avewright/proofwiki-math huggingface dataset:
Theorem: Factors of Composition Series for Prime Power Group

Let $G$ be a group such that $\order G = p^n$ where $p$ is a prime number.
Then $G$ has a composition series in which each factor group is cyclic of order $p$.

Proof:
From Composition Series of Group of Prime Power Order, $G$ has a sequence of subgroups:
:$\set e = G_0 \subset G_1 \subset \ldots \subset G_n = G$
such that $\order {G_k} = p^k$, $G_k \lhd G_{k + 1}$ and $G_{k + 1} / G_k$ is cyclic and of order $p$.
From Prime Group is Simple it follows that $G_{k + 1} / G_k$ is a simple group for all $k$.
From Prime Group is Cyclic it follows that $G_{k + 1} / G_k$ is cyclic.
The result follows from Condition for Composition Series.
{{qed}}
Theorem: Continuity of Linear Functionals

Let $V$ be a normed vector space, and let $L$ be a linear functional on $V$.
Then the following four statements are equivalent:
:$(1): \quad L$ is continuous
:$(2): \quad L$ is continuous at $\mathbf 0_V$
:$(3): \quad L$ is continuous at some point
:$(4): \quad L$ is bounded: $\exists c > 0: \forall v \in H: \size {L v} \le c \norm v$

Proof:
=== $(1)$ iff $(2)$ ===
Theorem: Equivalence of Definitions of Norm of Linear Functional

Let $V$ be a normed vector space, and let $L$ be a bounded linear functional on $V$.
Define the following norms of $L$:
{{begin-eqn}}
{{eqn | n = 1
      | l = \norm L_1
      | r = \sup \set {\size {L h}: \norm h \le 1}
}}
{{eqn | n = 2
      | l = \norm L_2
      | r = \sup \set {\size {L h}: \norm h = 1}
}}
{{eqn | n = 3
      | l = \norm L_3
      | r = \sup \set {\dfrac {\size {L h} } {\norm h}: h \in H \setminus \set {\mathbf 0} }
}}
{{eqn | n = 4
      | l = \norm L_4
      | r = \inf \set {c > 0: \forall h \in H: \size {L h} \le c \norm h}
}}
{{end-eqn}}
Then:
:$\norm L_1 = \norm L_2 = \norm L_3 = \norm L_4$

Proof:
We have:
:$\set {v \in V : \norm v = 1} \subseteq \set {v \in V : \norm v \le 1} \subseteq V$
So it follows from the definition of the supremum that 
:$\norm L_2 \le \norm L_1 \le \norm L_3$
Next we show that $\norm L_2 = \norm L_3$:
{{begin-eqn}}
{{eqn | l = \norm L_2
      | r = \sup \set {\norm {L v}: \norm h = 1}
}}
{{eqn | r = \sup \set {\norm {L v'}: v \in V \setminus \set {\mathbf 0}, \ v' = v / \norm v}
      | c = as $\norm {v / \norm v} = \norm v / \norm v = 1$ for all $v \in V \setminus \set {\mathbf 0}$
}}
{{eqn | r = \sup \set {\norm {\frac 1 {\norm v} L v}: v \in V \setminus \set {\mathbf 0} }
      | c = as $L$ is linear
}}
{{eqn | r = \sup \set {\frac {\norm {L v} } {\norm v}: v \in V \setminus \set {\mathbf 0} }
      | c = {{NormAxiomVector|2}}
}}
{{eqn | r = \norm L_3
}}
{{end-eqn}}
Therefore:
:$\norm L_1 = \norm L_2 = \norm L_3$
Moreover, if $\size {L v} \le c \norm v$ for all $v \in V \setminus \set {\mathbf 0}$, then we have:
:$\norm L_3 = \sup \set {\dfrac {\size {L v} } {\norm v}: v \in V \setminus \set {\mathbf 0} } \le c$
Taking the infimum over all such $c$ this reads:
:$\norm L_3 \le \norm L_4$
{{AimForCont}} $c_0 := \norm L_4 > \norm L_3$.
Then by the definitions of these two norms, this means that there exists $\epsilon > 0$ such that for every $v \in V \setminus \set {\mathbf 0}$:
:$\dfrac {\size {L v} } {\norm v} + \epsilon \le c_0$
But this in turn implies that for every $v \in V \setminus \set {\mathbf 0}$:
:$\size {L v} \le c_0 \norm v - \epsilon \norm v = \paren {c_0 - \epsilon} \norm v$
This contradicts the fact that $c_0$ is the least such number satisfying this inequality.
Therefore:
:$\norm L_3 = \norm L_4$
and the proof is complete.
{{qed}}
Theorem: Riesz Representation Theorem (Hilbert Spaces)

