nyuuzyou/wikis huggingface dataset:
Non-Abelian Group of Order p Cubed has Exactly One Normal Subgroup of Order p

== Theorem ==
Let $p$ be a prime number.
Let $G$ be a non-abelian group of order $p^3$.
Then $G$ contains exactly one normal subgroup of order $p$.
== Proof ==
From Center of Group of Prime Power Order is Non-Trivial, $\map Z G$ is not the trivial subgroup.
From Quotient of Group by Center Cyclic implies Abelian, $G / \map G Z$ cannot be cyclic and non-trivial.
Thus $\order {G / \map G Z}$ cannot be $p$ and so must be $p^2$.
Thus $\order {\map G Z} = p$.
Let $N$ be a normal subgroup of $G$ of order $p$.
Then from Normal Subgroup of p-Group of Order p is Subset of Center:
:$N \subseteq \map G Z$
Thus there is no normal subgroup of order $p$ different from $\map G Z$.
Hence the result.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $21$
Category:P-Groups
Non-Abelian Order 10 Group has Order 5 Element

== Theorem ==
Let $G$ be a non-abelian group of order $10$.
Then $G$ has at least one element of order $5$.
== Proof 1 ==
== Proof 2 ==
Category:Order of Group Elements
Category:Groups of Order 10
Category:Non-Abelian Order 10 Group has Order 5 Element
Non-Abelian Order 10 Group has Order 5 Element/Proof 1

== Theorem ==
== Proof ==
By Lagrange's Theorem, all the elements of $G$ have orders $1$, $2$, $5$ or $10$.
From Identity is Only Group Element of Order 1, $9$ elements of $G$ have orders greater than $1$.
From Cyclic Group is Abelian, $G$ is not cyclic.
If $g \in G$ was of order $10$ then $g$ would generate the cyclic group $C_{10}$.
Thus $G$ has no element of order $10$.
So all elements of $G$ except the identity have orders $2$ or $5$.
 $G$ has no element of order $5$.
Then all elements are of order $2$.
Then $G$ is a Boolean group.
Then by Boolean Group is Abelian, $G$ is abelian.
This contradicts the assertion that $G$ is non-abelian.
Hence the result by Proof by Contradiction.
== Sources ==
* : $\S 44$. Some consequences of Lagrange's Theorem: Worked Examples: $1$
Category:Non-Abelian Order 10 Group has Order 5 Element
Non-Abelian Order 10 Group has Order 5 Element/Proof 2

== Theorem ==
== Proof ==
As $10 = 2 \times 5$, $G$ is a non-abelian group of order $2 p$, where $p$ is an odd prime.
Hence this is an instance of the result Non-Abelian Order 2p Group has Order p Element.
Category:Non-Abelian Order 10 Group has Order 5 Element
Non-Abelian Order 2p Group has Order p Element

== Theorem ==
Let $p$ be an odd prime.
Let $G$ be a non-abelian group of order $2 p$.
Then $G$ has at least one element of order $p$.
== Proof ==
By Lagrange's Theorem, all the elements of $G$ have orders $1$, $2$, $p$ or $2p$.
From Identity is Only Group Element of Order 1, $2 p - 1$ elements of $G$ have orders greater than $1$.
From Cyclic Group is Abelian, $G$ is not the cyclic group $2 p$.
If $g \in G$ was of order $2 p$ then $g$ would generate the cyclic group $C_{2 p}$.
Thus $G$ has no element of order $2 p$.
So all elements of $G$ except the identity have orders $2$ or $p$.
 $G$ has no element of order $p$.
Then all elements are of order $2$.
Then $G$ is a Boolean group.
Then by Boolean Group is Abelian, $G$ is abelian.
This contradicts the assertion that $G$ is non-abelian.
Hence the result by Proof by Contradiction.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $23$
Category:Order of Group Elements
Category:Groups of Order 2 p
Non-Abelian Order 8 Group has Order 4 Element

