avewright/proofwiki-math huggingface dataset:
Theorem: First Sylow Theorem

Let $p$ be a prime number.
Let $G$ be a group such that:
:$\order G = k p^n$
where:
:$\order G$ denotes the order of $G$
:$p$ is not a divisor of $k$.
Then $G$ has at least one Sylow $p$-subgroup.

Proof:
Let $\order G = k p^n$ such that $p \nmid k$.
Let $\mathbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements.
Let $N = \order {\mathbb S}$.
Now $N$ is the number of ways $p^n$ elements can be chosen from a set containing $p^n k$ elements.
From Cardinality of Set of Subsets, this is given by:
:$N = \dbinom {p^n k} {p^n} = \dfrac {\paren {p^n k} \paren {p^n k - 1} \cdots \paren {p^n k - i} \cdots \paren {p^n k - p^n + 1} } {\paren {p^n} \paren {p^n - 1} \cdots \paren {p^n - i} \cdots \paren 1}$
From Binomial Coefficient involving Power of Prime:
:$\dbinom {p^n k} {p^n} \equiv k \pmod p$
Thus:
:$N \equiv k \pmod p$
Now let $G$ act on $\mathbb S$ by the rule:
:$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$
That is, $g * S$ is the left coset of $S$ by $g$.
From Group Action on Sets with k Elements, this is a group action.
Now, let $\mathbb S$ have $r$ orbits under this action.
From Set of Orbits forms Partition, the orbits partition $\mathbb S$.
Let these orbits be represented by $\set {S_1, S_2, \ldots, S_r}$, so that:
{{begin-eqn}}
{{eqn | l = \mathbb S
      | r = \Orb {S_1} \cup \Orb {S_2} \cup \ldots \cup \Orb {S_r}
      | c = 
}}
{{eqn | l = \size {\mathbb S}
      | r = \size {\Orb {S_1} } + \card {\Orb {S_2} } + \ldots + \size {\Orb {S_r} }
      | c = 
}}
{{end-eqn}}
If each orbit had length divisible by $p$, then $p \divides N$.
But this can not be the case, because, as we have seen:
:$N \equiv k \pmod p$
So at least one orbit has length which is ''not'' divisible by $p$.
Let $S \in \set {S_1, S_2, \ldots, S_r}$ be such that $\size {\Orb S)} = m: p \nmid m$.
Let $s \in S$.
It follows from Group Action on Prime Power Order Subset that:
:$\Stab S s = S$
and so:
:$\size {\Stab S} = \size S = p^n$
From Stabilizer is Subgroup:
:$\Stab S \le G$
Thus $\Stab S$ is the subgroup of $G$ with $p^n$ elements of which we wanted to prove the existence.
{{Qed}}
Theorem: First Sylow Theorem

Let $p$ be a prime number.
Let $G$ be a group such that:
:$\order G = k p^n$
where:
:$\order G$ denotes the order of $G$
:$p$ is not a divisor of $k$.
Then $G$ has at least one Sylow $p$-subgroup.