Let $H$ be a Hilbert space.
Let $L$ be a bounded linear functional on $H$.
Then there is a unique $h_0 \in H$ such that:
:$\forall h \in H: L h = \innerprod h {h_0}$

Proof:
If $L \equiv 0$ identically, then:
:$L h = 0 = \innerprod h 0$
and the theorem holds.
By Kernel of Bounded Linear Transformation is Closed Linear Subspace:
:$M := \ker L$ is a closed linear subspace of $H$.
Then we can decompose $H$ as a direct sum:
:$H \cong M \oplus M^\perp$
where $M^\perp$ denotes the orthocomplement of $M$.
From:
:$L \not \equiv 0$
we have:
:$M^\perp \ne \set 0$
Choose a $z \in M^\perp$ with norm $1$.
By linearity of $L$, for any $h \in H$:
{{begin-eqn}}
{{eqn | l = L \paren {z L h - h L z}
      | r = L z L h - L h L z
}}
{{eqn | r = 0
}}
{{end-eqn}}
So:
:$z L h - h L z \in \ker L = M$
Then:
{{begin-eqn}}
{{eqn | l = L h
      | r = L h \innerprod z z
      | c = as $\norm z = 1$
}}
{{eqn | r = \innerprod {z L h} z
      | c = linearity in the first argument
}}
{{eqn | r = \innerprod {z L h - h L z + h L z} z
      | c = adding and subtracting $h L z$ in the first argument
}}
{{eqn | r = \innerprod {z L h - h L z} z + \innerprod {h L z} z
      | c = linearity in the first argument
}}
{{eqn | r = \innerprod {h L z} z
      | c = $z L h - h L z \in M, z \in M^\perp$
}}
{{eqn | r = \innerprod h {z \paren {L z}^*}
      | c = conjugate symmetry
}}
{{end-eqn}}
Thus:
:$L h = \innerprod h {h_0}$
for $h_0 = z \paren {L z}^*$.
To show uniqueness, assume $h_0$ and $h_1$ both satisfy the above equation for all $h \in H$:
{{begin-eqn}}
{{eqn | l = \innerprod h {h_0}
      | r = \innerprod h {h_1}
}}
{{eqn | ll= \leadsto
      | l = \innerprod h {h_0} - \innerprod h {h_1}
      | r = 0
}}
{{eqn | r = \innerprod h {h_0 - h_1}
      | c = additivity in the second argument
}}
{{end-eqn}}
The result follows from setting $h = h_0 - h_1$ and invoking the positive definiteness of the inner product.
{{qed}}
Theorem: Orthonormal Subset of Hilbert Space Extends to Basis

Let $H$ be a Hilbert space, and let $S$ be an orthonormal subset of $H$.
Then there exists a basis for $H$ that contains $S$ as a subset.

Proof:
Consider the set $\OO$ of orthonormal subsets $S'$ of $H$ that contain $S$:
:$\OO := \set{ S' \subseteq H: S \subseteq S', \text{$S'$ is orthonormal} }$
In particular, $S \in \OO$ so that $\OO$ is non-empty.
Give $\OO$ the subset ordering.
For a chain $\CC \subseteq \OO$, we assert that $\bigcup \CC \in \OO$.
For each $c \in \bigcup \CC$, there is a $C \in \CC$ such that $c \in C$.
Hence $\norm c = 1$.
Moreover, for $c, c' \in \bigcup \CC$, there is a $C \in \CC$ such that $c, c' \in C$ because $\CC$ is a chain.
Then since $C \in \OO$, it follows that $c \perp c'$.
Hence $\bigcup \CC \in \OO$.
Therefore, the conditions of Zorn's Lemma apply to $\OO$.
Let $O$ be a maximal element of $\OO$.
Then $O$ is a maximal orthonormal subset of $H$.
That is, $O$ is a basis containing $S$.
{{qed}}
{{AoC|Zorn's Lemma}}
Theorem: Gram-Schmidt Orthogonalization

Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space over $\R$ or $\C$.
Let $S = \set {v_n: n \in \N_{>0} }$ be a linearly independent subset of $V$. 
Then there exists an orthonormal subset $E = \set {e_n: n \in \N_{>0} }$ of $V$ such that:
:$\forall k \in \N: \span \set {v_1 , \ldots , v_k} = \span \set {e_1 , \ldots ,e_k}$
where $\span$ denotes linear span.

