avewright/proofwiki-math huggingface dataset:
Theorem: Cyclic Groups of Same Order are Isomorphic

Two cyclic groups of the same order are isomorphic to each other.

Proof:
Let $G_1$ and $G_2$ be cyclic groups, both of finite order $k$.
Let $G_1 = \gen a, G_2 = \gen b$.
Then, by the definition of a cyclic group:
:$\order a = \order b = k$
Also, by definition:
:$G_1 = \set {a^0, a^1, \ldots, a^{k - 1} }$
and:
:$G_2 = \set {b^0, b^1, \ldots, b^{k - 1} }$
Let us set up the obvious bijection:
:$\phi: G_1 \to G_2: \map \phi {a^n} = b^n$
The next task is to show that $\phi$ is an isomorphism.
Note that $\map \phi {a^n} = b^n$ holds for all $n \in \Z$, not just where $0 \le n < k$, as follows:
Let $n \in \Z: n = q k + r, 0 \le r < k$, by the Division Theorem.
Then, by Element to Power of Remainder:
:$a^n = a^r, b^n = b^r$
Thus:
:$\map \phi {a^n} = \map \phi {a^r} = b^r = b^n$
Now let $x, y \in G_1$.
Since $G_1 = \gen a$, it follows that:
:$\exists s, t \in \Z: x = a^s, y = a^t$
Thus:
{{begin-eqn}}
{{eqn | l = \map \phi {x y}
      | r = \map \phi {a^s a^t}
      | c = 
}}
{{eqn | r = \map \phi {a^{s + t} }
      | c = 
}}
{{eqn | r = b^{s + t}
      | c = 
}}
{{eqn | r = b^s b^t
      | c = 
}}
{{eqn | r = \map \phi {a^s} \map \phi {a^t}
      | c = 
}}
{{eqn | r = \map \phi x \map \phi y
      | c = 
}}
{{end-eqn}}
So $\phi$ is a homomorphism.
As $\phi$ is bijective, $\phi$ is an isomorphism from $G_1$ to $G_2$.
Thus $G_1 \cong G_2$ and the result is proved.
{{Qed}}
Theorem: Order of Subgroup of Cyclic Group

Let $C_n = \gen g$ be the cyclic group of order $n$ which is generated by $g$ whose identity is $e$.
Let $a \in C_n: a = g^i$.
Let $H = \gen a$.
Then:
:$\order H = \dfrac n {\gcd \set {n, i} }$
where:
:$\order H$ denotes the order of $H$
:$\gcd \set {n, i}$ denotes the greatest common divisor of $n$ and $i$.

Proof:
The fact that $H$ is cyclic follows from Subgroup of Cyclic Group is Cyclic.
We need to show that $H$ has $\dfrac n d$ elements.
Let $\order H = k$.
By Non-Trivial Group has Non-Trivial Cyclic Subgroup:
:$k = \order a$
where $\order a$ denotes the order of $a$.
That is:
:$a^k = e$
We have that $a = g^i$.
So:
{{begin-eqn}}
{{eqn | l = a^k
      | r = e
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {g^i}^k
      | r = e = g^n
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = n
      | o = \divides
      | r = i k
      | c = {{Defof|Order of Group Element}}
}}
{{end-eqn}}
We now need to calculate the smallest $k$ such that:
: $n \divides i k$
where $\divides$ denotes divisibility.
That is, the smallest $t \in \N$ such that $n t = i k$.
Let $d = \gcd \set {n, i}$.
Thus:
:$t = \dfrac {k \paren {i / d} } {n / d}$
From Integers Divided by GCD are Coprime, $\dfrac n d$ and $\dfrac i d$ are coprime.
Thus from Euclid's Lemma:
:$\dfrac n d \divides k$
As $a \divides b \implies a \le b$, the smallest value of $k$ such that $\dfrac k {\paren {n / d} } \in \Z$ is $\dfrac n d$.
Hence the result.
{{Qed}}
Theorem: Number of Powers of Cyclic Group Element

Let $G$ be a cyclic group of order $n$, generated by $g$.
Let $d \divides n$.
Then the element $g^{n/d}$ has $d$ distinct powers.