== Theorem ==
Let $G$ be a non-abelian group of order $8$.
Then $G$ has at least one element of order $4$.
== Proof ==
Let $e \in G$ be the identity of $G$.
Let $g \in G$ be an arbitrary element of $G$ such that $g \ne e$.
From Identity is Only Group Element of Order 1, only $e$ has order $1$.
Thus from Order of Element Divides Order of Finite Group:
:$\order g \in \set {2, 4, 8}$
Suppose $\order g = 8$.
Then $G$ is cyclic.
So by Cyclic Group is Abelian, $G$ would be abelian.
So $g$ cannot be of order $8$.
Suppose all elements of $G \setminus \set e$ are of order $2$.
Then by definition $G$ is a Boolean group.
By Boolean Group is Abelian, $G$ would be abelian. 
So the only option left is for at least one element of $G$ to be of order $4$.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $18$
Category:Groups of Order 8
Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group

== Theorem ==
Let $G$ be a group with the following properties:
:$(1): \quad G$ is non-abelian.
:$(2): \quad G$ is of order $8$.
:$(3): \quad G$ has precisely one element of order $2$.
Then $G$ is isomorphic to the quaternion group $Q$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \kappa$
Category:Quaternion Group
Category:Groups of Order 8
Category:Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group
Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 1

== Lemma ==
:$\set {\pm 1}$ is a normal subgroup of $G$.
== Proof ==
So:
:$g \circ \paren {-1} \circ g^{-1} = 1$
or:
:$g \circ \paren {-1} \circ g^{-1} = -1$
Suppose $g \circ \paren {-1} \circ g^{-1} = 1$.
Then:
which is a contradiction.
So it is the other case:
:$g \circ \paren {-1} \circ g^{-1} = -1$
Therefore:
:$\forall g \in G: g \circ \set {\pm 1} \circ g^{-1} = \set {\pm 1}$
By definition, $\set {\pm 1}$ is a normal subgroup.
Category:Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group
Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 2

== Lemma ==
:$\paren {\pm a}^2 = \paren {\pm b}^2 = \paren {\pm c}^2 = -1$
== Proof ==
, $a$ is checked.
The proofs for other $5$ elements are similar.
So $a^2 = 1$ or $a^2 = -1$.
As the order of $a = 4$:
:$a^2 \ne 1$
Hence:
:$a^2 = -1$
as required.
Category:Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group
Non-Abelian Simple Finite Groups are Infinitely Many

== Theorem ==
There exist infinitely many types of group which are non-abelian and finite.
== Proof ==
We have that Alternating Group is Simple except on 4 Letters.
So for all $n \in \N$ such that $n \ne 4$, the alternating group $A_n$ is a simple group.
We also have that $A_n$ is non-abelian for all $n > 3$.
Hence the result.
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.11$
Category:Simple Groups
Non-Archimedean Division Ring iff Non-Archimedean Completion

