nyuuzyou/wikis huggingface dataset:
Group of Units Ring of Integers Modulo p^2 is Cyclic

== Theorem ==
Let $p$ be a prime.
Let $\struct {\Z / p^2 \Z, +, \times}$ be the ring of integers modulo $p^2$.
Let $U = \struct {\paren {\Z / p^2 \Z}^\times, \times}$ denote the group of units of $\struct {\Z / p^2 \Z, +, \times}$.
Then $U$ is cyclic.
== Proof ==
The case $p = 2$ follows from Isomorphism between Group of Units Ring of Integers Modulo $2^n$ and $C_2 \times C_{2^{n - 2} }$.
Next, we suppose $p > 2$.
From Ring of Integers Modulo Prime is Field and Group of Units of Finite Field is Cyclic, $\paren {\Z / p \Z}^\times$ is cyclic.
Let $\eqclass g p$ be a generator of $\paren {\Z / p \Z}^\times$ for some $g \in \Z$.
Let $d$ be the order of $\eqclass g {p^2}$ in $\paren {\Z / p^2 \Z}^\times$.
Then:
:$g^d \equiv 1 \pmod {p^2}$
so:
:$g^d \equiv 1 \pmod p$
The order of $\eqclass g p$ in $\paren {\Z / p \Z}^\times$ is:
:$\order {\paren {\Z / p \Z}^\times} = p - 1$
We get:
By Lagrange's Theorem
We need to prove that:
:$\eqclass g {p^2}$ generates $\paren {\Z / p^2 \Z}^\times \implies \paren {\Z / p^2 \Z}^\times$ is cyclic.
If not, then:
By $(1)(2)(3)$:
:$d = p - 1$
so:
Consider:
:$g' = g + p$
Let $d'$ be the order of $\eqclass {g'} {p^2}$ in $\paren {\Z / p^2 \Z}^\times$.
Then:
:${g'}^{d'} \equiv 1 \pmod {p^2}$
so
:${g'}^{d'} \equiv 1 \pmod p$
Note that:
:$\eqclass {g'} p = \eqclass g p$. 
Then the order of $\eqclass {g'} p$ in $\paren {\Z / p \Z}^\times$ is:
:$\order {\paren {\Z / p \Z}^\times} = p - 1$
We get
By Lagrange's Theorem
By Binomial Theorem for Integral Index:
so:
:${g'}^{p - 1} \not \equiv 1 \pmod {p^2}$
Therefore:
By $(5)$, $(6)$ and $(7)$:
:$d' = p \paren{p - 1}$
so:
:$\eqclass {g'} {p^2}$ generates $\paren {\Z / p^2 \Z}^\times \implies \paren {\Z / p^2 \Z}^\times$ is cyclic.
Category:Ring of Integers Modulo m
Group of Units in Unital Banach Algebra is Open

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra.
Let $\map G A$ be the group of units of $A$.
Then $\map G A$ is open in $A$. 
== Proof ==
Let $x \in \map G A$.
We find an open neighborhood of $x$ contained in $\map G A$. 
Clearly $x^{-1} \ne \mathbf 0_A$, so $\norm {x^{-1} } > 0$. 
We have, for $y \in A$:
So for $y \in A$ satisfying: 
:$\ds \norm {x - y} < \frac 1 {\norm {x^{-1} } }$
we have: 
:$\norm {1 - x^{-1} y} < 1$ 
From Element of Unital Banach Algebra Close to Identity is Invertible, we have that $x^{-1} y$ is invertible.
That is, $x^{-1} y \in \map G A$. 
Since $\map G A$ is a group, from Group of Units is Group, we have: 
:$y = x \paren {x^{-1} y} \in \map G A$ 
So we have: 
:$\map {B_{\norm {x^{-1} }^{-1} } } x \subseteq \map G A$
where $\map {B_{\norm {x^{-1} }^{-1} } } x$ is the open ball centered at $x$ with radius $\norm {x^{-1} }^{-1}$.
Since $x \in \map G A$ was arbitrary, we have that $\map G A$ is open.
== Sources ==
* : $4.4$: The group of units
Category:Unital Banach Algebras
Group of Units is Group

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity.
Then the set of units of $\struct {R, +, \circ}$ forms a group under $\circ$.
Hence the justification for referring to the group of units of $\struct {R, +, \circ}$.
== Proof ==
Follows directly from Invertible Elements of Monoid form Subgroup of Cancellable Elements.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains
* : $\S 55.5$ Special types of ring and ring elements
Category:Rings with Unity
Group of Units of Field

