nyuuzyou/wikis huggingface dataset:
Group is Abelian iff Opposite Group is Itself

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\struct {G, *}$ be the opposite group to $\struct {G, \circ}$.
$\struct {G, \circ}$ is an Abelian group :
:$\struct {G, \circ} = \struct {G, *}$
== Proof ==
By definition of opposite group:
:$(1): \quad \forall a, b \in G : a \circ b = b * a$
=== Necessary Condition ===
Let $\struct {G, \circ}$ be Abelian. 
Then:
=== Sufficient Condition ===
Let $\struct {G, \circ} = \struct {G, *}$.
Then:
Thus by definition $\struct {G, \circ}$ is an Abelian group.
Category:Abelian Groups
Category:Opposite Groups
Group is Abelian iff it has Cross Cancellation Property

== Theorem ==
Let $G$ be a group.
Then the following are equivalent:
:$(1): \quad G$ is abelian
:$(2): \quad G$ has the cross cancellation property
== Proof ==
Let us suppress the operation of $G$ for brevity.
=== $(2) \implies (1)$ ===
Suppose that $G$ has the cross cancellation property.
Then, for all $x, y \in G$:
Thus, $G$ is abelian.
=== $(1) \implies (2)$ ===
Conversely, suppose $G$ is abelian.
Let $a, b, c \in G$ be such that $a b = c a$.
Since $G$ is abelian, $c a = a c$.
We conclude that:
:$a b = c a = a c$
Thus, by left cancellation, $b = c$.
Category:Abelian Groups
Group is Abelian iff it has Middle Cancellation Property

== Theorem ==
Let $G$ be a group.
Then the following are equivalent:
:$(1): \quad G$ is abelian
:$(2): \quad G$ satisfies the middle cancellation property
== Proof ==
Let us suppress the operation of $G$ for brevity.
=== $(2) \implies (1)$ ===
Suppose that $G$ satisfies the middle cancellation property.
Then, for all $g, h \in G$:
Thus $G$ is abelian.
=== $(1) \implies (2)$ ===
Conversely, suppose $G$ is abelian.
Then, for all $a, b, c, d, x \in G$:
Thus the middle cancellation property holds in $G$.
Category:Abelian Groups
Group is B-Algebra Iff All Elements Self-Inverse

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity element is $e$.
Then $\struct {G, \circ}$ is also a $B$-algebra :
:$\forall g \in G: g = g^{-1}$
That is, all elements of $G$ are self-inverse.
== Proof ==
Let $\struct {G, \circ}$ be a group such that $\forall g \in G: g = g^{-1}$.
From Group Induces B-Algebra, the algebraic structure $\struct {G, *}$ where $*$ is defined as:
:$\forall a, b \in G: a * b := a \circ b^{-1}$
is a $B$-algebra.
But as $b = b^{-1}$, it follows that:
:$\forall a, b \in G: a * b := a \circ b$
and so $\struct {G, \circ}$ is itself a $B$-algebra.
Now suppose it is not the case that $\forall g \in G: g = g^{-1}$.
Suppose $\struct {G, \circ}$ were a $B$-algebra.
By , the identity element $e$ would be the $0$, as:
:$\forall a \in G: a \circ 0 = a$
As an Identity is Unique, there can be no other element in $G$ which can satisfy the conditions to be the $0$ of a $B$-algebra.
Now as not every element is self-inverse:
:$\exists x \in G: x \circ x \ne e$
Then by it follows that $\struct {G, \circ}$ is not a $B$-algebra.
Hence the result.
Category:Group Theory
Category:B-Algebras
Group is Cancellable Monoid

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Then $\struct {G, \circ}$ is a cancellable monoid.
== Proof ==
By definition, a group is a monoid.
From Group Operation is Cancellable, $\circ$ is a cancellable operation in $G$.
Hence the result by definition of cancellable monoid.
== Sources ==
* : Chapter $1$: Basic Definitions and Results
Category:Cancellable Monoids
Category:Group Theory
Group is Connected iff Subgroup and Quotient are Connected