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\struct {R', \norm {\, \cdot \,}' }$ be a normed division ring completion of $\struct {R, \norm {\, \cdot \,} }$
Then:
:$\norm {\, \cdot \,}$ is non-archimedean $\norm {\, \cdot \,}'$ is non-archimedean.
== Proof ==
By the definition of a normed division ring completion then:
:$(1): \quad$ there exists a distance-preserving ring monomorphism $\phi: R \to R'$.
:$(2): \quad \struct {R', \norm {\, \cdot \,}' }$ is a complete metric space.
:$(3): \quad \phi \sqbrk R$ is a dense subspace in $\struct {R', \norm {\, \cdot \,}' }$.
By Normed Division Ring is Dense Subring of Completion, $\phi \sqbrk R$ is a dense normed division subring of $R'$ and $\phi: R \to \map \phi R$ is an isometric isomorphism.
By Isometrically Isomorphic Non-Archimedean Division Rings, $\struct {R, \norm {\, \cdot \,} }$ is non-archimedean $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is.
So it remains to show that $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ non-archimedean $\struct {R', \norm {\, \cdot \,}' }$ is.
=== Necessary Condition ===
Let $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ be non-archimedean.
Let $x', y' \in R'$.
By the definition of a dense subset:
:$\map \cl {\phi \sqbrk R} = R'$
By Closure of Subset of Metric Space by Convergent Sequence then:
:there exists a sequence $\sequence {x_n'} \subseteq \phi \sqbrk R$ that converges to $x'$, that is, $\ds \lim_{n \mathop \to \infty} x_n' = x'$
:there exists a sequence $\sequence {y_n'} \subseteq \phi \sqbrk R$ that converges to $y'$, that is, $\ds \lim_{n \mathop \to \infty} y_n' = y'$
By Sum Rule for Sequences in Normed Division Ring then:
:$\ds \lim_{n \mathop \to \infty} x_n' + y_n' = x' + y'$
By modulus of limit then:
:$\ds \lim_{n \mathop \to \infty} \norm {x_n'}' = \norm {x'}'$
:$\ds \lim_{n \mathop \to \infty} \norm {y_n'}' = \norm {y'}'$
:$\ds \lim_{n \mathop \to \infty} \norm {x_n' + y_n'}' = \norm {x' + y'}'$
By Maximum Rule for Real Sequences then:
:$\ds \lim_{n \mathop \to \infty} \max \set {\norm {x_n'}', \norm {y_n'}' } = \max \set {\norm {x'}', \norm {y'}'}$
Since $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is non-archimedean then:
:$\forall n: \norm {x_n' + y_n'}' \le \max \set {\norm {x_n'}', \norm {y_n'}' }$
By Inequality Rule for Real Sequences then:
:$\ds \norm {x' + y'}' = \lim_{n \mathop \to \infty} \norm {x_n' + y_n'}' \le \lim_{n \mathop \to \infty} \max \set {\norm {x_n'}', \norm {y_n'}' } = \max \set {\norm {x'}', \norm {y'}'}$
The result follows.
=== Sufficient Condition ===
Let $\struct {R', \norm {\, \cdot \,}' }$ be non-archimedean.
By Subring of Non-Archimedean Division Ring then $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is non-archimedean.
== Sources ==
* : $\S 3.2$: Completions
Category:Normed Division Rings
Category:Non-Archimedean Norms
Non-Archimedean Division Ring is Totally Disconnected

== Theorem ==
Let $\struct {R, \norm{\,\cdot\,} }$ be a non-Archimedean normed division ring.
Let $\tau$ be the topology induced by the norm $\norm{\,\cdot\,}$.
Then the topological space $\struct {R, \tau}$ is totally disconnected.
== Proof ==
Let $S$ be a subset of $R$ such that:
:$\exists x, y \in S: x \ne y$
Let $r \in \R_{>0} : r = \norm {x - y}$
Consider the open ball $\map {B_r} x$ such that:
:$x \in \map {B_r} x$
:$y \notin \map {B_r} x$
By Open Balls are Clopen then $\map {B_r} x$ is both open and closed.
By Complement of Clopen Set is Clopen then $R \setminus \map {B_r} x$ is open.
Let:
:$A = S \cap \map {B_r} x$
and
:$B = S \cap \paren{R \setminus \map {B_r} x}$
We have:
:$x \in A \subseteq \map {B_r} x$
and
:$y \in B \subseteq \paren{R \setminus \map {B_r} x}$
By definition of separated sets:
:$A$ and $B$ are separated sets
As $S = A \cup B$ then:
:$S$ is the union of two non-empty separated sets of $\struct{R,\tau}$
By definition of connected subset:
:$S$ is not a connected subset.
Since $S$ was an arbitrary subset of $\struct{R,\tau}$ with two distinct elements:
:no set $S$ with two distinct elements is connected.
From Singleton is Connected in Topological Space:
:the only connected subsets of $\struct{R,\tau}$ are the singletons
By definition of totally disconnected:
:$\struct{R,\tau}$ is totally disconnected
== Sources ==
* : $\S 2.3$ Topology: Proposition $2.3.8$
* : $\S 2.1$ Elementary topological properties: Proposition $2.14$
Category:Normed Division Rings
Category:Non-Archimedean Norms
Category:Totally Disconnected Spaces
Non-Archimedean Norm iff Non-Archimedean Metric

