = \set {u_1, u_2, \ldots, u_r}$ and $V = \set {v_1, v_2, \ldots, v_r}$ be elements of $B_r$.
Then there exists a permutation $\rho \in S_n$ such that:
:$\map \rho {u_k} = v_k$
for all $k \in \set {1, 2, \ldots, r}$.
Thus:
:$\rho \sqbrk U = V$
Thus $B_r$ is the orbit of all $U \in B_r$.
Hence the result by definition of transitive group action.
{{qed}}
Theorem: Coset Product on Non-Normal Subgroup is not Well-Defined

Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$ which is not normal.
Let $a, b \in G$.
Then it is not necessarily the case that the coset product:
:$\paren {a \circ H} \circ \paren {b \circ H} = \paren {a \circ b} \circ H$
is well-defined.

Proof:
Proof by Counterexample:
Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:
{{:Symmetric Group on 3 Letters/Cayley Table}}
Consider the subgroups of $S_3$:
{{:Symmetric Group on 3 Letters/Subgroups}}
Let $H = \set {e, \tuple {12} }$.
From Normal Subgroups in Symmetric Group on 3 Letters, $H$ is not normal.
Let $A = \set {\tuple {123}, \tuple {23} }$.
We have:
{{begin-eqn}}
{{eqn | l = \tuple {123} H 
      | r = \set {\tuple {123}, \tuple {23} }
      | c = 
}}
{{eqn | l = \tuple {23} H 
      | r = \set {\tuple {123}, \tuple {23} }
      | c = 
}}
{{end-eqn}}
and so $A$ is the left coset of $H$ by both $\tuple {123}$ and $\tuple {23}$.
Let $B = \set {\tuple {132}, \tuple {13} }$.
{{begin-eqn}}
{{eqn | l = \tuple {132} H 
      | r = \set {\tuple {132}, \tuple {13} }
      | c = 
}}
{{eqn | l = \tuple {13} H 
      | r = \set {\tuple {132}, \tuple {13} }
      | c = 
}}
{{end-eqn}}
and so $B$ is the left coset of $H$ by both $\tuple {132}$ and $\tuple {13}$.
Now consider:
{{begin-eqn}}
{{eqn | l = A \circ B
      | r = \tuple {123} H \tuple {132} H
      | c = 
}}
{{eqn | r = \tuple {123} \tuple {132} H
      | c = 
}}
{{eqn | r = e H
      | c = 
}}
{{eqn | r = H
      | c = 
}}
{{end-eqn}}
But:
{{begin-eqn}}
{{eqn | l = A \circ B
      | r = \tuple {23} H \tuple {13} H
      | c = 
}}
{{eqn | r = \tuple {23} \tuple {13} H
      | c = 
}}
{{eqn | r = \tuple {132} H
      | c = 
}}
{{eqn | r = B
      | c = 
}}
{{eqn | o = \ne
      | r = H
      | c = 
}}
{{end-eqn}}
So two different evaluations of the coset product give two different result.
Hence by definition the coset product is not well-defined on $H$.
The result follows.
{{qed}}
Theorem: Klein Four-Group is Normal in A4

Let $A_4$ denote the alternating group on $4$ letters, whose Cayley table is given as:
{{:Alternating Group on 4 Letters/Cayley Table}}
The subsets of $A_4$ which form subgroups of $A_4$ are as follows:
Consider the order $4$ subgroup $V$ of $A_4$, presented by Cayley table:
:<nowiki>$\begin{array}{c|cccc}
\circ & e & t & u & v \\
\hline
e & e & t & u & v \\
t & t & e & v & u \\
u & u & v & e & t \\
v & v & u & t & e \\
\end{array}$</nowiki>
Then $V$ is normal in $A_4$.
Its index is:
:$\index {A_4} V = \dfrac {\order {A_4} } {\order V} = \dfrac {12} 4 = 3$
The (left) cosets of $V$ are:
:$V$
:$A := a V$
:$P := p V$
and the Cayley table of the quotient group $A_4 / V$ is given by:
:<nowiki>$\begin{array}{c|ccc}
\circ & V & A & P \\
\hline
V & V & A & P \\
A & A & P & V \\
P & P & V & A \\
\end{array}$</nowiki>
Note that while $A_4 / V$ is Abelian, $A_4$ is not.

