avewright/proofwiki-math huggingface dataset:
Theorem: Inverse in Group is Unique

Let $\struct {G, \circ}$ be a group.
Then every element $x \in G$ has exactly one inverse:
:$\forall x \in G: \exists_1 x^{-1} \in G: x \circ x^{-1} = e^{-1} = x^{-1} \circ x$
where $e$ is the identity element of $\struct {G, \circ}$.

Proof:
Let $x, y \in G$.
We already have, from the definition of inverse element, that:
:$\forall x \in G: \exists x^{-1} \in G: x \circ x^{-1} = e = x^{-1} \circ x$
By Group has Latin Square Property, there exists exactly one $a \in G$ such that $a \circ x = y$.
Similarly, there exists exactly one $b \in G$ such that $x \circ b = y$.
Substituting $e$ for $y$, it follows that $x^{-1}$ as defined above is unique.
{{qed}}
Theorem: Product Inverse in Ring is Unique

Let $\struct {R, +, \circ}$ be a ring with unity.
Let $x \in R$ be a unit of $R$.
Then the product inverse $x^{-1}$ of $x$ is unique.

Proof:
By definition of ring with unity, the algebraic structure $\struct {R, \circ}$ is a monoid.
The result follows from Inverse in Monoid is Unique.
{{qed}}
Theorem: Units of Gaussian Integers form Group

Let $U_\C$ be the set of units of the Gaussian integers:
:$U_\C = \set {1, i, -1, -i}$
where $i$ is the imaginary unit: $i = \sqrt {-1}$.
Let $\struct {U_\C, \times}$ be the algebraic structure formed by $U_\C$ under the operation of complex multiplication.
Then $\struct {U_\C, \times}$ forms a cyclic group under complex multiplication.

Proof:
By definition of the imaginary unit $i$:
{{begin-eqn}}
{{eqn | l = i^2
      | r = -1
}}
{{eqn | l = i^3
      | r = -i
}}
{{eqn | l = i^4
      | r = 1
}}
{{end-eqn}}
thus demonstrating that $U_\C$ is generated by $i$.
Thus $\struct {U_\C, \times}$ is by definition a cyclic group of order $4$.
{{qed}}
Theorem: Units of Gaussian Integers form Group

Let $U_\C$ be the set of units of the Gaussian integers:
:$U_\C = \set {1, i, -1, -i}$
where $i$ is the imaginary unit: $i = \sqrt {-1}$.
Let $\struct {U_\C, \times}$ be the algebraic structure formed by $U_\C$ under the operation of complex multiplication.
Then $\struct {U_\C, \times}$ forms a cyclic group under complex multiplication.

Proof:
From Gaussian Integer Units are 4th Roots of Unity:
: $\left\{{1, i, -1, -i}\right\}$ constitutes the set of the $4$th roots of unity.
The result follows from Roots of Unity under Multiplication form Cyclic Group.
{{qed}}
Theorem: Units of Gaussian Integers form Group

Let $U_\C$ be the set of units of the Gaussian integers:
:$U_\C = \set {1, i, -1, -i}$
where $i$ is the imaginary unit: $i = \sqrt {-1}$.
Let $\struct {U_\C, \times}$ be the algebraic structure formed by $U_\C$ under the operation of complex multiplication.
Then $\struct {U_\C, \times}$ forms a cyclic group under complex multiplication.

Proof:
From Units of Gaussian Integers, $U_\C$ is the set of units of the ring of Gaussian integers.
From Group of Units is Group, $\left({U_\C, \times}\right)$ forms a group.
It remains to note that:
{{begin-eqn}}
{{eqn | l = i^2
      | r = -1
}}
{{eqn | l = i^3
      | r = -i
}}
{{eqn | l = i^4
      | r = 1
}}
{{end-eqn}}
thus demonstrating that $U_\C$ is  cyclic.
{{qed}}
Theorem: Gaussian Integer Units form Multiplicative Subgroup of Complex Numbers

The group of Gaussian integer units under complex multiplication:
:$\struct {U_\C, \times} = \struct {\set {1, i, -1, -i}, \times}$
forms a subgroup of the multiplicative group of complex numbers.