Proof:
For all $k \in \N_{>0}$, define $u_k , e_k \in V$ inductively as:
{{begin-eqn}}
{{eqn | o = :=
      | l = u_k
      | r = \ds v_k - \sum_{i \mathop= 1}^{k-1} \innerprod {v_k}{e_i} e_i
}}
{{eqn | o = :=
      | l = e_k 
      | r = \dfrac {1}{\norm {u_k} } u_k
}}
{{end-eqn}}
where $\norm \cdot$ denotes the inner product norm on $V$.
We prove the theorem by induction for $k \in \N_{>0}$.
Theorem: Orthogonal Projection onto Closed Linear Span

Let $H$ be a Hilbert space with inner product $\innerprod \cdot \cdot$ and inner product norm $\norm \cdot$. 
Let $E = \set {e_1, \ldots, e_n}$ be an orthonormal subset of $H$.
Let $M = \vee E$, where $\vee E$ is the closed linear span of $E$. 
Let $P$ be the orthogonal projection onto $M$.
Then:
:$\forall h \in H: P h = \ds \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k$

Proof:
Let $h \in H$. 
Let: 
:$\ds u = \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k$ 
We have that:
:$u \in \map \span E$
and from the definition of closed linear span:
:$M = \paren {\map \span E}^-$
We therefore have, by the definition of closure: 
:$u \in M$ 
Let $v = h - u$ 
We want to show that $v \in M^\bot$. 
From Intersection of Orthocomplements is Orthocomplement of Closed Linear Span, it suffices to show that: 
:$v \in E^\bot$
Note that for each $l$ we have: 
:$\innerprod v {e_l} = \innerprod h {e_l} - \innerprod u {e_l}$
since the inner product is linear in its first argument. 
We have: 
{{begin-eqn}}
{{eqn	| l = \innerprod u {e_l} 
	| r = \innerprod {\sum_{k \mathop = 1}^n \innerprod h {e_k} e_k} {e_l}
}}
{{eqn	| r = \sum_{k \mathop = 1}^n \innerprod {\innerprod h {e_k} e_k} {e_l}
	| c = linearity of inner product in first argument
}}
{{eqn	| r = \sum_{k \mathop = 1}^n \innerprod h {e_k} \innerprod {e_k} {e_l}
	| c = linearity of inner product in first argument
}}
{{eqn	| r = \innerprod h {e_l} \innerprod {e_l} {e_l}
	| c = {{Defof|Orthonormal Subset}}
}}
{{eqn	| r = \innerprod h {e_l} \norm {e_l}^2
	| c = {{Defof|Inner Product Norm}}
}}
{{eqn	| r = \innerprod h {e_l}
	| c = since $\norm {e_l} = 1$
}}
{{end-eqn}}
so:
:$\innerprod v {e_l} = 0$ 
That is: 
:$v \in E^\bot$
so, by Intersection of Orthocomplements is Orthocomplement of Closed Linear Span, we have: 
:$v \in M^\bot$
We can therefore decompose each $h \in H$ as: 
:$h = u + v$
with $u \in M$ and $v \in M^\bot$. 
So we have: 
{{begin-eqn}}
{{eqn	| l = P h 
	| r = \map P {u + v}
}}
{{eqn	| r = \map P u + \map P v
	| c = Orthogonal Projection is Linear Transformation
}}
{{eqn	| r = v
	| c = Kernel of Orthogonal Projection, Fixed Points of Orthogonal Projection
}}
{{eqn	| r = \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k
}}
{{end-eqn}}
for each $h \in H$. 
{{qed}}
Theorem: Between two Rational Numbers exists Irrational Number

Let $a, b \in \Q$ where $a < b$. 
Then:
:$\exists \xi \in \R \setminus \Q: a < \xi < b$