== Theorem ==
Let $G$ be a topological group.
Let $H \le G$ be a subgroup.
:$(1):\quad$ $G$ is connected
:$(2):\quad$ $H$ is connected and the left quotient space $G / H$ is connected
:$(3):\quad$ $H$ is connected and the right quotient space $G / H$ is connected.
== Proof ==
Category:Topological Groups
Category:Connected Spaces
Category:Quotient Spaces
Group is Finite iff Finite Number of Subgroups

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Then $G$ is finite $\struct {G, \circ}$ has a finite number of subgroups.
== Proof ==
=== Necessary Condition ===
Suppose that $\struct {G, \circ}$ is a finite group.
Let $\struct {H, \circ}$ be a subgroup of $\struct {G, \circ}$.
$H \subseteq G$ by definition.
Therefore:
:$H \in \powerset G$
where $\powerset G$ denotes the power set of $G$.
By Power Set of Finite Set is Finite, $\powerset G$ is finite.
It is seen that the set of all subgroups form a subset of $\powerset G$.
The result then follows from Subset of Finite Set is Finite.
=== Sufficient Condition ===
Suppose that $\struct {G, \circ}$ is a group with only a finite number of subgroups.
It is noted that an Infinite Group has Infinite Number of Subgroups.
The result then follows from the Rule of Transposition.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 41 \eta$
Category:Finite Groups
Group is Generated by Itself

== Theorem ==
Let $G$ be a group.
Then:
:$G = \gen G$
where $\gen G$ denotes the group generated by $G$.
== Proof ==
By definition of generated subgroup, $\gen G$ is the smallest subgroup of $G$ containing $G$.
Hence the result by Group is Subgroup of Itself.
== Sources ==
* : $\S 5.3$. Subgroup generated by a subset: Example $97$
Category:Generators of Groups
Group is Hausdorff iff Discrete Subgroups are Closed

== Theorem ==
A topological group is Hausdorff its discrete subgroups are closed.
== Proof ==
Follows directly from:
: Discrete Subgroup of Hausdorff Group is Closed
: Topological Group is Hausdorff iff Identity is Closed
== Also see ==
* Group is Hausdorff iff has Closed Discrete Subgroup
Category:Topological Groups
Group is Hausdorff iff has Closed Discrete Subgroup

== Theorem ==
A topological group is Hausdorff it has a closed discrete subgroup.
== Proof ==
=== Necessary Condition ===
Follows directly from Topological Group is Hausdorff iff Identity is Closed.
=== Sufficient Condition ===
Let $G$ be a Hausdorff topological group.
Let $H \leq G$ be a closed discrete subgroup.
Let $e$ be the identity of $G$.
By Set in Discrete Topology is Clopen, $\left\{{e}\right\}$ is closed in $H$.
By Closed Set in Topological Subspace: Corollary, $\left\{{e}\right\}$ is closed in $G$.
By Topological Group is Hausdorff iff Identity is Closed, $G$ is Hausdorff.
== Also see ==
* Group is Hausdorff iff Discrete Subgroups are Closed
Category:Topological Groups
Category:Hausdorff Spaces
Group is Normal in Itself

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Then $\struct {G, \circ}$ is a normal subgroup of itself.
== Proof ==
First, by Group is Subgroup of Itself, $\struct {G, \circ}$ is a subgroup of itself.
To show $\struct {G, \circ}$ is normal in $G$:
:$\forall a, g \in G: a \circ g \circ a^{-1} \in G$
as $G$ is closed by definition.
Hence the result.
== Sources ==
* : $\S 6.6$. Normal subgroups: Example $123$
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures
* : $\S 1.10$: Example $35$
* : $1$: Introduction to Finite Group Theory: $1.7$
* : $\S 49.1$ Normal subgroups
* : Chapter $7$: Normal subgroups and quotient groups: Definition $7.3$: Remark
Category:Normal Subgroups
Group is Quasigroup