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with zero $0$.
Let $d$ be the metric induced by $\norm {\,\cdot\,}$.
Then:
:$\norm {\, \cdot \,}$ is a non-Archimedean norm $d$ is a non-Archimedean metric.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
== Sources ==
* : $\S 2.3$: Topology: Lemma $2.3.2$
Category:Normed Division Rings
Category:Metric Spaces
Category:Non-Archimedean Norms
Non-Archimedean Norm iff Non-Archimedean Metric/Necessary Condition

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,}}$ be a non-Archimedean normed division ring.
Let $d$ be the metric induced by $\norm {\,\cdot\,}$.
Then $d$ is a non-Archimedean metric.
== Proof ==
By Metric Induced by Norm on Normed Division Ring is Metric then $d$ satisfies the metric space axioms $(\text M 1)$ to $(\text M 4)$.
To complete the proof, all that remains is to show that $d$ is non-Archimedean.
Let $x, y, z \in R$.
== Sources ==
* : $\S 2.3$: Topology
Category:Normed Division Rings
Category:Metric Spaces
Category:Non-Archimedean Norms
Non-Archimedean Norm iff Non-Archimedean Metric/Sufficient Condition

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with zero $0$.
Let $d$ be the metric induced by $\norm {\, \cdot \,}$.
Let $d$ be non-Archimedean.
Then:
:$\norm {\, \cdot \,}$ is a non-Archimedean norm.
== Proof ==
Let $x, y \in R$.
Hence $\norm {\,\cdot\,}$ is non-Archimedean.
== Sources ==
* : $\S 2.3$: Topology
Category:Normed Division Rings
Category:Metric Spaces
Category:Non-Archimedean Norms
Non-Associative Division Algebra with Real Scalars has Dimension of Power of 2

== Theorem ==
Let $A$ be a non-associative division algebra with real scalars.
Then the dimension of $A$ is a power of $2$.
== Proof ==
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
Category:Non-Associative Algebras
Category:Division Algebras
Non-Associative Division Algebra with Real Scalars has Dimension of Power of 2/Historical Note

== Historical Note on Non-Associative Division Algebra with Real Scalars has Dimension of Power of 2 ==
Non-Associative Division Algebra with Real Scalars has Dimension of Power of 2 was demonstrated by in $1940$.
== Sources ==
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
Category:Historical Notes
Non-Cancellable Elements of Semigroup form Subsemigroup

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $T \subseteq S$ be the subset of $S$ containing the elements of $S$ which are specifically not cancellable in $\struct {S, \circ}$.
Then $\struct {T, \circ}$ forms a subsemigroup of $S$.
== Proof ==
Recall the definition of cancellable element:
An element $x \in \struct {S, \circ}$ is cancellable :
:$\forall a, b \in S: x \circ a = x \circ b \implies a = b$
:$\forall a, b \in S: a \circ x = b \circ x \implies a = b$
From the Subsemigroup Closure Test it is sufficient to demonstrate that:
:$\forall x, y \in T: x \circ y \in T$
Let $x, y \in T$.
Then:
:$\exists a, b \in S: a \circ y = b \circ y \text { but } a \ne b$
or:
:$\exists a, b \in S: y \circ a = y \circ b \text { but } a \ne b$
, suppose that:
:$\exists a, b \in S: y \circ a = y \circ b \text { but } a \ne b$
Hence:
So we have:
:$\paren {x \circ y} \circ a = \paren {x \circ y} \circ b$
but $a \ne b$.
Hence $x \circ y$ is not cancellable.
As $x$ and $y$ are arbitrary, the result follows.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.11$
Category:Cancellability
Category:Subsemigroups
Non-Closed Set of Real Numbers is not Compact

== Theorem ==
Let $\R$ be the set of real numbers considered as an Euclidean space.
Let $S \subseteq \R$ be non-closed in $\R$.
Then $S$ is not a compact subspace of $\R$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
Category:Compact Spaces
Category:Analysis
Category:Non-Closed Set of Real Numbers is not Compact
Non-Closed Set of Real Numbers is not Compact/Proof 1