Proof:
Let $\Bbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements.
Now let $G$ act on $\Bbb S$ by the rule:
:$\forall S \in \Bbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$
From Set of Orbits forms Partition, the orbits partition $\mathbb S$.
Let these orbits be represented by $\set {S_1, S_2, \ldots, S_n}$, so that:
:$(1): \quad \size {\mathbb S} = \size {\Orb {S_1} } + \card {\Orb {S_2} } + \ldots + \size {\Orb {S_r} }$
Thus each $\Orb {S_i}$ is the orbit under $*$ of some $S_i$ whose order is $p^n$.
By the Orbit-Stabilizer Theorem:
:$(2): \quad \order {\Orb {S_i} } = \dfrac {\order G} {\order {\Stab {S_i} } }$ for all $i \in \set {1, 2, \ldots, n}$
where $\Stab {S_i}$ is the stabilizer of $S_i$ under $*$.
Let $s \in S_i$ and $x \in \Stab {S_i}$.
Then $s x \in S_i$ because $S_i x = S_i$.
Let $s$ be fixed while $x$ runs over $\Stab {S_i}$.
Then:
:$\forall s \in S_i: s \Stab {S_i} \subseteq S_i$
Therefore $S_i$ is the union of all the left cosets $s \Stab {S_i}$.
These left cosets may not all be distinct.
Let $r_i$ be the number of distinct left cosets of $s \Stab {S_i}$.
Then:
:$\card {S_i} = r_i \card {\Stab {S_i} }$
because:
: distinct left cosets are disjoint
: all have the same cardinality $\card {\Stab {S_i} }$.
Thus:
:$(3): \quad \card {\Stab {S_i} } \divides \card {S_i} = p^n$
Because the only divisors of $p^n$ are smaller powers of $p$:
:$\card {\Stab {S_i} } = p^{n_i}$
for some $n_i \le n$, for $i = 1, 2, \ldots, r$.
We can write:
:$g = \order G = k p^n$
where $p$ does not divide $k$.
From $(2)$ and $(3)$:
:$\card {S_i} = k p^{d_i}$
where $d_i = n - n_i$ for $i = 1, 2, \ldots, r$.
From Cardinality of Set of Subsets:
:$\card {\Bbb S} = \dbinom {k p^n} {p_n}$
We have from Binomial Coefficient involving Power of Prime that:
:$\dbinom {k p^n} {p_n} \equiv k \pmod p$
But $p$ does not divide $k$.
Hence $p$ does not divide $\card {\Bbb S}$.
Hence by $(1)$ and $(4)$:
:$p \nmid k \paren {p^{d_1} + p^{d_2} + \cdots p^{d_r} }$
If $d_i > 0$ for all $i = 1, 2, \ldots, r$, then $p$ would divide this integer.
So there must be some $i$ such that $d_i = 0$.
Then, for this $i$:
:$n_i = n$
That is:
:$\card {\Stab {S_i} } = p^n$
Thus $\Stab {S_i}$ is the subgroup of $G$ of order $p_n$ that was asserted.
{{qed}}
Theorem: H-Cobordism Theorem

Let $X^n, Y^n$ be two simply connected manifolds.
Let $n \in \N: n \ge 5$ and $\exists W$ such that $W$ is an h-cobordism between $X$ and $Y$.
Then $\exists \psi: W \to X \times \closedint 0 1$ such that $\psi$ is a diffeomorphism.
In particular, $X$ and $Y$ are diffeomorphic.

Proof:
{{ProofWanted|In progress}}
Category:Cobordisms
a0li2st5wmkmtvb6ao38mv19ri10w03
Theorem: Group has Subgroups of All Prime Power Factors

Let $p$ be a prime.
Let $G$ be a finite group of order $n$.
If $p^k \divides n$ then $G$ has at least one subgroup of order $p^k$.

Proof:
From Composition Series of Group of Prime Power Order, a $p$-group has subgroups corresponding to every divisor of its order.
Thus, taken with the First Sylow Theorem, a finite group has a subgroup corresponding to every prime power divisor of its order.
{{qed}}
Theorem: Normalizer of Sylow p-Subgroup

Let $P$ be a Sylow $p$-subgroup of a finite group $G$.
Let $\map {N_G} P$ be the normalizer of $P$.
Then any $p$-subgroup of $\map {N_G} P$ is contained in $P$.
In particular, $P$ is the unique Sylow $p$-subgroup of $\map {N_G} P$.

Proof:
Let $Q$ be a $p$-subgroup of $N = \map {N_G} P$.
Let $\order Q = p^m, \order P = p^n$.
By Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup:
:$P \lhd \map {N_G} P$
thus by Subset Product with Normal Subgroup as Generator:
:$\gen {P, Q} = P Q$
Thus by Order of Subgroup Product:
:$P Q \le G: \order {P Q} = p^{n + m - s}$
where $\order {P \cap Q} = p^s$.
Since $n$ is the highest power of $p$ dividing $\order G$, this is possible only when $m \le s$.
Since $P \cap Q \le Q, s \le m$ thus we conclude that $m = s$ and therefore $P \cap Q = Q$.
Thus from Intersection with Subset is Subset:
:$Q \subseteq P$
In particular, if $Q$ is a Sylow $p$-subgroup of $\map {N_G} P$, then $Q = P$.
{{qed}}
Theorem: Second Sylow Theorem

Let $P$ be a Sylow $p$-subgroup of the finite group $G$.
Let $Q$ be any $p$-subgroup of $G$.
Then $Q$ is a subset of a conjugate of $P$.