Proof:
Let $d = b - a$.
As $a, b \in \Q: a < b$ it follows from Rational Numbers form Ordered Integral Domain that $d \in \Q: d > 0$.
From Square Root of 2 is Irrational, $\sqrt 2$ is not a rational number, so it is an element of $\R \setminus \Q$.
From Square Number Less than One, for any given real number $x$:
:$x^2 < 1 \implies x \in \openint {-1} 1$
Let $k = \dfrac {\sqrt 2} 2$.
Then from Lemma 1:
:$k \in \R \setminus \Q$
As $k^2 = \dfrac 1 2$, it follows that $0 < k < 1$.
Let $\xi = a + k d$.
Then, since $a, d \in \Q$ and $k \in \R \setminus \Q$, it follows from Lemma 1 and Lemma 2 that:
:$\xi \in \R \setminus \Q$
$d > 0$ and $k > 0$, so:
:$\xi = a + k d > a + 0 \cdot 0 = a$
$k < 1$, so:
:$\xi = a + k d < a + 1 \cdot d < a + \paren {b - a} = b$
We thus have:
:$\xi \in \R \setminus \Q: \xi \in \openint a b$
Hence the result.
{{qed}}
Theorem: Between two Rational Numbers exists Irrational Number

Let $a, b \in \Q$ where $a < b$. 
Then:
:$\exists \xi \in \R \setminus \Q: a < \xi < b$

Proof:
From Between two Real Numbers exists Rational Number, there exists $x \in \Q$ such that:
:$a - \sqrt2 < x < b - \sqrt2$
Since Square Root of 2 is Irrational, by the Lemma:
:$x + \sqrt 2$ is irrational.
But:
:$a < x + \sqrt 2 < b$
which is what we wanted to show.
{{qed}}
Theorem: Abelian Group is Simple iff Prime

Let $G$ be a non-trivial abelian group.
Then $G$ is simple {{iff}} $G$ is a prime group.

Proof:
First we note that the trivial group is (trivially) simple, but not prime because $1$ is not prime.
Hence the specification of $G$ as being non-trivial.
Theorem: Cyclic Group is Simple iff Prime

Let $G$ be a cyclic group.
Then $G$ is simple {{iff}} $G$ is a prime group.

Proof:
Let $G$ be a cyclic group.
From Cyclic Group is Abelian it follows that $G$ is an abelian group.
The result follows from Abelian Group is Simple iff Prime.
{{qed}}
Category:Simple Groups
Category:Cyclic Groups
ihiodpl0ircj9fl5b5ya623qbow3qh2
Theorem: Factors of Solvable Group are Prime

Let $G$ be a solvable group.
Let $\HH$ be a composition series of $G$.
Then all factor groups of $\HH$ must be prime.

Proof:
By definition, a composition series is a normal series whose factor groups are all simple.
A solvable group, by definition, is one which has a composition series whose factor groups are all cyclic.
From Cyclic Group is Simple iff Prime, it follows that all the factor groups of a composition series of a solvable group must all be prime.
{{qed}}
Theorem: Linear Subspace is Convex Set

Let $V$ be a vector space over $\R$ or $\C$, and let $L$ be a linear subspace of $V$.
Then $L$ is a convex set.

Proof:
Let $x, y \in L, t \in \closedint 0 1$ be arbitrary.
Then, as $L$ is by definition closed under addition and scalar multiplication, it follows immediately that
:$t x + \paren {1 - t} y \in L$
Hence $L$ is convex.
{{qed}}
Theorem: Singleton is Convex Set

Let $V$ be a vector space over $\R$ or $\C$, and let $v \in V$.
Then the singleton $S = \set v$ is a convex set.

Proof:
For any $x, y \in S$, we have $x = y = v$.
It follows that:
:$\forall t \in \closedint 0 1: t x + \paren {1 - t} y = v \in S$
Hence $S$ is a convex set.
{{qed}}
Theorem: Group is Solvable iff Normal Subgroup and Quotient are Solvable

Let $G$ be a finite group.
Let $H$ be a normal subgroup of $G$.
Then $G$ is solvable {{iff}}:
:$(1): \quad H$ is solvable
and
:$(2): \quad G / H$ is solvable
where $G / H$ is the quotient group of $G$ by $H$.