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Then $\struct {G, \circ}$ is a quasigroup.
== Proof 1==
By the definition of a quasigroup it must be shown that:
:for all $g \in G$, the left and right regular representations of $\lambda_g$ and $\rho_g$ are permutations on $G$.
This follows immediately from Regular Representations in Group are Permutations.
== Proof 2==
Follows immediately from Structure is Group iff Semigroup and Quasigroup.
Category:Group Theory
Category:Quasigroups
Group is Solvable iff Normal Subgroup and Quotient are Solvable

== Theorem ==
Let $G$ be a finite group.
Let $H$ be a normal subgroup of $G$.
Then $G$ is solvable :
:$(1): \quad H$ 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.
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 75$. Solvable Groups
Category:Normal Subgroups
Category:Quotient Groups
Category:Solvable Groups
Group is Subgroup of Itself

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Then:
:$\struct {G, \circ} \le \struct {G, \circ}$
That is, a group is always a subgroup of itself.
== Proof ==
By Set is Subset of Itself, we have that:
:$G \subseteq G$
Thus $\struct {G, \circ}$ is a group which is a subset of $\struct {G, \circ}$, and therefore a subgroup of $\struct {G, \circ}$.
== Sources ==
* : $\S 5.2$. Subgroups: Example $90$
* : $\S 1.9$: Subgroups: Example $25$
* : Chapter $\text{II}$: Groups: Subgroups
* : Chapter $2$: Subgroups and Cosets: $\S 35$
* : $\S 36.1$: Subgroups
* : Chapter $4$: Subgroups: Definition $4.1$
Category:Subgroups
Group of Gaussian Integer Units/Cayley Table

== Cayley Table for Group of Gaussian Integer Units ==
The group of Gaussian integer units:
:$\struct {U_\C, \times}$
can be described completely by showing its Cayley table:
:$\begin{array}{r|rrrr}
\times & 1 & i & -1 & -i \\
\hline
1 & 1 & i & -1 & -i \\
i & i & -1 & -i & 1 \\
-1 & -1 & -i & 1 & i \\
-i & -i & 1 & i & -1 \\
\end{array}$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Example $6.2$
* : $\S 34$. Examples of groups: $(6) \ \text{(i)}$
Category:Examples of Cayley Tables
Group of Order 105 has Normal Cyclic Subgroup of Index 3

== Theorem ==
Let $G$ be a group of order $105$.
Then $G$ has a normal cyclic subgroup $N$ such that:
:$\index G N = 3$
where $\index G N$ denotes the index of $N$ in $G$.
== Proof ==
Let $G$ be a group of order $105$ whose identity is $e$.
From Group of Order 105 has Normal Sylow 5-Subgroup or Normal Sylow 7-Subgroup, $G$ has either:
:exactly one normal Sylow $5$-subgroup
or:
:exactly one normal Sylow $7$-subgroup.
Suppose $G$ has exactly one normal Sylow $5$-subgroup, which we will denote $P$.
Then $G / P$ is the quotient group of $G$ by $P$ and is of order $21$.
From Group of Order $p q$ has Normal Sylow $p$-Subgroup, $G / P$ has exactly one normal Sylow $7$-subgroup which we denote $N / P$.
From the Correspondence Theorem, $N$ is a normal subgroup of $G$ of order $7 \times 5 = 35$.
It follows from Group of Order 35 is Cyclic Group that $N$ is cyclic.
Now suppose that $G$ has exactly one normal Sylow $7$-subgroup, which we will denote $Q$.
Then $G / Q$ is the quotient group of $G$ by $Q$ and is of order $15$.
From Group of Order $p q$ has Normal Sylow $p$-Subgroup, $G / Q$ has exactly one normal Sylow $5$-subgroup which we denote $M / Q$.
From the Correspondence Theorem, $m$ is a normal subgroup of $G$ of order $5 \times 7 = 35$.
It follows from Group of Order 35 is Cyclic Group that $M$ is cyclic.
Both cases have been covered, and the existence of a normal cyclic subgroup of order $35$ has been proved.
The result follows by definition of the index of a subgroup.
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: Exercise $2$
Category:Groups of Order 105
Category:Examples of Normal Subgroups
Group of Order 105 has Normal Sylow 5-Subgroup or Normal Sylow 7-Subgroup