Proof:
Let $P$ be a Sylow $p$-subgroup of $G$.
Let $\mathbb S$ be the set of all distinct $G$-conjugates of $P$:
:$\mathbb S = \set {g P g^{-1}: g \in G}$
Let $h * S$ be the conjugacy action:
:$\forall h \in P, S \in \mathbb S: h * S = h S h^{-1}$
From Conjugacy Action on Subgroups is Group Action, this is a group action for $S \le G$.
To show it is closed for $S \in \mathbb S$:
{{begin-eqn}}
{{eqn | l = S
      | o = \in
      | r = \mathbb S
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists g \in G
      | l = S
      | r = g P g^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = h * S
      | r = h \paren {g P g^{-1} } h^{-1}
      | c = 
}}
{{eqn | r = \paren {h g} P \paren {h g}^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = h * S
      | o = \in
      | r = \mathbb S
      | c = 
}}
{{end-eqn}}
So, consider the orbits and stabilizers of $\mathbb S$ under this group action.
Since $\forall S \in \mathbb S: \Stab S \le P$, we have that:
:$\size {\Stab S} \divides \order P$
Therefore, by the Orbit-Stabilizer Theorem, these orbit lengths are all congruent to either $0$ or $1$ modulo $p$, since $P$ is a Sylow $p$-subgroup of $G$.
Note that this will imply, as we shall mark later on:
:$\size {\mathbb S} \equiv 1 \pmod p$
Now, $h * P = h P h^{-1} = P$, so:
:$\Orb P = \set P$
We now show that $P$ is the only element of $\mathbb S$ such that $\size {\Orb S} = 1$.
If $g P g^{-1}$ has one element in its orbit, then:
:$\forall x \in P: x \paren {g P g^{-1} } x^{-1} = g P g^{-1}$
Thus $\forall x \in P$ we have that:
:$g^{-1} x g \in \map {N_G} P$
From Order of Conjugate Element equals Order of Element, we have that:
:$\order {g^{-1} x g} = \order x$
Thus $P_1 = g^{-1} P g$ is a $p$-subgroup of $\map {N_G} P$.
As $P$ and $P_1$ have the same number of elements, $P_1$ is a Sylow $p$-subgroup of $\map {N_G} P$.
Hence $P_1 = P$ by Normalizer of Sylow p-Subgroup, so $g P g^{-1} = P$.
Thus $P$ is the only element of $\mathbb S$ whose orbit has length $1$.
From Stabilizer of Coset Action on Set of Subgroups, $P = \map {N_G} P$.
Thus, for any $g \notin P$, $\size {\Orb {g P g^{-1} } }$ under conjugation by elements of $P$ has orbit greater than $1$.
Hence:
: $\size {\mathbb S} \equiv 1 \pmod p$
as promised.
Next we consider orbits of $\mathbb S$ under conjugation by elements of $Q$.
Since every orbit has length a power of $p$, the above conclusion shows there is at least one orbit of length $1$.
So there is an element $g$ such that:
:$\forall x \in Q: x \paren {g P g^{-1} } x^{-1} = g P g^{-1}$
As previously:
:$g^{-1} Q g \subseteq \map {N_G} P$
So by Normalizer of Sylow p-Subgroup:
:$g^{-1} Q g \subseteq P$
Thus $Q \subseteq g P g^{-1}$ as required.
{{qed}}
Theorem: Third Sylow Theorem

All the Sylow $p$-subgroups of a finite group are conjugate.

Proof:
Suppose $P$ and $Q$ are Sylow $p$-subgroups of $G$.
By the Second Sylow Theorem, $Q$ is a subset of a conjugate of $P$.
But since $\order P = \order Q$, it follows that $Q$ must equal a conjugate of $P$.
{{qed}}
Theorem: Third Sylow Theorem

All the Sylow $p$-subgroups of a finite group are conjugate.