Proof:
Follows directly from Order of Subgroup of Cyclic Group:
:$\order {\gen {g^{n/d} } } = \dfrac n {\gcd \set {n, n/d} } = d$
Thus from List of Elements in Finite Cyclic Group:
:$\gen {g^{n/d} } = \set {e, g^{n/d}, \paren {g^{n/d} }^2, \ldots, \paren {g^{n/d} }^{d - 1} }$
and the result follows.
{{qed}}
Theorem: Subgroup of Finite Cyclic Group is Determined by Order

Let $G = \gen g$ be a cyclic group whose order is $n$ and whose identity is $e$.
Let $d \divides n$, where $\divides$ denotes divisibility.
Then there exists exactly one subgroup $G_d = \gen {g^{n / d} }$ of $G$ with $d$ elements.

Proof:
Let $G$ be generated by $g$, such that $\order g = n$.
From Number of Powers of Cyclic Group Element, $g^{n/d}$ has $d$ distinct powers.
Thus $\gen {g^{n / d} }$ has $d$ elements.
Now suppose $H$ is another subgroup of $G$ of order $d$.
Then by Subgroup of Cyclic Group is Cyclic, $H$ is cyclic.
Let $H = \gen y$ where $y \in G$.
Thus $\order y = d$.
Thus $\exists r \in \Z: y = g^r$.
Since $\order y = d$, it follows that $y^d = g^{r d} = e$.
From Equal Powers of Finite Order Element:
: $n \divides r d$
Thus:
{{begin-eqn}}
{{eqn | q = \exists k \in \N
      | l = k n
      | r = r d
      | c = 
}}
{{eqn | r = k \paren {\dfrac n d} d
      | c = 
}}
{{eqn | ll= \leadsto
      | l = r
      | r = k \paren {\dfrac n d}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac n d
      | o = \divides
      | r = r
      | c = 
}}
{{end-eqn}}
Thus $y$ is a power of $g^{n / d}$.
Hence $H$ is a subgroup of $\gen {g^{n / d} }$.
Since both $H$ and $\gen {g^{n / d} }$ have order $d$, they must be equal.
{{qed}}
Theorem: Cyclic Group Elements whose Powers equal Identity

Let $G$ be a cyclic group whose identity is $e$ and whose order is $n$.
Let $d \divides n$.
Then there exist exactly $d$ elements $x \in G$ satisfying the equation $x^d = e$.
These are the elements of the group $G_d$ generated by $g^{n / d}$: 
:$G_d = \gen {g^{n / d} }$

Proof:
From the argument in Subgroup of Finite Cyclic Group is Determined by Order, it follows that $x$ satisfies the equation $x^d = e$ {{iff}} $x$ is a power of $g^{n/d}$.
Thus there are $d$ solutions to this equation.
{{qed}}
Theorem: Prime Group is Cyclic

Let $p$ be a prime number.
Let $G$ be a group whose order is $p$.
Then $G$ is cyclic.

Proof:
Let $a \in G: a \ne e$ where $e$ is the identity of $G$.
From Group of Prime Order p has p-1 Elements of Order p, $a$ has order $p$.
Hence by definition, $a$ generates $G$.
Hence also by definition, $G$ is cyclic.
{{qed}}
Theorem: Group of Order less than 6 is Abelian

All groups with less than $6$ elements are abelian.

Proof:
Let $G$ be a non-abelian group.
From Non-Abelian Group has Order Greater than 4, the order of $G$ must be at least $5$.
But $5$ is a prime number.
By Prime Group is Cyclic it follows that a group of order $5$ is cyclic.
By Cyclic Group is Abelian this group is abelian.
Hence the result.
{{qed}}
Theorem: Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order

Let $C_n$ be the cyclic group of order $n$.
Let $C_n = \gen a$, that is, that $C_n$ is generated by $a$.
Then:
:$C_n = \gen {a^k} \iff k \perp n$
That is, $C_n$ is also generated by $a^k$ {{iff}} $k$ is coprime to $n$.