Proof:
By Units of Gaussian Integers form Group, $\struct {U_\C, \times}$ forms a group.
Each of the elements of $U_\C$ is a complex number, and non-zero, and therefore $U_\C \subseteq \C \setminus \set 0$.
The result follows by definition of subgroup.
{{qed}}
Theorem: Permutation Group is Subgroup of Symmetric Group

Let $S$ be a set.
Let $\struct {\map \Gamma S, \circ}$ be the symmetric group on $S$, where $\circ$ denotes the composition operation.
Let $\struct {H, \circ}$ be a set of permutations of $S$ which forms a group under $\circ$.
Then $\struct {H, \circ}$ is a subgroup of $\struct {\map \Gamma S, \circ}$.

Proof:
Follows directly from the definition of subgroup:
$H$ is a subset of $\map \Gamma S$, and $\struct {H, \circ}$ is a group.
Hence the result.
{{qed}}
Theorem: Units of Field of Complex Numbers form Group

:$\C^\times = \C \setminus \set 0$
where $\C^\times$ denotes the group of units of $\C$.

Proof:
By Complex Numbers form Field, $\C$ is a field.
From Group of Units of Field it follows that:
: $\C^\times = \C \setminus \set 0$
{{qed}}
Category:Examples of Groups
Category:Complex Numbers
tss7spdnxtbd4mireozplygblsf6ij5
Theorem: Determinant of Block Diagonal Matrix

Let $\mathbf A$ be a block diagonal matrix of order $n$.
Let $\mathbf A_1, \ldots, \mathbf A_k$ be the square matrices on the diagonal:
:<nowiki>$\ds \mathbf A = \begin {bmatrix}
\mathbf A_1 & 0 & \cdots & 0 \\
0 & \mathbf A_2 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \mathbf A_k 
\end {bmatrix}$</nowiki>
Then the determinant $\map \det {\mathbf A}$ of $\mathbf A$ satisfies:
:$\ds \map \det {\mathbf A} = \prod_{i \mathop = 1}^k \map \det {\mathbf A_i}$

Proof:
{{tidy}}
{{MissingLinks}}
To prove this fact, we need to prove additional helper propositions.
{{ExtractTheorem}}
$\textbf{Claim 1}$
The determinant of the block-diagonal matrix of type $M = \begin{pmatrix}
A & 0 \\
0 & I \\
\end{pmatrix}$ or $M = \begin{pmatrix}
I & 0 \\
0 & A \\
\end{pmatrix}$ equals $\map \det A$.
$\textbf{Proof}$
For this we utilize mathematical induction.
$\textbf{Base case}$ - $k = 1$ and $I = I_k = 1$.
Then, $\map \det M = 1 \map \det A + 0 + \ldots + 0$, because the first column (or last, depending on the location of $I$ block) has all zeros except the first (or last) element.
$\textbf{Induction step}$
By the same reason - all except one elements of the column are zeros - we have $\map \det {M_k} = 1 \map \det {M_{k - 1} } + 0 + \ldots + 0$, where $M_k$ is the block matrix discussed above wiith $I_k$ block.
Thus, by induction:
:$\map \det {M_k} = \map \det {M_{k - 1} } = \ldots = \map \det {M_1} = \map \det A$
{{qed|lemma}}
$\textbf{Claim 2}$ about the determinant of upper-triangular block matrix
$M = \begin{pmatrix}
A & B \\
0 & D \\
\end{pmatrix}$, $\map \det M = \map \det A \map \det D$
$\textbf{Proof}$
$M = \begin{pmatrix}
A & B \\
0 & D \\
\end{pmatrix} = \begin{pmatrix}
I & 0 \\
0 & D \\
\end{pmatrix} \begin{pmatrix}
I & B \\
0 & I \\
\end{pmatrix} \begin{pmatrix}
A & 0 \\
0 & I \\
\end{pmatrix}$. Thus, using the properties $\map \det {A B} = \map \det A \map \det B$, $\map \det I = 1$ and $\det \begin{pmatrix}
I & B \\
0 & I \\
\end{pmatrix} = 1$, because this is just triangular matrix with all ones on the diagonal.
So, we get $\map \det M = \map \det D \map \det A$
{{qed|lemma}}
From the above propositions one can see that for  $ A = \begin{pmatrix}
A_1 & 0 \\
0 & A_2 \\
\end{pmatrix}$, which is the special case of the upper-triangular matrix:
:$\map \det A = \map \det {A_1} \map \det {A_2}$.
Since $A$ is diagonal, $A_1$ and $A_2$ are also diagonal and their determinants equal to the product of corresponding diagonal blocks.
Thus:
:$\map \det {A_1} = \map \det {A_{1, 1} } \map \det {A_{1, 2} }$
and:
:$\map \det {A_2} = \map \det {A_{2, 1} } \map \det {A_{2, 2} }$
which imply:
:$\map \det A = \map \det {A_{1, 1} } \map \det {A_{1, 2} } \map \det {A_{2, 1} } \map \det {A_{2, 2} }$
Following this recursion argument:
:$\ds \map \det {\mathbf A} = \prod_{i \mathop = 1}^k \map \det {\mathbf A_i}$
{{qed}}
Theorem: Multiplicative Group of Positive Rationals is Non-Cyclic