Proof:
Let $G$ be a finite group of order $p^n m$, where $p \nmid m$ and $n > 0$.
Let $H$ be a Sylow $p$-subgroup of $G$.
We have that:
:$\order H = p^n$
:$\index G H = m$
Let $S_1, S_2, \ldots, S_m$ denote the left cosets of $G \pmod H$.
We have that $G$ acts on $G / H$ by the rule:
:$g * S_i = g S_i$.
Let $H_i$ denote the stabilizer of $S_i$.
By the Orbit-Stabilizer Theorem:
:$\order {H_i} = p^n$
while:
:$S_i = g H \implies g H g^{-1} \subseteq H_i$
Because $\order {g H g^{-1} } = \order H = \order {H_i}$, we have:
:$g H g^{-1} \subseteq H_i$
Let $H'$ be a second Sylow $p$-subgroup of $G$.
Then $H'$ acts on $G / H$ by the same rule as $G$.
Since $p \nmid m$, there exists at least one orbit under $H'$ whose cardinality is not divisible by $p$.
Suppose that $S_1, S_2, \ldots, S_r$ are the elements of an orbit where $p \nmid r$.
Let $K = H' \cap H_1$.
Then $K$ is the stabilizer of $S_1$ under the action of $H'$.
Therefore:
:$\index {H'} K = r$
However:
:$\order {H'} = p^n$
and:
:$p \nmid r$
from which it follows that:
:$r = 1$
and:
:$K = H'$
Therefore:
:$\order K = \order {H'} = \order {H_1} = p^n$
and:
:$H' = K = H_1$
Thus $H'$ and $H$ are conjugates.
Theorem: Fourth Sylow Theorem

The number of Sylow $p$-subgroups of a finite group is congruent to $1 \pmod p$.

Proof:
Let $G$ be a finite group of order $p^n m$, where $p \nmid m$ and $n > 0$.
Let $r$ be the number of Sylow $p$-subgroups of $G$.
Let $H$ be a Sylow $p$-subgroup of $G$.
We have that:
:$\order H = p^n$
:$\index G H = m$
Let $S_1, S_2, \ldots, S_m$ denote the elements of the left coset space of $G / H$.
We have that $H$ acts on $G / H$ by the rule:
:$g * S_i = g S_i$
for $S_i \in G / H$.
Unless $H = G$ and $r = 1$, there is more than $1$ orbit.
{{explain|Why? See Orbits of Group Action on Sets with Power of Prime Size}}
We have that $H$ is the stabilizer of the coset $H$, which must be one of $S_1, S_2, \ldots, S_m$.
Let $S_1, S_2, \ldots, S_k$ be the elements of $G / H$ whose stabilizer is $H$.
From the Orbit-Stabilizer Theorem and from $\order H = p^n$ we see there are $2$ cases:
:$(1): \quad$ The orbit of $S_i$ contains $p^t$ elements where $0 < t < n$
:$(2): \quad$ The orbit of $S_i$ contains only the element $S_i$.
$(2)$ occurs {{iff}} $S_i$ is one of the cosets $S_1, S_2, \ldots, S_k$ whose stabilizer is $H$.
So counting the elements of $G / H$, we see that:
:$m = k + u p$
or:
:$m \equiv k \pmod p$
From the Fifth Sylow Theorem, we have:
:$m \equiv k r \pmod p$
and so:
:$k r \equiv k \pmod p$
from which it follows:
:$r \equiv 1 \pmod p$
because $k \not \equiv 0 \pmod p$.
Hence the result.
{{qed}}
Theorem: Sylow p-Subgroup is Unique iff Normal

A group $G$ has exactly one Sylow $p$-subgroup $P$ {{iff}} $P$ is normal.

Proof:
If $G$ has precisely one Sylow $p$-subgroup, it must be normal from Unique Subgroup of a Given Order is Normal.
Suppose a Sylow $p$-subgroup $P$ is normal.
Then it equals its conjugates.
Thus, by the Third Sylow Theorem, there can be only one such Sylow $p$-subgroup.
{{qed}}
Theorem: Intersection of Normal Subgroup with Sylow P-Subgroup

Let $P$ be a Sylow $p$-subgroup of a finite group $G$.
Let $N$ be a normal subgroup of $G$.
Then $P \cap N$ is a Sylow $p$-subgroup of $N$.