Proof:
{{ProofWanted|Straightforward but tedious, unless someone has a short cut better than testing all the products}}
:$\index {A_4} V = 3$ follows from Lagrange's Theorem.
Theorem: Klein Four-Group is Normal in A4

Let $A_4$ denote the alternating group on $4$ letters, whose Cayley table is given as:
{{:Alternating Group on 4 Letters/Cayley Table}}
The subsets of $A_4$ which form subgroups of $A_4$ are as follows:
Consider the order $4$ subgroup $V$ of $A_4$, presented by Cayley table:
:<nowiki>$\begin{array}{c|cccc}
\circ & e & t & u & v \\
\hline
e & e & t & u & v \\
t & t & e & v & u \\
u & u & v & e & t \\
v & v & u & t & e \\
\end{array}$</nowiki>
Then $V$ is normal in $A_4$.
Its index is:
:$\index {A_4} V = \dfrac {\order {A_4} } {\order V} = \dfrac {12} 4 = 3$
The (left) cosets of $V$ are:
:$V$
:$A := a V$
:$P := p V$
and the Cayley table of the quotient group $A_4 / V$ is given by:
:<nowiki>$\begin{array}{c|ccc}
\circ & V & A & P \\
\hline
V & V & A & P \\
A & A & P & V \\
P & P & V & A \\
\end{array}$</nowiki>
Note that while $A_4 / V$ is Abelian, $A_4$ is not.

Proof:
By Order of Conjugate of Subgroup, conjugate subgroups have the same order.
Therefore, any conjugation of $V$ has $4$ elements.
By Subgroups of $A_4$, the only order-$4$ subgroup of $A_4$ is $V$.
Hence, any conjugation of $V$ in $A_4$ is $V$ itself.
By Subgroup equals Conjugate iff Normal, $V$ is normal in $A_4$.
{{qed}}
Theorem: Coset of Trivial Subgroup is Singleton

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $E := \struct {\set e, \circ}$ denote the trivial subgroup of $\struct {G, \circ}$.
Let $g \in G$.
Then the left coset and right coset of $E$ by $g$ is $\set g$.

Proof:
{{begin-eqn}}
{{eqn | l = g \circ \set e
      | r = \set {g \circ x: x \in \set e}
      | c = {{Defof|Left Coset}}
}}
{{eqn | r = \set {g \circ e}
      | c = 
}}
{{eqn | r = \set g
      | c = 
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \set e \circ g
      | r = \set {x \circ g: x \in \set e}
      | c = {{Defof|Right Coset}}
}}
{{eqn | r = \set {e \circ g}
      | c = 
}}
{{eqn | r = \set g
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Inverse Elements of Right Transversal is Left Transversal

Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $S \subseteq G$ be a right transversal for $H$ in $G$.
Let $T$ be the set defined as:
:$T := \set {x^{-1}: x \in S}$
where $x^{-1}$ is the inverse of $x$ in $G$.
Then $T$ is a left transversal for $H$ in $G$.

Proof:
Let $g \in G$.
We show that $g H$ contains exactly $1$ element of $T$.
Since $S$ is a right transversal:
:$\exists ! x \in S: x \in H g^{-1}$
By Right Cosets are Equal iff Element in Other Right Coset:
:$H x = H g^{-1}$
By Right Cosets are Equal iff Left Cosets by Inverse are Equal:
:$x^{-1} H = g H$
We have from definition $x^{-1} \in T$.
The result follows from uniqueness of $x$.
{{qed}}
Theorem: Condition for Subset of Group to be Right Transversal

Let $G$ be a group.
Let $H$ be a subgroup of $G$ whose index in $G$ is $n$:
:$\index G H = n$
Let $S \subseteq G$ be a subset of $G$ of cardinality $n$.
Then $S$ is a right transversal for $H$ in $G$ {{iff}}:
:$\forall x, y \in S: x \ne y \implies x y^{-1} \notin H$

Proof:
From {{Defof|Right Transversal}}, $S$ is a right transversal for $H$ in $G$ {{iff}} every right coset of $H$ contains exactly one element of $S$.
Since there are $n$ right cosets of $H$  and $S$ has cardinality $n$, if $S$ is not a right transversal for $H$ in $G$, at least one right coset of $H$ contain more than one element of $S$.
Thus the contrapositive of the theorem is as follows:
:At least one right coset of $H$ contain more than one element of $S$ {{iff}}:
::$\exists x, y \in S: x \ne y \land x y^{-1} \in H$
This is a consequence of Elements in Same Right Coset iff Product with Inverse in Subgroup.
{{qed}}
Theorem: Group Action on Coset Space is Transitive

Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Let $*: G \times G / H \to G / H$ be the action on the (left) coset space:
:$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$
Then $G$ is a transitive group action.