Let $\struct {\Q_{>0}, \times}$ be the multiplicative group of positive rational numbers.
Then $\struct {\Q_{>0}, \times}$ is not a cyclic group.

Proof:
{{AimForCont}} $\struct {\Q_{>0}, \times}$ is cyclic.
Then $\struct {\Q_{>0}, \times}$ has a generator $x$ such that $x > 1$.
It would follow that:
:$\Q = \set {\ldots, x^{-2}, x^{-1}, 1, x, x^2, \ldots}$
where the elements are arranged in ascending order.
But then consider $y \in \Q: y = \dfrac {1 + x} 2$.
So $1 < y < x$ and so $y \notin \Q$.
From this contradiction it is concluded that there can be no such single generator of $\struct {\Q_{>0}, \times}$.
Therefore, by definition, $\struct {\Q_{>0}, \times}$ is not a cyclic group.
{{qed}}
Theorem: General Linear Group to Determinant is Homomorphism

Let $\GL {n, \R}$ be the general linear group over the field of real numbers.
Let $\struct {\R_{\ne 0}, \times}$ denote the multiplicative group of real numbers.
Let $\det: \GL {n, \R} \to \struct {\R_{\ne 0}, \times}$ be the mapping:
:$\mathbf A \mapsto \map \det {\mathbf A}$
where $\map \det {\mathbf A}$ is the determinant of $\mathbf A$.
Then $\det$ is a group homomorphism.

Proof:
From Determinant of Matrix Product:
: $\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \, \map \det {\mathbf B}$
which is seen to be a group homomorphism by definition.
{{qed}}
Theorem: Logarithm on Positive Real Numbers is Group Isomorphism

Let $\struct {\R_{>0}, \times}$ be the multiplicative group of positive real numbers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $b$ be any real number such that $b > 1$.
Let $\log_b: \struct {\R_{>0}, \times} \to \struct {\R, +}$ be the mapping:
:$x \mapsto \map {\log_b} x$
where $\log_b$ is the logarithm to base $b$.
Then $\log_b$ is a group isomorphism.

Proof:
From Sum of Logarithms we have:
:$\forall x, y \in \R_{>0}: \map {\log_b} {x y} = \map {\log_b} x + \map {\log_b} y$
That is $\log_b$ is a group homomorphism.
From Change of Base of Logarithm, $\log_b$ is a constant multiplied by the natural logarithm function.
Then we have that Logarithm is Strictly Increasing.
From Strictly Monotone Real Function is Bijective, it follows that $\log_b$ is a bijection.
So $\log_b$ is a bijective group homomorphism, and so a group isomorphism.
{{qed}}
Theorem: Kernel is Trivial iff Monomorphism/Group

Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a group homomorphism.
Let $\map \ker \phi$ be the kernel of $\phi$.
Then $\phi$ is a group monomorphism {{iff}} $\map \ker \phi$ is trivial.