Proof:
Since $N \lhd G$, we see that:
:$\gen {P, N} = P N$
from Subset Product with Normal Subgroup as Generator.
Since $P \cap N \le P$, it follows that:
:$\order {P \cap N} = p^k$
where $k > 0$.
By Order of Subgroup Product:
:$\order {P N} \order {P \cap N} = \order P \order N$
Hence from Lagrange's Theorem:
:$\index N {P \cap N} = \index {P N} P$
By Tower Law for Subgroups:
:$\index G P = \index G {P N} \index {P N} P$
Thus:
:$\index {P N} P \divides \index G P$
where $\divides$ denotes divisibility.
Thus:
:$p \nmid \index {P N} P$
so:
:$p \nmid \index N {P \cap N}$
Thus $P \cap N$ is a Sylow $p$-subgroup of $N$.
{{qed}}
Theorem: Quotient of Sylow P-Subgroup

Let $P$ be a Sylow $p$-subgroup of a finite group $G$.
Let $N$ be a normal subgroup of $G$.
Then $P N / N$ is a Sylow $p$-subgroup of $G / N$.

Proof:
We have that $P \le G$ and $ N \lhd G$.
So by the Second Isomorphism Theorem for Groups:
:$P N / N \cong P / \paren {P \cap N}$
We have that:
:$P N / N = \set {p N : p \in P}$
and so every element of $P N / N$ has order a power of $p$.
Hence $P N / N$ is a $p$-subgroup of $G / N$.
From Intersection of Normal Subgroup with Sylow P-Subgroup, we have that:
:$p \nmid \index G {P N}$
So $P N / N$ is a Sylow $p$-subgroup of $G / N$.
{{qed}}
Theorem: Fifth Sylow Theorem

The number of Sylow $p$-subgroups of a finite group is a divisor of their common index.

Proof:
Let $G$ be a finite group of order $p^n m$, where $p \nmid m$ and $n > 0$.
Let $r$ be the number of Sylow $p$-subgroups of $G$.
It is to be shown that $r \divides m$.
Let $H$ be a Sylow $p$-subgroup of $G$.
We have that:
:$\order H = p^n$
:$\index G H = m$
Let $S_1, S_2, \ldots, S_m$ denote the elements of the left coset space of $G / H$.
We have that $G$ acts on $G / H$ by the rule:
:$g * S_i = g S_i$
for $S_i \in G / H$.
There is only one orbit under this action, namely the whole of $G / H$.
{{explain|Why? See Orbits of Group Action on Sets with Power of Prime Size}}
Therefore the stabilizer of each $S_i$ is a subgroup of $G$ of index $m$ and order $p^n$.
In other words, each $S_i$ has a Sylow $p$-subgroup as a stabilizer.
Now it is shown that each Sylow $p$-subgroup is the stabilizer of one or more of the cosets $S_1, S_2, \ldots, S_m$.
We have that $H$ is the stabilizer of the coset $H$, which must be one of $S_1, S_2, \ldots, S_m$.
Let $S_1, S_2, \ldots, S_k$ be the elements of $G / H$ whose stabilizer is $H$.
By the Third Sylow Theorem, any other Sylow $p$-subgroup of $H$ is a conjugates $g H g^{-1}$ of $H$.
Thus it is seen that $g H g^{-1}$ is a stabilizer of the cosets $g S_1, g S_2, \ldots, g S_k$.
So each of the $r$ distinct Sylow $p$-subgroups of $G$ is the stabilizer of exactly $k$ elements of $G / H$.
Thus:
:$m = k r$
and so:
:$r \divides m$
as required.
{{qed}}
Theorem: Finite Submodule of Function Space

Let $\struct {G, +}$ be a group whose identity is $e$.
Let $R$ be a ring.
Let $\struct {G, +, \circ}_R$ be an $R$-module.
Let $S$ be a set.
Let $G^S$ the set of all mappings $f: S \to G$.
Let $G^{\paren S}$ be the set of all mappings $f: S \to G$ such that $\map f x = e$ for all but finitely many elements $x$ of $S$.
Then:
:$\struct {G^{\paren S}, +', \circ}_R$ is a submodule of $\struct {G^S, +, \circ}_R$
where $+'$ is the operation induced on $G^{\paren S}$ by $+$.