== Theorem ==
Let $k^\times$ denote the group of units of a field $\struct {k, +, \times}$.
Then:
:$k^\times = k \setminus \set 0$
where:
:$\setminus$ denotes set difference
:$\set 0$ denotes the set containing only the zero of $k$.
That is, $0$ is the only element of $k$ which does not have a multiplicative inverse in $k$.
== Proof ==
$0$ is not invertible in $k$ since $0 a = 0$ for all $a \in k$.
Thus $0 \notin k^\times$.
Consider now $a \in k$ such that $a \ne 0$.
From it follows that there exists $a^{-1}$ such that:
:$a \times a^{-1} = 1 = a^{-1} \times a$
where $1$ is the unity of $k$.
So:
:$a \in k^\times$
Hence we have demonstrated that $k^\times$ equals $k$ without $0$, that is:
:$k^\times = k \setminus 0$
Category:Field Theory
Group of Units of Submonoid is Subgroup

== Theorem ==
Let $T$ be a monoid.
Let $S$ be a submonoid of $T$.
Let $\map G T$ and $\map G S$ be the groups of units of $T$ and $S$ respectively.
Then $\map G S \subseteq \map G T$ and $\map G S$ is a subgroup of $\map G T$.
== Proof ==
Let $x \in \map G S$.
Then $x \in S$ and there exists $y \in S$ such that $x y = y x = e$.
Since $S \subseteq T$, we have $y \in T$. 
So $x \in \map G T$.
So we have $\map G S \subseteq \map G T$.
From Group of Units is Group, $\map G S$ is a group. 
So $\map G S$ is a subgroup of $\map G T$.
Category:Groups of Units
Group whose Derived Subgroup is Trivial is Abelian

== Theorem ==
Let $G$ be a group with identity $e$.
Let $G'$ be the derived subgroup of $G$.
Let $G'$ be the trivial group $\set e$.
Then $G$ is abelian.
== Proof ==
We assume that the derived subgroup of $G$ is $\set e$.
 $G$ is non-abelian.
Hence by definition:
:$\exists g, h \in G: g h \ne h g$
Multiplying both sides by $g^{-1} h^{-1}$:
:$\exists g, h \in G: g^{-1} h^{-1} g h \ne e$
But from the definition of the derived subgroup:
:$G' := \set {g^{-1} h^{-1} g h : g, h \in G}$
Hence:
:$G' \ne \set e$
So by definition, $G'$ is not a trivial group.
By Proof by Contraposition it follows that $G$ is abelian.
== Also see ==
* Derived Subgroup of Abelian Group is Trivial
Category:Derived Subgroups
Category:Abelian Groups
Category:Trivial Group
Group whose Order equals Order of Element is Cyclic

== Theorem ==
Let $G$ be a finite group of order $n$.
Let $g \in G$ have order $n$.
Then $G$ is a cyclic group which is generated by $g$.
== Proof ==
The subgroup of $G$ generated by $g$ is:
:$\gen g = \set {g^0, g^1, g^2, \ldots, g^{n - 1} }$
This contains $n$ elements.
Thus:
:$\gen g = G$
and the result follows by definition of cyclic group.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 9$: Cyclic Groups
Category:Order of Group Elements
Category:Order of Groups
Category:Cyclic Groups
Group with Normal Series with Solvable Factor Groups is Solvable

== Theorem ==
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.
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 75 \beta$
Category:Subgroups
Category:Solvable Groups
Group with One Sylow Subgroup per Prime Divisor is Solvable

== Theorem ==
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.
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 75 \epsilon$
Category:Solvable Groups
Category:Prime Groups
Group with Order Less than 60 is Solvable

== Theorem ==
Every group whose order is less than $60$ is solvable.
== Proof ==
A direct consequence of Burnside's Theorem: the smallest number with $3$ distinct prime factors is $2 \times 3 \times 5 = 60$.
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 75 \gamma$
Category:Solvable Groups
Group with Zero Element is Trivial