Proof:
As $H \lhd G$ we can construct the normal series:
:$(A): \quad \set e \lhd H \lhd G$
By Finite Group has Composition Series, $(A)$ can be refined to a composition series for $G$:
:$(B): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$
Suppose $G_k = H$.
Then we can construct the composition series:
:$(C): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_k = H$
and:
:$(D): \quad \set e = G_k / H \lhd G_{k + 1} / H \lhd \cdots \lhd G_n / H = G / H$
Furthermore, by the Third Isomorphism Theorem:
:$\dfrac {G_{i + 1} / H} {G_i / H} \cong \dfrac {G_{i + 1} } {G_i}$
for all $k \le i \le n$.
So each factor of the composition series for $G$ is a factor of either:
:the composition series for $H$
or:
:the composition series for $G / H$.
The result follows.
{{qed}}
Theorem: Finite Abelian Group is Solvable

Let $G$ be a finite abelian group.
Then $G$ is solvable.

Proof:
Proof by strong induction on the order of $G$:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:All abelian groups of order $n$ and below are solvable.
Theorem: Intersection of Convex Sets is Convex Set (Vector Spaces)

Let $V$ be a vector space over $\R$ or $\C$.
Let $\CC$ be a family of convex subsets of $V$.
Then the intersection $\ds \bigcap \CC$ is also a convex subset of $V$.

Proof:
Let $x, y \in \ds \bigcap \CC$.
Then by definition of set intersection, $\forall C \in \CC: x, y \in C$.
The convexity of each $C$ yields:
:$\forall t \in \closedint 0 1: t x + \paren {1 - t} y \in C$
Therefore, these elements are also in $\ds \bigcap \CC$, by definition of set intersection.
Hence $\ds \bigcap \CC$ is also convex.
{{qed}}
Theorem: Prime Power Group is Solvable

Let $G$ be a group whose order is $p^n$ where $p$ is a prime number and $n$ is a positive integer.
Then $G$ is solvable.

Proof:
A direct consequence of Factors of Composition Series for Prime Power Group and the definition of solvable group.
{{qed}}
Category:P-Groups
Category:Solvable Groups
31xinsqnol2xxwyvwlv1v8vg7gcgpq1
Theorem: Subgroup of Solvable Group is Solvable

Let $G$ be a solvable group.
Let $H$ be a subgroup of $G$.
Then $H$ is solvable.

Proof:
Let $\set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$ be a normal series for $G$ with abelian quotients.
For every $i = 1, 2, \ldots, n$ we have:
:$\paren {H \cap G_i} \cap G_{i - 1} = H \cap G_{i - 1}$
From the Second Isomorphism Theorem for Groups:
:$\dfrac {\paren {H \cap G_i} G_{i - 1} } {G_{i - 1} } \cong \dfrac {H \cap G_i} {\paren {H \cap G_i} \cap G_{i - 1} } = \dfrac {H \cap G_i} {H \cap G_{i - 1} }$
where $\cong$ denotes group isomorphism.
In particular, $H \cap G_{i - 1}$ is a normal subgroup of $H \cap G_i$.
We have that:
:$\paren {H \cap G_i} G_{i - 1} \subseteq G_i$
and so from the Correspondence Theorem:
:$\dfrac {\paren {H \cap G_i} G_{i - 1} } {G_{i - 1} } \le G_i / G_{i - 1}$
We have that $G_i / G_{i - 1}$ is abelian.
Thus from Subgroup of Abelian Group is Abelian:
:$\dfrac {\paren {H \cap G_i } G_{i - 1} } {G_{i - 1} }$ is abelian.
Hence $\dfrac {H \cap G_i} {H \cap G_{i - 1} }$ is abelian.
Therefore, the series :
:$\set e = H \cap G_0 \lhd H \cap G_1 \lhd \cdots \lhd H \cap G_n = H$
is a normal series with abelian factor groups for $H$.
Therefore $H$ is solvable.
{{qed}}
Theorem: Subgroup of Solvable Group is Solvable

Let $G$ be a solvable group.
Let $H$ be a subgroup of $G$.
Then $H$ is solvable.

Proof:
Firstly we, know that a group is solvable {{iff}} its derived series:
:$\map D G = \sqbrk {G, G} \ , \ \map {D^i} G = \sqbrk {\map {D^{i - 1} } G, \map {D^{i - 1} } G}$
becomes trivial after finite iteration.
Meaning:
{{handwaving|trivial}}
$\map {D^j} G = \set 1$ 
for some finite $j$.
Now it is trivial that:
{{handwaving|trivial}}
:$\map D H \le \map D G$
since $H$ is smaller than $G$.
Further since $\map {D^i} H$ is dominated by $\map {D^i} G$, it too has to become trivial after a finite amount of steps.
{{handwaving|trivial}}
{{qed}}
Theorem: Subgroup of Solvable Group is Solvable

Let $G$ be a solvable group.
Let $H$ be a subgroup of $G$.
Then $H$ is solvable.