== Theorem ==
Let $G$ be a group of order $105$.
Then $G$ has either:
:exactly one normal Sylow $5$-subgroup
or:
:exactly one normal Sylow $7$-subgroup.
== Proof ==
Let $G$ be a group of order $105$ whose identity is $e$.
We have that:
:$105 = 3 \times 5 \times 7$
From the First Sylow Theorem, $G$ has at least one Sylow $3$-subgroup, Sylow $5$-subgroup and Sylow $7$-subgroup.
Let:
:$n_5$ denote the number of Sylow $5$-subgroups of $G$
:$n_7$ denote the number of Sylow $7$-subgroups of $G$.
$5$ and $7$ appear in $105$ with multiplicity $1$.
Hence any such Sylow $p$-subgroups are prime groups.
From Sylow p-Subgroup is Unique iff Normal:
:if $n_5 = 1$ then the unique Sylow $5$-subgroup is normal
and:
:if $n_7 = 1$ then the unique Sylow $7$-subgroup is normal.
It remains to be shown that either $n_5 = 1$ or $n_7 = 1$.
By the Fourth Sylow Theorem:
:$n_5 \equiv 1 \pmod 5$
and from the Fifth Sylow Theorem:
:$n_5 \divides 105$
where $\divides$ denotes divisibility.
It follows that $n_5 \in \set {1, 21}$.
By the Fourth Sylow Theorem:
:$n_7 \equiv 1 \pmod 7$
and from the Fifth Sylow Theorem:
:$n_7 \divides 105$
where $\divides$ denotes divisibility.
It follows that $n_7 \in {1, 15}$.
If either $n_5 = 1$ or $n_7 = 1$ the proof is finished.
Suppose $n_7 = 15$.
As all of these Sylow $7$-subgroups are prime, the intersection of any $2$ of them is $\set e$.
Thus, these $15$ Sylow $7$-subgroups contribute $6 \times 15 = 90$ distinct elements to $G$.
This leaves $15$ elements still to be accounted for.
Suppose $n_5 = 21$.
By a similar argument, these $21$ Sylow $5$-subgroups contribute $4 \times 21 = 84$ distinct elements to $G$.
None of these elements, apart from $e$, can also be elements of one of the Sylow $7$-subgroups.
But there are only $15$ elements available.
So it is not possible for both $n_7 = 15$ and $n_5 = 21$.
So either $n_7 = 1$ or $n_5 = 1$.
The result follows.
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: Exercise $2$
Category:Groups of Order 105
Category:Sylow p-Subgroups
Category:Examples of Normal Subgroups
Group of Order 15 has Cyclic Subgroups of Order 3 and Order 5

== Theorem ==
Let $G$ be a group whose order is $15$.
Then $G$ has
:a cyclic subgroup of order $3$
and:
:a cyclic subgroup of order $5$.
== Proof ==
Let $G$ be a group of order $15$.
We have that $15 = 3 \times 5$.
Thus from the First Sylow Theorem:
:$G$ has at least one subgroup $H_3$ of order $3$
and:
:$G$ has at least one subgroup $H_5$ of order $5$.
From Prime Group is Cyclic, all such subgroups of order $3$ and order $5$ are cyclic.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Example $11.5$
Category:Groups of Order 15
Group of Order 15 is Cyclic Group

== Theorem ==
Let $G$ be a group whose order is $15$.
Then $G$ is cyclic.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $11$: The Sylow Theorems: Example $11.3$
Category:Groups of Order 15
Category:Finite Cyclic Groups
Category:Group of Order 15 is Cyclic Group
Category:Groups of Order p q
Group of Order 15 is Cyclic Group/Proof 1