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\struct {G, \circ}$ have a zero element.
Then $\struct {G, \circ}$ is the trivial group.
== Proof ==
Let $e \in G$ be the identity element of $G$.
Let $z \in G$ be a zero element.
Let $x \in G$ be any arbitrary element of $\struct {G, \circ}$.
Then:
So whatever $x \in G$ is, it has to be the identity element of $G$.
So $G$ can contain only that one element, and is therefore the trivial group.
== Sources ==
* : $\S 4.5$. Examples of groups: Example $83$
Category:Group Theory
Category:Trivial Group
Category:Zero Elements
Grouped Data/Examples/Heights of People

== Examples of Grouped Data ==
For $200$ people, their exact height $x$ in centimetres is measured.
The observations are grouped as follows:
$\begin {array} {r|l} \hline
\text {Height $x$ (cm)} & \text {Count} \\
\hline
x < 165 & 6 \\
165 \le x < 170 & 39 \\
170 \le x < 175 & 93 \\
175 \le x < 180 & 44 \\
180 \le x < 185 & 15 \\
x \ge 185 & 3 \\
\hline \end {array}$
== Sources ==
* 
* 
Category:Examples of Grouped Data
Groups of Order 21

== Theorem ==
There exist exactly $2$ groups of order $21$, up to isomorphism:
:$(1): \quad C_{21}$, the cyclic group of order $21$
:$(2): \quad$ the group whose group presentation is:
:::$\gen {x, y: x^7 = e = y^3, y x y^{-1} = x^2}$
== Proof ==
Let $G$ be of order $21$.
From Group of Order $p q$ has Normal Sylow $p$-Subgroup, $G$ has exactly one Sylow $7$-subgroup, which is normal.
Let this Sylow $7$-subgroup of $G$ be denoted $P = \gen {x: x^7 = 1}$.
From the First Sylow Theorem, $G$ also has at least one Sylow $3$-subgroup.
Thus there exists $y \in G$ of order $3$.
As $P$ is normal:
:$y x y^{-1} = x^i$
for some $i \in \set {0, 1, \ldots, 6}$.
Thus:
So $x^1 = x^{i^3}$ and so:
:$i^3 \equiv 1 \pmod 7$
and so:
:$7 \divides \paren {i^3 - 1}$
where $\divides$ indicates divisibility.
Let us consider the $7$ possible values of $i$ in turn.
Thus $i \bmod 7 \in \set {1, 2, 4}$.
Suppose $i \equiv 1 \pmod 7$.
Then:
Hence:
and:
It follows that:
:$\order {x y} = 21$
where $\order {x y}$ denotes the order of $x y$.
Thus $G$ is cyclic.
Suppose that $i \equiv 2 \pmod 7$.
Thus, let $y$ be an element of order $3$ for which $y x y^{-1} = x^2$.
Then $z = y^2$ is an element of order $3$ for which $z x z^{-1} = x^4$.
Thus the group as defined here where $i = 2$ is isomorphic to the group as defined here where $i = 4$.
Thus, apart from $C_{21}$, the other group of order $21$ has the group presentation:
:$\gen {x, y: x^7 = e = y^3, y x y^{-1} = x^2 }$
=== Matrix Representation of Non-Abelian Instance ===
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: $(2)$ Groups of order $21$
Category:Groups of Order 21
Groups of Order 21/Matrix Representation of Non-Abelian Instance

== Theorem ==
Let $G$ be the group of order $21$ whose group presentation is:
:$\gen {x, y: x^7 = e = y^3, y x y^{-1} = x^2}$
Then $G$ can be instantiated by the following pair of matrices over $\Z_7$:
:$X = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \qquad Y = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}$
== Proof ==
We calculate the powers of $X$ and $Y$ in turn:
and so on to:
Thus we have:
:$X^7 = \mathbf I$
where $\mathbf I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is the identity matrix.
Then:
Thus we have:
:$Y^3 = \mathbf I$
and:
:$Y^{-1} = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}$.
Then:
and the result is apparent.
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: $(2)$ Groups of order $21$
Category:Groups of Order 21
Groups of Order 2p