Proof:
Let $H \le G$ and $G$ be solvable with normal series:
:$\set e = G_0 \lhd G_1 \lhd \dots \lhd G_m = G$
such that $G_{i + 1} / G_i$ is abelian for all $i$.
Define $N_i = G_i \cap H$.
We show that these $N_i$ will form a normal series with abelian factors.
;Normality:
Let $x \in N_i$ and $y \in N_{i + 1}$.
Then $y x y^{-1} \in N$ since $N$ as a group is closed.
{{explain|The $N$ above seems to be $H$.}}
We also have $y x y^{-1} \in G_i$ since $G_i$ is normal in $G_{i + 1}$.
Hence $N_i$ is invariant under conjugation and therefore normal.
;Abelian Factors: 
Note that:
{{begin-eqn}}
{{eqn | l = N_{i + 1} \cap G_i
      | r = \paren {G_{i + 1} \cap H} \cap G_i
      | c = Definition of $N_{i + 1}$
}}
{{eqn | r = H \cap \paren {G_{i + 1} \cap G_i}
      | c = Intersection is Commutative, Intersection is Associative
}}
{{eqn | r = H \cap G_i
      | c = Intersection with Subset is Subset: $G_i \subseteq G_{i + 1}$
}}
{{eqn | r = N_i
      | c = Definition of $N_i$
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \dfrac {N_{i + 1} } {N_i}
      | r = \dfrac {N_{i + 1} } {N_{i + 1} \cap G_i}
      | c = 
}}
{{eqn | o = \cong
      | r = \dfrac {N_{i + 1} G_i} {G_i}
      | c = Second Isomorphism Theorem for Groups
}}
{{eqn | o = \le
      | r = \dfrac {G_{i + 1} } {G_i}
      | c =
}}
{{end-eqn}}
{{explain|Justify the step above}}
so the quotient $\dfrac {N_{i + 1}} {N_i}$ is isomorphic to a subgroup of the abelian group $\dfrac{G_{i + 1}} {G_i}$.
Hence it is abelian, proving the theorem.
{{qed}}
Theorem: Group with Normal Series with Solvable Factor Groups is Solvable

Let $G$ be a solvable group.
Let $\HH$ be a normal series for $G$.
Let all the factor groups of $\HH$ be solvable.
Then $G$ is solvable.

Proof:
To be proved by construction of a composition series of $G$ from $\HH$.
Suppose $\HH$ is a normal series for $G$:
:$\HH: \quad \set e = G_0 \subset G_1 \subset \dots \subset G_n = G$,
where $G_{k + 1} / G_k$ is solvable for $k = 0, 1, \dots, n - 1$.
If every factor of $\HH$ is cyclic of prime order then we are done.
Otherwise, pick a factor $G_{k + 1} / G_k$ that is not cyclic of prime order.
Then there exists a composition series $\SS_{k}$ for $G_{k + 1} / G_k$:
:$\SS_k: \quad G_k / G_k = H_0 / G_k \subset H_1 / G_k \subset \dots \subset H_m / G_k = G_{k + 1} / G_k$,
where $\paren {H_{i + 1} / G_k} / \paren {H_i / G_k}$ is cyclic of prime order.
The Third Isomorphism Theorem induces the sequence:
:$\HH_k: \quad G_k = H_0 \subset H_1 \subset \dots \subset H_m = G_{k + 1}$,
where $H_{i + 1} / H_i$ is also cyclic of prime order.
This shows that $\HH_k$ is complete.
We may refine the original series $\HH$ with $\HH_k$:
:$\HH': \quad \set e = G_0 \subset G_1 \subset \dots \subset G_k = H_0 \subset \dots \subset H_m = G_{k + 1} \subset \dots \subset G_n = G$.
Repeating this process for all factors of $\HH$ that are not cyclic of prime order to obtain a composition series $\GG$ for $G$.
By this construction, every factor of $\GG$ is cyclic.
The result follows from the definition of solvable groups.
{{qed}}
Theorem: Group with Order Less than 60 is Solvable

Every group whose order is less than $60$ is solvable.