== Theorem ==
== Proof ==
Consider the complement of $S$ in $\R$:
:$S' = \relcomp \R S = \R \setminus S$
As $S$ is not closed, by definition $S'$ is not open.
Thus by definition there exists $x \in S'$ such that:
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \notin S'$
where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball of $x$.
That is:
:$\forall \epsilon \in \R_{>0}: \exists y \in S: y \in \map {B_\epsilon} x$
That is, for all positive $\epsilon$ there exists a $y$ within $\epsilon$ of $x$.
From Union of Open Sets of Metric Space is Open:
:$O_\epsilon := \openint \gets {x - \epsilon} \cup \openint {x + \epsilon} \to$
is an open set of $\R$.
Let $\CC = \set {O_\epsilon: \epsilon \in \R_{>0} }$.
Then:
:$\ds \bigcup \CC = \R \setminus \set x \supseteq S$
as $x \notin S$.
So $\CC$ is an open cover of $S$.
Let $\FF \subseteq \CC$ be a finite subset of $\CC$.
Then:
:$\ds \bigcup \FF = \openint \gets {x - \epsilon'} \cup \openint {x + \epsilon'} \to$
for some $\epsilon' \in \R_{>0}$.
But, as has been seen, there exists $y \in S$ closer to $x$ than $\epsilon'$.
Thus $\ds \bigcup \FF$ does not cover $S$.
This construction demonstrates an open cover $\CC$ of $S$ which has no finite subcover.
So, by definition, $S$ is not a compact subspace of $\R$.
Category:Non-Closed Set of Real Numbers is not Compact
Non-Closed Set of Real Numbers is not Compact/Proof 3

== Theorem ==
== Proof ==
By the rule of transposition, it suffices to show that if $S$ is a compact subspace of $\R$, then $S$ is closed.
Consider the relative complement of $S$ in $\R$:
:$T = \relcomp \R S = \R \setminus S$
It remains to be shown that $T$ is open in $\R$.
Let $x \in T$.
For all strictly positive real numbers $\epsilon \in \R_{>0}$, it follows from Union of Open Sets of Metric Space is Open that:
:$O_\epsilon := \openint \gets {x - \epsilon} \cup \openint {x + \epsilon} \to$
is open in $\R$.
Let $\CC = \set {O_\epsilon: \epsilon \in \R_{>0} }$.
Then:
:$\ds \bigcup \CC = \R \setminus \set x \supseteq S$
as $x \notin S$.
So $\CC$ is an open cover for $S$.
Let $\GF$ be a finite subcover of $\CC$ for $S$.
Then:
:$\ds \bigcup \GF = \openint \gets {x - \delta} \cup \openint {x + \delta} \to \supseteq S$
for some strictly positive real number $\delta \in \R_{>0}$.
By the rule of transposition, it follows that:
:$\openint {x - \delta} {x + \delta} \subseteq T$
Hence, $T$ is open in $\R$.
Category:Non-Closed Set of Real Numbers is not Compact
Non-Commutative Finite-Dimensional Associative Division Algebra over Real Numbers is Set of Quaternions