== Theorem ==
Let $p$ be a prime number.
Let $G$ be a group.
Let the order of $G$ be $2 p$.
Then $G$ is either:
:the cyclic group $C_{2 p}$
or:
:the dihedral group $D_p$.
== Proof ==
When $p = 2$, the result follows from Groups of Order 4.
Let $p$ be an odd prime.
From Sylow p-Subgroups of Group of Order 2p, $G$ has exactly $1$ normal subgroup $P$ of order $p$.
$p$ is prime number.
So from Prime Group is Cyclic, $P$ is a cyclic group.
Let $P = \gen x$ for some $x \in G$.
By the First Sylow Theorem there exists at least one subgroup of $G$ of order $2$.
Hence:
:$\exists y \in G: y^2 = e$
It follows that the elements of $G$ are known:
:$G = \set {e, x, \ldots, x^{p - 1}, y, y x, y x^{p - 1} }$
Then:
Thus:
:the even powers of $y x$ are powers of $x$
:the odd powers of $y x$ are of the form $y x^j$ for some $j \in \Z_{>0}$.
By Order of Element Divides Order of Finite Group:
:$\order {y x} \divides 2 p$
where:
:$\order {y x}$ denotes the order of $y x$
:$\divides$ denotes divisibility.
We have that:
:$y x \ne e$
and so by Identity is Only Group Element of Order 1:
:$\order {y x} \ne 1$
Thus:
:$\order {y x} \in \set {2, p, 2 p}$
Suppose $i \ne -1$ in $(1)$ above.
Then:
:$\paren {y x}^2 \ne e$
and so:
:$\order {y x} \ne 2$
Because odd powers of $y x$ are of the form $y x^j$:
:$\paren {y x}^p \ne e$
and so:
:$\order {y x} \ne p$
It follows that:
:$\order {y x} = 2 p$
and from Group whose Order equals Order of Element is Cyclic, $G$ is cyclic.
Thus, when $i \ne -1$:
:$G = \gen {y x}$
and so is cyclic.
Thus by Cyclic Group is Abelian:
:$y x = x y$
When $i = -1$ in $(1)$ above, we have that:
leading to the group presentation of $G$:
:$G = \gen {x, y: x^p = e = y^2, y x = x^{-1} y}$
which is the Group Presentation of Dihedral Group $D_p$.
 
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 59 \beta$
* : Chapter $12$: Applications of Sylow Theory: $(1)$ Groups of order $2p$
Category:Groups of Order 2 p
Category:Finite Cyclic Groups
Category:Dihedral Groups
Groups of Order 30

== Theorem ==
Let $G$ be a group of order $30$.
Then $G$ is one of the following:
:The cyclic group $C_{30}$
:The dihedral group $D_{15}$
:The group direct product $C_5 \times D_3$
:The group direct product $C_3 \times D_5$
== Proof ==
First we introduce a lemma:
=== Lemma ===
From the lemma, it remains to be shown that the group presentations:
:$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$
:$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$
give the groups $C_5 \times D_3$ and $C_3 \times D_5$.
=== Direct Product $C_5 \times D_3$ ===
=== Direct Product $C_3 \times D_5$ ===
== Sources ==
* : Chapter $13$: Direct products: Theorem $13.8$
Category:Groups of Order 30
Groups of Order 30/C 3 x D 5

== Theorem ==
Let $G$ be a group of order $30$.
Let $G$ have the group presentation:
:$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$
Then $G$ is isomorphic to the group direct product of the cyclic group $C_3$ and the dihedral group $D_5$:
:$G \cong C_3 \times D_5$
== Proof ==
Let $G$ be defined by its group presentation:
:$G = \gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$
Let $z$ denote $x^5$.
Then:
So $z$ commutes with $y$.
As $z$ is a power of $x$, $z$ also commutes with $x$.
Hence by definition of center:
:$z \in \map Z G$
It follows that $\gen z$ is a normal subgroup of order $3$.
Let $N$ be the subgroup of $G$ generated by $x^3$ and $y$.
Note that:
Hence the generator of $N$ satisfies:
:$\paren{x^3}^5 = e = y^2$
and:
:$y x^3 y^{-1} = x^{-3}$
Let $w := x^3$.
Then $N$ is generated by $w$ and $y$ where:
:$w^5 = 1 = y^2$
and:
:$w y = y w^4 = y w^{-1}$
and it is seen that $N$ is isomorphic to the dihedral group $D_5$.
It is now to be shown that $G$ is an internal group direct product of $N$ and $\gen z$.
We have that $N \cap \gen z = \set e$.
From the Internal Direct Product Theorem, we need to prove only that $N$ is a normal subgroup of $G$.
We have that:
:$\order N = 10$
where $\order N$ denotes the order of $N$.
We also have that $N$ is a subgroup of its normalizer $\map {N_G} N$.
Hence by Lagrange's Theorem:
:$10 \divides \order {\map {N_G} N}$
where $\divides$ denotes divisibility.
Again by Lagrange's Theorem:
:$\order {\map {N_G} N} \divides 30$
We have:
demonstrating that $x$ is conjugate to $w$.
Then:
demonstrating that $x$ is conjugate to $y$.
Thus $x \in \map {N_G} N$ and so $\order {\map {N_G} N} > 10$.
As $10 \divides \order {\map {N_G} N}$ and $\order {\map {N_G} N} \divides 30$, it follows that:
:$\map {N_G} N = G$
and so $N$ is normal in $G$.
Thus:
: $N$ and $\gen z$ are normal in $G$
: $N \cap \gen z = \set e$
: $N \gen z = G$
and it therefore follows from the Internal Direct Product Theorem that:
:$G = C_3 \times D_5$
== Sources ==
* : Chapter $13$: Direct products: Exercise $2$
Category:Groups of Order 30
Groups of Order 30/C 3 x D 5/Mistake