Proof:
It is established in Action of Group on Coset Space is Group Action that $*$ is a group action.
It remains to be shown that:
:$\forall g' H \in G / H: \Orb {g' H} = G / H$
where $\Orb {g' H}$ denotes the orbit of $g' H \in G / H$ under $*$.
Let $a H, b H \in G / H$ such that $a H \ne b H$.
We have that:
:$\exists x \in G: x a = b$
and so:
:$x * a H = \paren {x a} H = b H$
and so:
:$b H \in \Orb {a H}$
As both $a$ and $b$ are arbitrary, the result follows.
{{qed}}
Theorem: Stabilizer of Coset under Group Action on Coset Space

Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Let $*: G \times G / H \to G / H$ be the action on the (left) coset space:
:$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$
Then the stabilizer of $a H$ under $*$ is given by:
:$\Stab {a H} = a H a^{-1}$

Proof:
It is established in Action of Group on Coset Space is Group Action that $*$ is a group action.
Then:
{{begin-eqn}}
{{eqn | l = \Stab {a H}
      | r = \set {g \in G: g * a H = a H}
      | c = {{Defof|Stabilizer}}
}}
{{eqn | r = \set {g \in G: \paren {g a} H = a H}
      | c = 
}}
{{eqn | r = \set {g \in G: g H = a H a^{-1} }
      | c = 
}}
{{eqn | r = a H a^{-1}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Index of Subgroup equals Index of Conjugate

Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Then:
:$\index G H = \index G {a H a^{-1} }$
where $\index G H$ denotes the index of $H$ in $G$.

Proof:
{{ProofWanted|tired}}
Theorem: Normality Relation is not Transitive

Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $K$ be a normal subgroup of $N$.
Then it is not necessarily the case that $K$ is a normal subgroup of $G$.

Proof:
Proof by Counterexample:
Let $A_4$ denote the alternating group on $4$ letters, whose Cayley table is given as:
{{:Alternating Group on 4 Letters/Cayley Table}}
From Normality of Subgroups of Alternating Group on 4 Letters:
:$K := \set {e, t, u, v}$ is a normal subgroup of $A_4$
:$T := \set {e, t}$ is not a normal subgroup of $A_4$.
But by Subgroup of Abelian Group is Normal:
:$T$ is a normal subgroup of $K$.
Thus we have:
:$T \lhd K$, $K \lhd A_4$
but:
:$T \not \lhd A_4$
{{qed}}
Theorem: Normality Relation is not Transitive

Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $K$ be a normal subgroup of $N$.
Then it is not necessarily the case that $K$ is a normal subgroup of $G$.

Proof:
Proof by Counterexample:
Let $D_4$ denote the dihedral group $D_4$.
Let $D_4$ be presented in matrix representation:
{{:Dihedral Group D4/Matrix Representation/Formulation 1}}
Its Cayley table is given as:
{{:Dihedral Group D4/Matrix Representation/Formulation 1/Cayley Table}}
Consider the subgroup $H$ whose underlying set is:
:$H = \set {\mathbf I, \mathbf A, -\mathbf I, -\mathbf A}$
From Subgroup of Index 2 is Normal, $H$ is normal in $D_4$.
Consider the subgroup $H$ whose underlying set is:
:$K = \set {\mathbf I, \mathbf A} = \gen {\mathbf A}$
From Subgroup of Index 2 is Normal, $K$ is normal in $H$.
It remains to be demonstrated that $K$ is not normal in $D_4$.
From the Cayley table:
:$\mathbf C \mathbf A \mathbf C^{-1} = \mathbf B \mathbf C = -\mathbf A \notin K$
Hence $K$ is not normal in $D_4$.
{{qed}}
Theorem: Stabilizer is Normal iff Stabilizer of Each Element of Orbit

Let $\struct {G, \circ}$ be a group.
Let $S$ be a set.
Let $*: G \times S \to S$ be a group action.
Let $x \in S$.
Let $\Stab x$ denote the stabilizer of $x$ under $*$.
Let $\Orb x$ denote the orbit of $x$ under $*$.
Then $\Stab x$ is normal in $G$ {{iff}} $\Stab x$ is also the stabilizer of every element in $\Orb x$.

Proof:
{{tidy}}