Proof:
Let $\struct {G, +, \circ}_R$ be an $R$-module and $S$ be a set.
We need to show that $\struct {G^{\paren S}, +'}$ is a group.
Let $f, g \in G^{\paren S}$.
Let:
:$F = \set {x \in S: \map f x \ne e}$
:$G = \set {x \in S: \map g x \ne e}$
From the definition of $f$ and $g$, both $F$ and $G$ are finite.
Since $e + e = e$ by definition of identity element, it follows that if:
:$\map f x + \map g x \ne e$
then necessarily $\map f x \ne e$ or $\map g x \ne e$.
That is:
:$\map {\paren {f +' g} } x = \map f x + \map g x \ne e \implies x \in F \cup G$
But as $F$ and $G$ are both finite, it follows that $F \cup G$ is also finite.
Hence $f +' g \in G^{\paren S}$ and $\struct {G^{\paren S}, +'}$ is closed.
Now let $f \in G^{\paren S}$.
Let $f^*$ be the pointwise inverse of $f$.
Thus:
:$\map {f^*} x = -\paren {\map f x}$
Again, let $F = \set {x \in S: \map f x \ne e}$.
It follows directly that:
:$x \in S \setminus F \implies \map {f^*} x = e$
Hence:
:$\map {f^*} x \ne e \implies x \in F$
and hence:
:$f^* \in G^{\paren S}$
So by the Two-Step Subgroup Test, it follows that $\struct {G^{\paren S}, +'}$ is a subgroup of $\struct {G^S, +}$.
Hence $\struct {G^{\paren S}, +', \circ}_R$ is an $R$-module.
The result follows.
{{qed}}
Theorem: Elements of Module with Equal Images under Linear Transformations form Submodule

Let $G$ and $H$ be $R$-modules.
Let $\phi$ and $\psi$ be linear transformations from $G$ into $H$.
Then the set $S = \set {x \in G: \map \phi x = \map \psi x}$ is a submodule of $G$.

Proof:
Let $x, y \in S$.
Let $\lambda \in R$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {x + y}
      | r = \map \phi x + \map \phi y
      | c = {{Defof|Linear Transformation}}
}}
{{eqn | r = \map \psi x + \map \psi y
      | c = $x, y \in S$
}}
{{eqn | r = \map \psi {x + y}
      | c = {{Defof|Linear Transformation}}
}}
{{eqn | l = \map \phi {\lambda \circ x}
      | r = \lambda \circ \map \phi x
      | c = {{Defof|Linear Transformation}}
}}
{{eqn | r = \lambda \circ \map \psi x
      | c = $x \in S$
}}
{{eqn | r = \map \psi {\lambda \circ x}
      | c = {{Defof|Linear Transformation}}
}}
{{end-eqn}}
Hence $x + y, \lambda \circ x \in S$.
By Submodule Test, $S$ is a submodule of $G$.
{{qed}}
Theorem: Poincaré Conjecture

Let $\Sigma^m$ be a smooth $m$-manifold.
Let $\Sigma^m$ satisfy:
:$H_0 \struct {\Sigma; \Z} = 0$
and:
:$H_m \struct {\Sigma; \Z} = \Z$
Then $\Sigma^m$ is homeomorphic to the $m$-sphere $\Bbb S^m$.

Proof:
The proof proceeds on several dimensional-cases. Note that the case $m = 3$ is incredibly intricate, and that a full proof would be impractical to produce here.  An outline of the $m = 3$ case will be given instead.
Theorem: Generated Submodule is Linear Combinations

Let $G$ be a unitary $R$-module.
Let $S \subseteq G$.
Then the submodule $H$ generated by $S$ is the set of all linear combinations of $S$.

Proof:
First the extreme case:
The smallest submodule of $G$ containing $\O$ is $\set {e_G}$.
By definition of linear combination of empty set, $\set {e_G}$ is the set of all linear combinations of $\O$.
Now the general case:
Let $\O \subset S \subseteq G$.
Let $L$ be the set of all linear combinations of $S$.
Since $G$ is a unitary $R$-module, every element $x \in S$ is the linear combination $1_R x$, so $S \subseteq L$.
But $L$ is closed for addition and scalar multiplication, so is a submodule.
Thus $H \subseteq L$.
But as every linear combination of $S$ clearly belongs to any submodule of $G$ which contains $S$, we also have $L \subseteq H$.
{{qed}}
Theorem: Empty Set is Linearly Independent

The empty set is a linearly independent set.