== Theorem ==
== Proof ==
We have that $15 = 3 \times 5$.
Thus:
:$15$ is square-free
:$5 \equiv 2 \pmod 3$
:$3 \equiv 3 \pmod 5$
The conditions are fulfilled for Condition for Nu Function to be 1.
Thus $\map \nu {15} = 1$ and so all groups of order $15$ are cyclic.
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.4$
Category:Group of Order 15 is Cyclic Group
Group of Order 15 is Cyclic Group/Proof 2

== Theorem ==
== Proof ==
From Number of Sylow p-Subgroups in Group of Order 15:
:the number of Sylow $3$-subgroups is in the set $\set {1, 4, 7, \ldots}$
:the number of Sylow $5$-subgroups is in the set $\set {1, 6, 11, \ldots}$.
From the Fifth Sylow Theorem
:the number of Sylow $3$-subgroups is a divisor of $15$
:the number of Sylow $5$-subgroups is a divisor of $15$.
Combining the above:
:there is a unique Sylow $3$-subgroup in $G$, which we will denote $P$
:there is a unique Sylow $5$-subgroup in $G$, which we will denote $Q$.
Let $x \in G$ be of order $3$.
Then by Non-Trivial Group has Non-Trivial Cyclic Subgroup, $\gen x$ has $3$ elements.
Thus $\gen x$ is a Sylow $3$-subgroup and so:
:$\gen x = P$
and so $x$ is one of the $2$ non-identity elements of $P$.
Similarly, let $y \in G$ be of order $5$.
Then by a similar argument:
:$\gen y = Q$
and so $y$ is one of the $4$ non-identity elements of $Q$.
So $G$ has:
:$1$ element of order $1$ (that is, the identity)
:$2$ elements of order $3$
:$4$ elements of order $5$
which leaves $8$ elements whose order is still to be determined.
From Order of Element Divides Order of Finite Group, they are all of order $1$, $3$, $5$ or $15$.
As the elements of order $1$, $3$ and $5$ have been accounted for, they must all be of order $15$.
So $G$ has $8$ distinct elements of order $15$.
Hence $G$ must be cyclic.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Example $11.12$
Category:Group of Order 15 is Cyclic Group
Group of Order 15 is Cyclic Group/Proof 3

== Theorem ==
== Proof ==
 $G$ is non-abelian.
Let $n_3$ denote the number of elements of $G$ of order $3$.
From Number of Elements of Order p in Group of Order pq is Multiple of q, $n_3$ is a multiple of $5$.
From Number of Order p Elements in Group with m Order p Subgroups, $n_3$ is a multiple of $2$.
Therefore $n_3$ is a multiple of $10$.
Let $n_5$ denote the number of elements of $G$ of order $5$.
From Number of Elements of Order p in Group of Order pq is Multiple of q, $n_5$ is a multiple of $3$.
From Number of Order p Elements in Group with m Order p Subgroups, $n_5$ is a multiple of $4$.
Therefore $n_5$ is a multiple of $12$.
Together with the identity element which has order $1$, that makes $1 + 12 a + 10 b = 15$ for some positive integers $a$ and $b$.
This is impossible.
Hence by Proof by Contradiction it follows that $G$ must be abelian.
Since $15$ is a product of $2$ distinct primes, by Abelian Group of Semiprime Order is Cyclic, $G$ is cyclic.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $20$
Category:Group of Order 15 is Cyclic Group
Group of Order 27 has Subgroup of Order 3

== Theorem ==
Let $G$ be a group whose identity element is $e$.
Let $G$ be of order $27$.
Then $G$ has at least one subgroup of order $3$.
== Proof ==
Let $x \in G \setminus \set e$.
From Identity is Only Group Element of Order 1:
:$\order x > 1$
where $\order x$ denotes the order of $x$.
From Lagrange's Theorem, $\order x$ is $3$, $9$ or $27$.
Thus one of the following applies:
Hence one of the following holds:
where $\gen x$ denotes the subgroup generated by $x$.
Thus one of these: $\gen x$, $\gen {x^3}$ or $\gen {x^9}$ is a subgroup of $G$ of order $3$.
== Sources ==
* : $\S 44$. Some consequences of Lagrange's Theorem
Category:Order of Group Elements
Category:Groups of Order 27
Group of Order 30 has Normal Cyclic Subgroup of Order 15