Proof:
=== Necessary Condition ===
Let $k \perp n$.
Then by Integer Combination of Coprime Integers:
:$\exists u, v \in \Z: 1 = u k + v n$
So $\forall m \in \Z$, we have:
{{begin-eqn}}
{{eqn | l =a^m
      | r = a^{m \paren 1}
      | c = Integer Multiplication Identity is One
}}
{{eqn | r = a^{m \paren {u k + v n} }
      | c = as shown above
}}
{{eqn | r = a^{m \paren {u k} + m v n}
      | c = Integer Multiplication Distributes over Addition
}}
{{eqn | r = a^{m \paren {u k} } a^{m v n}
      | c = Product of Powers
}}
{{eqn | r = a^{m \paren {u k} } e
      | c = as $a^{m v n} = e$
}}
{{eqn | r = a^{m \paren {u k} }
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = a^{\paren {m u} k}
      | c = Integer Multiplication is Associative
}}
{{eqn | r = a^{k \paren {m u} }
      | c = Integer Multiplication is Commutative
}}
{{eqn | r = \paren {a^k}^{m u}
      | c = Power of Power
}}
{{end-eqn}}
Thus $a^k$ generate $C_n$.
{{qed|lemma}}
{{improve|Do we really need all the above clutter? The arithmetic properties of the integers are really not under the microscope here -- is it really necessary to go blow by blow through all this?}}
Theorem: Order of Conjugate of Subgroup

Let $G$ be a group.
Let $H$ be a subgroup of $G$ such that $H$ is of finite order.
Then $\order {H^a} = \order H$.

Proof:
From the definition of Conjugate of Group Subet we have $H^a = a H a^{-1}$.
From Set Equivalence of Regular Representations:
:$\order {a H a^{-1} } = \order {a H} = \order H$
{{qed}}
Theorem: Subgroup of Index 2 is Normal

A subgroup of index $2$ is always normal.

Proof:
Suppose $H \le G$ such that $\index G H = 2$.
Thus $H$ has two left cosets (and two right cosets) in $G$.
If $g \in H$, then $g H = H = H g$.
If $g \notin H$, then $g H = G \setminus H$ as there are only two cosets and the cosets partition $G$.
For the same reason, $g \notin H \implies H g = G \setminus H$.
That is, $g H = H g$.
The result follows from the definition of normal subgroup.
{{qed}}
Theorem: Intersection of Normal Subgroups is Normal

Let $G$ be a group.
Let $I$ be an indexing set.
Let $\family {N_i}_{i \mathop \in I}$ be a non-empty indexed family of normal subgroups of $G$.
Then $\ds \bigcap_{i \mathop \in I} N_i$ is a normal subgroup of $G$.

Proof:
Let $\ds N = \bigcap_{i \mathop \in I} N_i$.
From Intersection of Subgroups is Subgroup, $N \le G$.
Suppose $H \in \set {N_i: i \in I}$.
We have that $N \subseteq H$.
Thus from Subgroup is Superset of Conjugate iff Normal:
:$a N a^{-1} \subseteq a H a^{-1} \subseteq H$
Thus $a N a^{-1}$ is a subset of each one of the subgroups in $\set {N_i: i \in I}$, and hence in their intersection $N$.
That is, $a N a^{-1} \subseteq N$.
The result follows by Subgroup is Superset of Conjugate iff Normal.
{{qed}}
Theorem: Union of Conjugacy Classes is Normal

Let $G$ be a group.
Let $H \le G$.
Then $H$ is normal in $G$ {{iff}} $H$ is a union of conjugacy classes of $G$.

Proof:
{{begin-eqn}}
{{eqn | l = H
      | o = \lhd
      | r = G
      | c = where $\lhd$ denotes that $H$ is normal in $G$
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall g \in G
      | l = g H g^{-1}
      | o = \subseteq
      | r = H
      | c = {{Defof|Normal Subgroup}}
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall x \in H: \forall g \in G
      | l = g x g^{-1}
      | o = \in
      | r = H
      | c =
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall x \in H
      | l = \conjclass x
      | o = \subseteq
      | r = H
      | c = where $\conjclass x$ is the conjugacy class of $x \in G$
}}
{{eqn | ll= \leadstoandfrom
      | l = H
      | r = \bigcup_{x \mathop \in H} \conjclass x
      | c =
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Unique Subgroup of a Given Order is Normal

Let a group $G$ have only one subgroup of a given order.
Then that subgroup is normal.