Proof:
There are no sequences at all of $n$ terms of the empty set for any $n > 0$.
Hence the result holds vacuously.
{{qed}}
Theorem: Subset of Module Containing Identity is Linearly Dependent

Let $G$ be a group whose identity is $e$.
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $H \subseteq G$ such that $e \in H$.
Then $H$ is a linearly dependent set.

Proof:
From Scalar Product with Identity, $\forall \lambda: \lambda \circ e = e$.
Let $H \subseteq G$ such that $e \in H$.
Consider any sequence $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ in $H$ which includes $e$.
So, let $a_j = e$ for some $j \in \closedint 1 n$.
Let $c \in R \ne 0_R$.
Consider the sequence $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ of elements of $R$ defined as:
:<nowiki>$\lambda_k = \begin {cases}
c & : k \ne j \\
0_R & : k= j
\end {cases}$</nowiki>
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \lambda_k \circ a_k
      | r = \lambda_1 \circ a_1 + \lambda_2 \circ a_2 + \cdots + \lambda_j \circ a_j + \cdots + \lambda_n \circ a_n
      | c = 
}}
{{eqn | r = 0_R \circ a_1 + 0_R \circ a_2 + \cdots + c \circ e +  \cdots + 0_R \circ a_n
      | c = 
}}
{{eqn | r = e + e + \cdots + e + \cdots + e
      | c = 
}}
{{eqn | r = e
      | c = 
}}
{{end-eqn}}
Thus there exists a sequence $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ in which not all $\lambda_k = 0_R$ such that:
:$\ds \sum_{k \mathop = 1}^n \lambda_k \circ a_k = e$
Hence the result.
{{qed}}
Theorem: Subset of Linearly Independent Set is Linearly Independent

A subset of a linearly independent set is also linearly independent.

Proof:
Let $G$ be an unitary $R$-module.
Then $\sequence {a_n}$ is a linearly independent sequence {{iff}} $\set {a_1, a_2, \ldots, a_n}$ is a linearly independent set of $G$.
So suppose that $\set {a_1, a_2, \ldots, a_n}$ is a linearly independent set of $G$.
Then clearly $\sequence {a_n}$ is a linearly independent sequence of $G$.
Conversely, let $\sequence {a_n}$ be a linearly independent sequence of $G$.
Let $\sequence {b_m}$ be a sequence of distinct terms of $\set {a_1, a_2, \ldots, a_n}$.
Let $\sequence {\mu_m}$ be a sequence of scalars such that $\ds \sum_{j \mathop = 1}^m \mu_j b_j = 0$.
For each $k \in \closedint 1 n$, let:
:<nowiki>$\lambda_k = \begin{cases}
\mu_j & : j \text { is the unique index such that } a_k = b_j \\
0 & : a_k \notin \set {b_1, b_2, \ldots, b_m}
\end{cases}$</nowiki>
Then:
:$\ds 0 = \sum_{j \mathop = 1}^m \mu_j b_j = \sum_{k \mathop = 1}^n \lambda_k a_k$
Thus:
:$\forall k \in \closedint 1 n: \lambda_k = 0$
As $\set {\mu_1, \ldots, \mu_m} \subseteq \set {\lambda_1, \ldots, \lambda_n}$, it follows that:
:$\forall j \in \closedint 1 m: \mu_j = 0$
and so $\sequence {b_m}$ has been shown to be a linearly independent sequence.
Hence the result.
{{Qed}}
Theorem: Superset of Linearly Dependent Set is Linearly Dependent

Let $S$ be a set of elements of a unitary module.
Let $S$ contain a subset $T$ which is linearly dependent.
Then $S$ is also linearly dependent.

Proof:
Let $S$ be a linearly independent set.
Let $T$ be a subset of $S$.
By Subset of Linearly Independent Set is Linearly Independent, $T$ is also linearly independent.
Thus by Proof by Contraposition, if $T$ is linearly dependent, then so must $S$ be.
{{qed}}
Theorem: Number of Ordered Bases from a Basis

Each basis of $n$ elements determines $n!$ ordered bases.

Proof:
Follows directly from Cardinality of Set of Bijections.
{{qed}}