== Source Work ==
:
:Chapter $13$: Direct products:
::Exercise $2$ (solution)
== Mistake ==
:Now let $N$ be the subgroup of $G$ generated by $x^3$ and $y$. Note that since
::$y x^3 y^{-1} = \paren {y x y^{-1} }^3 = \paren {x^4}^3 = x^{12} = x^{-3}$,
:$N$ is isomorphic to the dihedral group $\map D 3$. ...
:... To show that $x$ normalises $N$, note that
::$x x^3 x^{-1} = x^3$; and $x y x^{-1} = y x^4 x^{-1} = x y^3 \in N$.
== Correction ==
There are two mistakes here:
:$(1): \quad$ In the first section, it should say:
::::$N$ is isomorphic to the dihedral group $\map D 5$.
:$(2): \quad$ That second expression should read:
::::$x y x^{-1} = y x^4 x^{-1} = y x^3 \in N$
== Sources ==
* : Chapter $13$: Direct products: Exercise $2$
Category:Mistakes/A Course in Group Theory
Groups of Order 30/C 5 x D 3

== Theorem ==
Let $G$ be a group of order $30$.
Let $G$ have the group presentation:
:$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$
Then $G$ is isomorphic to the group direct product of the cyclic group $C_5$ and the dihedral group $D_3$:
:$G \cong C_5 \times D_3$
== Proof ==
Let $G$ be defined by its group presentation:
:$G = \gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$
Let $z$ denote $x^3$.
Then:
So $z$ commutes with $y$.
As $z$ is a power of $x$, $z$ also commutes with $x$.
Hence by definition of center:
:$z \in \map Z G$
It follows that $\gen z$ is a normal subgroup of order $5$.
Let $K$ be the subgroup of $G$ generated by $x^5$ and $y$.
Note that:
Hence the generator of $K$ satisfies:
:$\paren{x^5}^3 = e = y^2$
and:
:$y x^5 y^{-1} = x^{-5}$
Let $w := x^5$.
Then $K$ is generated by $w$ and $y$ where:
:$w^3 = 1 = y^2$
and:
:$w y = y w^2 = y w^{-1}$
and it is seen that $K$ is isomorphic to the dihedral group $D_3$.
It is now to be shown that $G$ is an internal group direct product of $K$ and $\gen z$.
We have that $K \cap \gen z = \set e$.
From the Internal Direct Product Theorem, we need to prove only that $K$ is a normal subgroup of $G$.
We have that:
:$\order K = 6$
where $\order K$ denotes the order of $K$.
We also have that $K$ is a subgroup of its normalizer $\map {N_G} K$.
Hence by Lagrange's Theorem:
:$6 \divides \order {\map {N_G} K}$
where $\divides$ denotes divisibility.
Again by Lagrange's Theorem:
:$\order {\map {N_G} K} \divides 30$
We have:
demonstrating that $x$ is conjugate to $w$.
Then:
demonstrating that $x$ is conjugate to $y$.
Thus $x \in \map {N_G} K$ and so $\order {\map {N_G} K} > 6$.
As $6 \divides \order {\map {N_G} K}$ and $\order {\map {N_G} K} \divides 30$, it follows that:
:$\map {N_G} K = G$
and so $K$ is normal in $G$.
Thus:
:$K$ and $\gen z$ are normal in $G$
:$K \cap \gen z = \set e$
:$K \gen z = G$
and it therefore follows from the Internal Direct Product Theorem that:
:$G = C_5 \times D_3$
== Sources ==
* : Chapter $13$: Direct products: Theorem $13.8$
Category:Groups of Order 30
Groups of Order 30/Lemma