Proof:
Let $H \le G$, where $\le$ denotes that $H$ is a subgroup of $G$.
Let $H$ be the only subgroup of $G$ whose order is $\order H$.
Let $g \in G$.
From Conjugate of Subgroup is Subgroup:
:$g H g^{-1} \le G$
From Order of Conjugate of Subgroup:
: $\order {g H g^{-1} } = \order H$
But $H$ is the only subgroup of $G$ of order $\order H$.
Hence any subgroup whose order is $\order H$ must in fact ''be'' $H$.
That is, $g H g^{-1} = H$.
The result follows from Subgroup equals Conjugate iff Normal.
{{Qed}}
Theorem: Group Generated by Normal Intersection is Normal

Let $I$ be an indexing set, and $\set {N_i: i \in I}$ be a set of normal subgroups of the group $G$.
Then $\family {N_i: i \in I}$ is a normal subgroup of $G$.

Proof:
By definition, $\family {N_i: i \in I}$ is the intersection of all the subgroups of $G$ which contain every $N_i$.
For each $H \le G$, the conjugate $g H g^{-1}$ contains each $g N_i g^{-1}$.
Since each $N_i \lhd G$, it follows that:
:$N_i \subseteq g H g^{-1}$
Thus it follows that:
:$\family {N_i: i \in I} \lhd G$
{{qed}}
Category:Normal Subgroups
9k9eyzmuyh28an00sf71extpirx5dhd
Theorem: Smallest Normal Subgroup containing Set

Let $S \subseteq G$ where $G$ is a group.
Then there exists a unique smallest normal subgroup of $G$ which contains $S$.

Proof:
Let $\Bbb S$ be the set of all normal subgroups of $G$ that contain $S$.
Since $S \subseteq G \lhd G$, $\Bbb S \ne \O$.
Let $N = \bigcap \Bbb S$, that is, the intersection of all elements of $\Bbb S$.
By Intersection of Normal Subgroups is Normal, $N \lhd G$.
By the definition of intersection, $S \subseteq N$.
By the method of construction, $N$ is the smallest such subgroup.
By the definition of "smallest", $N$ is unique.
{{qed}}
Category:Normal Subgroups
7804uvnc9t3ovii5v8p1ovp5xbme8pw
Theorem: Conjugate of Set with Inverse Closed for Inverses

Let $G$ be a group.
Let $S \subseteq G$.
Let $\hat S = S \cup S^{-1}$, where $S^{-1}$ is the set of all the inverses of all the elements of $S$.
Let $\tilde S = \set {a s a^{-1}: s \in \hat S, a \in G}$.
That is, $\tilde S$ is the set containing all the conjugates of the elements of $S$ and all their inverses.
Then:
: $\forall x \in \tilde S: x^{-1} \in \tilde S$

Proof:
Let $x \in \tilde S$.
That is:
:$\exists s \in \hat S: x = a s a^{-1}$
Then:
{{begin-eqn}}
{{eqn | l = x^{-1}
      | r = \paren {a s a^{-1} }^{-1}
      | c = 
}}
{{eqn | r = a s^{-1} a^{-1}
      | c = Power of Conjugate equals Conjugate of Power
}}
{{end-eqn}}
Since $s^{-1} \in \hat S$, it follows that $x^{-1} \in \tilde S$.
{{Qed}}
Category:Conjugacy
o56izk8q21jzbzwty9uvh72whozxfbh
Theorem: Conjugate of Set with Inverse is Closed

Let $G$ be a group.
Let $S \subseteq G$.
Let $\hat S = S \cup S$.
Let $\tilde S = \set {a s a^{-1}: s \in \hat S, a \in G}$.
Let $\map W {\tilde S}$ be the set of words of $\tilde S$.
Then:
:$\forall w \in \map W {\tilde S}: \forall a \in G: a w a^{-1} \in \map W {\tilde S}$