Proof:
Let $n \in \Z$ be the smallest integer such that there is a non-solvable group of order $n$.
{{AimForCont}} $n < 60$.
Let $G$ be a non-solvable group of order $n$.
Since $\order G < 60$, it follows from Simple Group of Order Less than 60 is Prime that either:
:$G$ is prime
or:
:$G$ is not simple.
Since {{hypothesis}} $G$ is not solvable, $G$ cannot be prime.
Thus $G$ is not simple.
Therefore there exists a non-trivial proper subgroup $N \trianglelefteq G$.
{{explain|Is this so? I believe it needs also to be asserted that $N$ is normal. If normality is expected to be inferred from the symbol, then would $\triangleleft$ be more appropriate? Subgroups are proper.}}
Since both $N$ and $G / N$ have order less than $n$, both groups are solvable.
Then the composition series $1 \trianglelefteq N \trianglelefteq G$ is a normal series with solvable composition factors.
So, by Group with Normal Series with Solvable Factor Groups is Solvable, $G$ is solvable.
This contradicts our assertion that $G$ is a non-solvable whose order is less than $60$.
It follows that our assumption that $n < 60$ is false.
That is, there is no non-solvable group whose order is less than $60$.
{{qed}}
Theorem: Direct Product of Solvable Groups is Solvable

Let $G$ and $H$ be groups which are solvable.
Then their (external) direct product $G \times H$ is also solvable.

Proof:
By Image of Canonical Injection is Normal Subgroup, $G \times \set {e_H}$ is a normal subgroup of $G \times H$.
Also, by Quotient Group of Direct Products, $\paren {G \times H} / \paren {G \times \set{e_H} }$ is isomorphic to $H$.
The result then follows from Group is Solvable iff Normal Subgroup and Quotient are Solvable.
{{qed}}
Theorem: Group with One Sylow Subgroup per Prime Divisor is Solvable

Let $G$ be a group of order $n$.
Suppose that, for each prime number $p$ which divides $n$, $G$ has exactly one $p$-Sylow subgroup.
Then $G$ is solvable.

Proof:
From Finite Group with One Sylow p-Subgroup per Prime Divisor is Isomorphic to Direct Product, 
:$G$ is isomorphic to the direct product of its $p$-Sylow subgroups.
From Prime Power Group is Solvable, each $p$-Sylow subgroup is solvable.
From Direct Product of Solvable Groups is Solvable, $G$ is solvable.
{{qed}}
Theorem: Characterization of Bases (Hilbert Spaces)

Let $\HH$ be a Hilbert space, and let $\EE$ be an orthonormal subset of $\HH$.
Then the following six statements are equivalent:
:$(1): \quad \EE$ is a basis for $\HH$
:$(2): \quad h \in \HH, h \perp \EE \implies h = \mathbf 0_\HH$, where $\perp$ denotes orthogonality
:$(3): \quad \vee \EE = \HH$, where $\vee \EE$ denotes the closed linear span of $E$
:$(4): \quad \forall h \in \HH: h = \ds \sum \set {\innerprod h e e: e \in \EE}$
:$(5): \quad \forall g, h \in \HH: \innerprod g h = \ds \sum \set {\innerprod g e \innerprod e h: e \in \EE}$
:$(6): \quad \forall h \in \HH: \norm h^2 = \ds \sum \set {\size {\innerprod h e}^2: e \in \EE}$
In the last three statements, $\ds \sum$ denotes a generalized sum.
Statement $(6)$ is commonly known as '''Parseval's identity'''.

Proof:
=== $(1)$ implies $(2)$ ===
Suppose that $\EE$ is a basis for $\HH$.
Suppose that $h \perp \EE$ for some $h \in \HH$ with $h \ne {\mathbf 0}_\HH$. 
Then $\innerprod h x = 0$ for all $x \in \EE$.
So $\ds \innerprod {\frac h {\norm h} } x = 0$ for all $x \in \EE$.
Hence:
:$\ds \EE \cup \set {\frac h {\norm h} }$ is a orthonormal set. 
Hence $\dfrac h {\norm h} \in \EE$.
However, we would then have $\dfrac h {\norm h} \perp \dfrac h {\norm h}$ giving:
:$\ds \innerprod {\frac h {\norm h} } {\frac h {\norm h} } = \frac {\innerprod h h} {\norm h} = \norm h = 0$ 
Then by {{NormAxiomVector|1}}, we have $h = {\mathbf 0}_\HH$, a contradiction. 
{{qed|lemma}}