== Theorem ==
Let $G$ be a group of order $30$.
Then $G$ is one of the following:
:The cyclic group $C_{30}$
:The dihedral group $D_{15}$
:Isomorphic to one of:
::$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$
::$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$
== Proof ==
By Group of Order 30 has Normal Cyclic Subgroup of Order 15, $G$ has a normal subgroup of order $15$ which is cyclic.
Let this normal cyclic order $15$ subgroup be denoted $N$:
:$N = \gen x$
Let $y$ be the generator for any Sylow $2$-subgroup of $G$.
Then:
Then:
and so:
:$i^2 - 1 \equiv 0 \pmod {15}$
Investigating the powers of $i$, case by case, searching for those which satisfy this congruence, yields:
:$i \in \set {1, 4, 11, 14}$
The case $i \equiv 1 \pmod {15}$ leads to the cyclic group $C_{30}$.
The case where $i \equiv {14} \equiv {-1} \pmod {15}$ leads to the dihedral group $D_{15}$.
The other two cases lead to:
:$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$
:$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: $(5)$ Groups of order $30$: Proposition $12.6$
Category:Groups of Order 30
Groups of Order 30/Mistake

== Source Work ==
:
:Chapter $13$: Direct products:
::Theorem $13.8$
== Mistake ==
:To show that $x$ normalises $K$, note that
::$x x^5 x^{-1} = x$, and $x y x^{-1} = x y^{-1} x^{-1} = y x^{11} x^{-1} = y x^{10} \in N$.''
== Correction ==
There are two mistakes here:
:$(1): \quad x x^5 x^{-1} = x$ makes no sense in this context. Here we have $x^5 = w$, which we are trying to show is the conjugate of $x$.
:::So it is apparent that what is meant is more likely something along the lines of:
::::$x w x^{-1} = x x^5 x^{-1} = x^5 = w$
:$(2): \quad$ That last $N$ looks as though it ought to be $K$, as there is no $N$ defined in this section.
== Sources ==
* : Chapter $13$: Direct products: Theorem $13.8$
Category:Mistakes/A Course in Group Theory
Groups of Order 4

== Theorem ==
There exist exactly $2$ groups of order $4$, up to isomorphism:
:$C_4$, the cyclic group of order $4$
:$K_4$, the Klein $4$-group.
== Proof ==
From Existence of Cyclic Group of Order n we have that one such group of order $4$ is the cyclic group of order $4$:
This is exemplified by the additive group of integers modulo $4$, whose Cayley table can be presented as:
From Group whose Order equals Order of Element is Cyclic, any group with an element of order $4$ is cyclic.
From Cyclic Groups of Same Order are Isomorphic, no other groups of order $4$ which are not isomorphic to $C_4$ can have an element of order $4$.
From Order of Element Divides Order of Finite Group, any other group of order $4$ must have elements of order $2$.
We have the Klein $4$-group, whose Cayley table can be presented as:
and is seen to have that property.
We have that Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic.
It remains to be shown that the Klein $4$-group is the only groups of order $4$ whose elements are all of order $2$ (except the identity).
Let the Cayley table be populated as far as can be directly established:
:$\begin{array}{c|cccc}
 & e & a & b & c \\
\hline
e & e & a & b & c \\
a & a & e & & \\
b & b & & e & \\
c & c & & & e \\
\end{array}$
Consider $a b$.
As $a^2 = e$, $a b \ne e$.
As $a e = a$, $a b \ne a$.
As $e b = b$, $a b \ne b$.
It follows that $a b = c$.
Hence we have:
:$\begin{array}{c|cccc}
 & e & a & b & c \\
\hline
e & e & a & b & c \\
a & a & e & c & \\
b & b & & e & \\
c & c & & & e \\
\end{array}$
and the rest of the table is completed by following the result that Group has Latin Square Property.
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $4$
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem
* : Chapter $2$: Subgroups and Cosets: $\S 41 \zeta$
* : Chapter $3$: Elementary consequences of the definitions: Groups with four elements
Category:Groups of Order 4