Proof:
Let $w \in \map W {\tilde S}$.
From the definition of $\map W {\tilde S}$, we have:
:$w = \paren {a_1 s_1 a_1^{-1} } \paren {a_2 s_2 a_2^{-1} } \cdots \paren {a_n s_n a_n^{-1} }, n \in \N^*, a_i \in G, s_1 \in \hat S, 1 \le i \le n$
Thus:
{{begin-eqn}}     
{{eqn | l = a w a^{-1}
      | r = a \paren {a_1 s_1 a_1^{-1} } \paren {a_2 s_2 a_2^{-1} } \cdots \paren {a_n s_n a_n^{-1} } a^{-1}
      | c = 
}}
{{eqn | r = a \paren {a_1 s_1 a_1^{-1} } a^{-1} a \paren {a_2 s_2 a_2^{-1} } a^{-1} \cdots a \paren {a_n s_n a_n^{-1} } a^{-1}
      | c = 
}}
{{eqn | r = \paren {\paren {a a_1} s_1 \paren {a a_1}^{-1} } \paren {\paren {a a_2} s_2 \paren {a a_2}^{-1} } \cdots \paren {\paren {a a_n} s_n \paren {a a_n}^{-1} }
      | c = 
}}
{{end-eqn}}
As $G$ is a group, all of the $a a_i \in G$.
The result follows.
{{qed}}
Category:Conjugacy
odhreesvly8w4ddlvnnefhw8bo994ej
Theorem: Generator of Normal Subgroup

Let $G$ be a group.
Let $S \subseteq G$.
Let $\hat S = S \cup S^{-1}$, where $S^{-1}$ is the set of all the inverses of all the elements of $S$.
Let $\tilde S = \set {a s a^{-1}: s \in \hat S, a \in G}$.
Let $\map W {\tilde S}$ be the set of words of $\tilde S$.
Let $N$ be the smallest normal subgroup of $G$ that contains $S$.
Then $N = \gen S = \map W {\tilde S}$.

Proof:
Let $N$ be the smallest normal subgroup of $G$ that contains $S$, where $S \subseteq G$.
$N$ must certainly include $\hat S$, because any group containing $s \in S$ must also contain $s^{-1}$. 
Therefore, $N$ must be the smallest normal subgroup containing $\hat S$.
Since $N \lhd G$, it follows that $\forall a \in G: \forall s \in \hat S: a s a^{-1} \in N$.
Thus, $\forall x \in \tilde S: x \in N$.
Thus $\tilde S \subseteq N$.
By the closure axiom, $N$ must also contain all products of any finite number of elements of $\tilde S$.
Thus $\map W {\tilde S} \subseteq N$.
Now we prove that $\map W {\tilde S} \lhd G$.
$\tilde S \ne \O$, as $e \paren {s s^{-1} } e^{-1} = e \in \tilde S$.
By Conjugate of Set with Inverse Closed for Inverses, $\tilde S$ is closed under taking inverses.
So from Set of Words Generates Group: Corollary:
:$\map W {\tilde S} \le G$
From Conjugate of Set with Inverse is Closed:
:$\forall w \in\map W {\tilde S}: \forall a \in G: a w a^{-1} \in \map W {\tilde S}$
From Subgroup is Normal iff Contains Conjugate Elements:
:$\map W {\tilde S} \lhd G$.
The result follows by the definition of the minimality of $N$.
{{qed}}
Category:Normal Subgroups
d7h19ngiiwmd4p497557dgh59n4l15r
Theorem: Subset Product with Normal Subgroup as Generator

Let $G$ be a group whose identity is $e$.
Let:
:$H$ be a subgroup of $G$
:$N$ be a normal subgroup of $G$.
Then:
:$N \lhd \gen {N, H} = N H = H N \le G$
where:
:$\le$ denotes subgroup
:$\lhd$ denotes normal subgroup
:$\gen {N, H}$ denotes a subgroup generator
:$N H$ denotes subset product.

Proof:
From Subset Product is Subset of Generator:
:$N H \subseteq \gen {N, H}$
From Subset Product with Normal Subgroup is Subgroup:
:$N H = H N \le G$
Then by the definition of a subgroup generator, $\gen {N, H}$ is the smallest subgroup containing $N H$ and so:
:$\gen {N, H} = N H = H N \le G$
From Normal Subgroup of Subset Product of Subgroups we have that:
:$N \lhd N H$
Hence the result.
{{qed}}
Theorem: Subset Product of Normal Subgroups is Normal

Let $\struct {G, \circ}$ be a group.
Let $N$ and $N'$ be normal subgroups of $G$.
Then $N N'$ is also a normal subgroup of $G$.