== Theorem ==
Up to isomorphism, the quaternions form the only non-commutative, finite-dimensional associative division algebra over the set of real numbers $\R$.
== Proof ==
== Also see ==
* Frobenius's Theorem
== Sources ==
* 
* 
Category:Quaternions
Category:Associative Algebras
Category:Division Algebras
Category:Finite-Dimensional Algebras
Category:Commutative Algebras
Non-Commutative Ring with Unity and 2 Ideals not necessarily Division Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_F$ and whose unity is $1_F$.
Let $\struct {R, +, \circ}$ specifically not be commutative.
Let $\struct {R, +, \circ}$ be such that the only ideals of $\struct {R, +, \circ}$ are $\set {0_R}$ and $R$ itself.
Then it is not necessarily the case that $\struct {R, +, \circ}$ is a division ring.
== Proof ==
Let $S$ be the set of square matrices of order $2$ over the real numbers $\R$.
$S$ is not a division ring, as for example:
:$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$
and so both $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ are proper zero divisors of $S$.
Let $J$ be an ideal of $S$ containing a non-zero matrix $A$ whose $\tuple {r, s}$th entry is $\lambda \ne 0$.
Let $E_{i j}$ be the matrix of order $2$ defined as:
:$e_{a b} = \begin{cases} 1 & : a = i, b = j \\ 0 & : \text {otherwise} \end{cases}$
Thus for example:
:$E_{1 2} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$
Then for each $i, j \in \set {1, 2}$, consider the matrix $\mathbf M = \dfrac 1 \lambda E_{i r} A E_{s j}$.
In Matrix Multiplication is Associative it is shown that, if $\mathbf A, \mathbf B, \mathbf C \in S$ and $\mathbf S = \mathbf A \mathbf B \mathbf C$:
:$\ds s_{m n} = \sum_{k \mathop = 1}^2 \sum_{l \mathop = 1}^2 a_{m l} b_{l k} c_{k n}$
Notice that each $E_{i j}$ only has one non-zero entry $e_{i j} = 1$.
From the equation above we see that only $m_{i j}$ is can be non-zero.
Moreover the summand is non-zero only when $l = r$ and $k = s$.
Hence:
:$m_{i j} = \dfrac 1 \lambda \times 1 \times a_{r s} \times 1 = 1$
and thus $\mathbf M = E_{i j}$.
It follows that:
:$\forall i, j \in \set {1, 2}: E_{i j} \in J$
It remains to be shown that $J$ is the whole of $S$. 
Since $J$ is an ideal, we have:
:$\forall r \in \R, i, j \in \set {1, 2}: r E_{i j} \in J$
:$\struct {J, +}$ is a subgroup of $\struct {S, +}$, in particular, $\struct {J, +}$ is closed under $+$.
Since every element of $S$ can be written as:
:$\begin{pmatrix} a & b \\ c & d \end{pmatrix} = a E_{1 1} + b E_{1 2} + c E_{2 1} + d E_{2 2}$
for some $a, b, c, d \in \R$, by the above every element of $S$ is an element of $J$.
Therefore we have $S = J$, and thus the only ideals of $S$ are $\set {\mathbf 0}$ and $S$ itself.
== Sources ==
* : Chapter $9$: Rings: Exercise $6 \ \text {(ii)}$
Category:Division Rings
Category:Rings with Unity
Category:Ideal Theory
Non-Cyclic Group of Order 55 has Order 5 Element and Order 11 Element

== Example of Order of Group Element ==
Let $G$ be a non-cyclic group whose order is $55$.
Then $G$ has:
:at least $1$ element of order $5$
:at least $1$ element of order $11$.
== Proof ==
By Order of Element Divides Order of Finite Group, all elements of $G$ have order in $\set {1, 5, 11, 55}$.
But as $G$ is non-cyclic, it can have no element of order $55$.
By Identity is Only Group Element of Order 1, $G$ has $54$ elements of order $5$ and $11$.
Let $m$ denote the number of subgroups of $G$ of order $5$.
Let $n$ denote the number of subgroups of $G$ of order $11$.
From Number of Order p Elements in Group with m Order p Subgroups, $G$ has:
:$4 m$ elements of order $5$
:$10 n$ elements of order $11$.
Then:
:$4 m + 10 n = 54$
As neither $4$ nor $10$ divide $54$, it follows that neither $n$ nor $m$ are equal to $0$.
Hence the result.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $19 \ \text {(i)}$
Category:Examples of Order of Group Elements
Non-Cyclic Group of Order p^2 has p+3 Subgroups

== Theorem ==
Let $p$ be a prime number.
Let $G$ be a non-cyclic group whose order is $p^2$.
Then $G$ has exactly $p + 3$ subgroups.
== Proof ==
By Order of Element Divides Order of Finite Group, all elements of $G$ have order in $\set {1, p, p^2}$.
But as $G$ is non-cyclic, it can have no element of order $p^2$.
By Identity is Only Group Element of Order 1, $G$ has $p^2 - 1$ elements of order $p$.
Let $m$ denote the number of subgroups of $G$ of order $p$.
From Number of Order p Elements in Group with m Order p Subgroups, $G$ has $m \paren {p - 1}$ elements of order $p$.
Then:
Together with the trivial subgroup and $G$ itself, that makes $p + 3$ subgroups.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $19 \ \text {(ii)}$
Category:Examples of Order of Group Elements