== Theorem ==
Let $G$ be of order $30$.
Then $G$ has a normal subgroup of order $15$ which is cyclic.
== Proof ==
By Group of Order 15 is Cyclic Group, any subgroup of $G$ of order $15$ is cyclic.
It remains to be proved that a subgroup of $G$ of order $15$ exists, and that it is normal.
Let $n_3$ denote the number of Sylow $3$-subgroups of $G$.
From the Fourth Sylow Theorem:
:$n_3 \equiv 1 \pmod 3$
and from the Fifth Sylow Theorem:
:$n_3 \divides 10$
where $\divides$ denotes divisibility.
It follows that $n_3 \in \set {1, 10}$.
Let $n_5$ denote the number of Sylow $5$-subgroups of $G$.
From the Fourth Sylow Theorem:
:$n_5 \equiv 1 \pmod 5$
and from the Fifth Sylow Theorem:
:$n_5 \divides 30$
It follows that $n_5 \in \set {1, 6}$.
Suppose $n_3 = 1$.
Let $P$ denote the unique Sylow $3$-subgroup of $G$.
From Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.
Thus we may form the quotient group $G / P$, which is of order $10$.
From Groups of Order 2p, $G / P$ has a unique Sylow $5$-subgroup, which we will denote $N / P$.
By the Correspondence Theorem, $N$ is a normal subgroup of $G$ with $15$ elements.
As noted, $N$ is cyclic.
Suppose $n_3 = 10$.
Let the Sylow $3$-subgroups of $G$ be denoted $P_1, P_2, \ldots, P_{10}$.
Each intersection $P_i \cap P_j$ for $i, j \in \set {1, 2, \ldots, 10}, i \ne j$ is the trivial subgroup of $G$:
:$P_i \cap P_j = \set e$
Thus $G$ contains:
:The identity element $e$
:$20$ elements of order $3$, of which $2$ each are in $P_1, P_2, \ldots, P_{10}$
:$9$ more elements of $G$ which are in subgroups of $G$ different from the Sylow $3$-subgroups, and so have orders different from $3$.
Suppose $n_5 = 6$.
By a similar argument to the above, these $6$ Sylow $5$-subgroups contribute $24$ elements of order $5$ to $G$.
This is not possible because there are only $9$ elements of $G$ so far unaccounted for.
So if $G$ has $10$ Sylow $3$-subgroups, it can have only $1$ Sylow $5$-subgroup, which we will denote $Q$.
From Sylow $p$-Subgroup is Unique iff Normal, $Q$ is normal.
Thus we may form the quotient group $G / Q$, which is of order $16$.
From Groups of Order 2p, $G / Q$ has a unique Sylow $3$-subgroup, which we will denote $N / Q$.
Again by the Correspondence Theorem, $N$ is a normal subgroup of $G$ with $15$ elements.
As noted, $N$ is cyclic.
Both cases have been accounted for:
:that $G$ has $1$ Sylow $3$-subgroup
:that $G$ has $10$ Sylow $3$-subgroups.
Both cases result in $G$ having a normal subgroup of order $15$ which is cyclic.
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: $(5)$ Groups of order $30$: Proposition $12.5$
Category:Groups of Order 30
Group of Order 30 is not Simple

== Theorem ==
Let $G$ be a group of order $30$.
Then $G$ is not simple.
== Proof ==
From Group of Order 30 has Normal Cyclic Subgroup of Order 15, $G$ has a normal subgroup of order $15$.
Hence the result, by definition of simple group.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 59 \epsilon$
Category:Groups of Order 30
Category:Simple Groups
Group of Order 35 is Cyclic Group/Proof 1

== Theorem ==
== Proof ==
We have that $35 = 5 \times 7$.
Then we have that $5$ and $7$ are primes such that $5 < 7$ and $5$ does not divide $7 - 1$.
Thus Group of Order $p q$ is Cyclic can be applied.
Category:Group of Order 35 is Cyclic Group