Proof:
From Subset Product with Normal Subgroup is Subgroup, we already have that $N N'$ is a subgroup of $G$.
Let $n n' \in N N'$, so that $n \in N, n' \in N'$.
Let $g \in G$.
From Subgroup is Normal iff Contains Conjugate Elements:
:$g n g^{-1}\in N, g n' g^{-1}\in N'$
So:
:$\paren {g n g^{-1} } \paren {g n' g^{-1} } = g n n' g^{-1} \in N N'$
So $N N'$ is normal.
{{qed}}
Theorem: Prime Group is Simple

Groups of prime order are simple.

Proof:
Follows directly from Prime Group has no Proper Subgroups: a group of prime order has only itself and the trivial group as subgroups.
From Trivial Subgroup and Group Itself are Normal, these subgroups are normal.
{{qed}}
Theorem: Prime Group has no Proper Subgroups

A nontrivial group $G$ has no proper subgroups except the trivial group {{iff}} $G$ is finite and its order is prime.

Proof:
=== Sufficient Condition ===
Suppose $G$ is finite and of prime order $p$.
From Lagrange's Theorem, the order of any subgroup of $G$ must divide the order $p$ of $G$.
From the definition of prime, any subgroups of $p$ can therefore only have order $1$ or $p$.
Hence $G$ can have only itself and the trivial group as subgroups.
{{qed|lemma}}
Theorem: Quotient Group of Cyclic Group

Let $G$ be a cyclic group which is generated by $g$.
Let $H$ be a subgroup of $G$.
Then $g H$ generates $G / H$.

Proof:
Let $G$ be a cyclic group generated by $g$.
Let $H \le G$.
We need to show that every element of $G / H$ is of the form $\left({g H}\right)^k$ for some $k \in \Z$.
Suppose $x H \in G / H$.
Then, since $G$ is generated by $g$, $x = g^k$ for some $k \in \Z$.
But $\left({g H}\right)^k = \left({g^k}\right) H = x H$.
So $g H$ generates $G / H$.
{{Qed}}
Theorem: Quotient Group of Cyclic Group

Let $G$ be a cyclic group which is generated by $g$.
Let $H$ be a subgroup of $G$.
Then $g H$ generates $G / H$.

Proof:
Let $H$ be a subgroup of the cyclic group $G = \gen g$.
Then by Homomorphism of Powers for Integers:
:$\forall n \in \Z: \map {q_H} {g^n} = \paren {\map {q_H} g}^n = \paren {g H}^n$
As $G = \set {g^n: n \in \Z}$, we conclude that:
:$G / H = q_H \sqbrk G = \set {\paren {g H}^n: n \in \Z}$
Thus, by Epimorphism from Integers to Cyclic Group, $g H$ generates $G / H$.
{{Qed}}
Theorem: Order of Element in Quotient Group

Let $G$ be a group, and let $H$ be a normal subgroup of $G$.
Let $G / H$ be the quotient group of $G$ by $H$.
The order of $a H \in G / H$ divides the order of $a \in G$.

Proof:
Let $G$ be a group with normal subgroup $H$.
Let $G / H$ be the quotient of $G$ by $H$.
From Quotient Group Epimorphism is Epimorphism, $G / H$ is a homomorphic image of $G$.
Let $q_H: G \to G / H$ given by $\map f a = a H$ be that quotient epimorphism.
Let $a \in G$ such that $a^n = e$ for some integer $n$.
Then, by the morphism property of $q_H$:
{{begin-eqn}}
{{eqn | l = \map {q_H} {a^n}
      | r = \paren {\map {q_H} a}^n
      | c = 
}}
{{eqn | r = \paren {a H}^n
      | c = 
}}
{{eqn | r = a^n H
      | c = {{Defof|Coset Product}}
}}
{{eqn | r = e H
      | c = by hypothesis
}}
{{eqn | r = H
      | c = 
}}
{{end-eqn}}
Hence $\order H$ divides $n$.
{{qed}}
Theorem: Intersection of Quotient Groups

Let $N \lhd G$ be a normal subgroup of $G$.
Let:
:$N \le A \le G$
:$N \le B \le G$
For a subgroup $H$ of $G$, let $\alpha$ be the bijection defined as:
:$\map \alpha H = \set {h N: h \in H}$
Then:
:$\map \alpha {A \cap B} = \map \alpha A \cap \map \alpha B$

Proof:
From the proof of the Correspondence Theorem:
:$\map \alpha H \subseteq G / N$
Then:
{{begin-eqn}}
{{eqn | l = \map \alpha A \cap \map \alpha B
      | r = \set {g N: g \in A \cap B}
      | c = 
}}
{{eqn | r = \paren {A \cap B} / N
      | c = 
}}
{{eqn | r = \map \alpha {A \cap B}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Generator of Quotient Groups

Let $N \lhd G$ be a normal subgroup of $G$.
Let:
:$N \le A \le G$
:$N \le B \le G$
For a subgroup $H$ of $G$, let $\alpha$ be the bijection defined as:
:$\map \alpha H = \set {h N: h \in H}$
Then:
:$\map \alpha {\gen {A, B} } = \gen {\map \alpha A, \map \alpha B}$
where $\gen {A, B}$ denotes the subgroup of $G$ generated by $\set {A, B}$.

Proof:
From the proof of the Correspondence Theorem:
:$\map \alpha H \subseteq G / N$
Then:
{{begin-eqn}}
{{eqn | l = \map \alpha {\gen {A, B} }
      | r = \set {h N: h \in \gen {A, B} }
      | c = 
}}
{{eqn | r = \set {h N \in \gen {A / N, B / N} }
      | c = 
}}
{{eqn | r = \gen {\map \alpha A, \map \alpha B}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Quotient of Group by Center Cyclic implies Abelian

Let $G$ be a group.
Let $\map Z G$ be the center of $G$.
Let $G / \map Z G$ be the quotient group of $G$ by $\map Z G$.
Let $G / \map Z G$ be cyclic.
Then $G$ is abelian, so $G = \map Z G$.
That is, the group $G / \map Z G$ cannot be a cyclic group which is non-trivial.

Proof:
Suppose $G / \map Z G$ is cyclic.
Then by definition:
:$\exists \tau \in G / \map Z G: G / \map Z G = \gen \tau$
Since $\tau$ is a coset by $\map Z G$:
:$\exists t \in G: \tau = t \map Z G$
Thus each coset of $\map Z G$ in $G$ is equal to $\paren {t \map Z G}^i = t^i \map Z G$ for some $i \in \Z$.
Now let $x, y \in G$.
Suppose $x \in t^m \map Z G, y \in t^n \map Z G$.
Then $x = t^m z_1, y = t^n z_2$ for some $z_1, z_2 \in \map Z G$.
Thus:
{{begin-eqn}}
{{eqn | l = x y
      | r = \paren {t^m z_1} \paren {t^n z_2 }
      | c = {{hypothesis}}
}}
{{eqn | r = t^m \paren {z_1 t^n}  z_2
      | c = Subset Product within Semigroup is Associative
}}
{{eqn | r = t^m \paren {t^n z_1} z_2
      | c = $z_1$ commutes with all $t \in G$ since it is in the center
}}
{{eqn | r = \paren {t^m t^n} \paren {z_1 z_2}
      | c = 
}}
{{eqn | r = t^{m + n} \paren {z_1 z_2}
      | c = Powers of Group Elements: Sum of Indices
}}
{{eqn | r = t^{n + m} \paren {z_2 z_1}
      | c = 
}}
{{eqn | r = \paren {t^n t^m} \paren {z_2 z_1}
      | c = Powers of Group Elements: Sum of Indices
}}
{{eqn | r = t^n \paren {t^m z_2} z_1
      | c = 
}}
{{eqn | r = t^n \paren {z_2 t^m} z_1
      | c = $z_2$ commutes with all $t \in G$ since it is in the center
}}
{{eqn | r = \paren {t^n z_2 } \paren {t^m z_1 }
      | c = Subset Product within Semigroup is Associative
}}
{{eqn | r = y x
      | c = {{hypothesis}}
}}
{{end-eqn}}
This holds for all $x, y \in G$, and thus $G$ is abelian.
Thus by definition of abelian group:
:$\map Z G = G$
Therefore Quotient of Group by Itself it follows that $G / \map Z G$ is the trivial group.
{{qed}}