Proof:
=== Necessary Condition ===
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a group monomorphism.
By Homomorphism to Group Preserves Identity, $e_S \in \map \ker \phi$.
If $\map \ker \phi$ contained another element $s \ne e_S$, then $\map \phi s = \map \phi {e_S} = e_T$ and $\phi$ would not be injective, thus not be a group monomorphism.
So $\map \ker \phi$ can contain only one element, and that must be $e_S$, which is therefore the trivial subgroup of $S$.
{{qed|lemma}}
Theorem: Kernel is Trivial iff Monomorphism/Ring

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $\map \ker \phi$ be the kernel of $\phi$.
Then $\phi$ is a ring monomorphism {{iff}} $\map \ker \phi = 0_{R_1}$.

Proof:
The proof for the ring monomorphism follows directly from:
:Ring Homomorphism of Addition is Group Homomorphism
and:
:Kernel is Trivial iff Group Monomorphism for the group monomorphism.
{{qed}}
Theorem: Exponential on Real Numbers is Group Isomorphism

Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {\R_{> 0}, \times}$ be the multiplicative group of positive real numbers.
Let $\exp: \struct {\R, +} \to \struct {\R_{> 0}, \times}$ be the mapping:
:$x \mapsto \map \exp x$
where $\exp$ is the exponential function.
Then $\exp$ is a group isomorphism.

Proof:
From Exponential of Sum we have:
:$\forall x, y \in \R: \map \exp {x + y} = \exp x \cdot \exp y$
That is, $\exp$ is a group homomorphism.
Then we have that Exponential is Strictly Increasing.
From Strictly Monotone Real Function is Bijective, it follows that $\exp$ is a bijection.
So $\exp$ is a bijective group homomorphism, and so a group isomorphism.
{{qed}}
Theorem: Exponential on Real Numbers is Group Isomorphism

Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {\R_{> 0}, \times}$ be the multiplicative group of positive real numbers.
Let $\exp: \struct {\R, +} \to \struct {\R_{> 0}, \times}$ be the mapping:
:$x \mapsto \map \exp x$
where $\exp$ is the exponential function.
Then $\exp$ is a group isomorphism.

Proof:
From Real Numbers under Addition form Group, $\struct {\R, +}$ is a group.
From Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group, $\struct {\R_{> 0}, \times}$ is a group.
We have that for all $y \in R_{> 0}$ there exists $x = \map \ln y \in R$ which satisfies $\map \exp x = y$.
{{explain|Link to a result backing up the above claim.}}
Thus $\exp$ is a surjection.
Then from Exponential on Real Numbers is Injection:
:$\exp$ is an injection.
Therefore, $\exp$ is a bijection.
Let $x, y \in R$.
From Exponential of Sum:
:$\map \exp {x + y} = \map \exp x \, \map \exp y$
So $\exp$ is a homomorphism and a bijection.
It follows by definition that $\exp: \struct {\R, +} \to \struct {\R_{> 0}, \times}$ is an isomorphism.
{{qed}}
Theorem: First Isomorphism Theorem/Groups

Let $\phi: G_1 \to G_2$ be a group homomorphism.
Let $\map \ker \phi$ be the kernel of $\phi$.
Then:
:$\Img \phi \cong G_1 / \map \ker \phi$
where $\cong$ denotes group isomorphism.

Proof:
Let $K = \map \ker \phi$.
By Kernel is Normal Subgroup of Domain, $G_1 / K$ exists.
We need to establish that the mapping $\theta: G_1 / K \to G_2$ defined as:
:$\forall x \in G_1: \map \theta {x K} = \map \phi x$
is well-defined.
That is, we need to ensure that:
:$\forall x, y \in G_1: x K = y K \implies \map \theta {x K} = \map \theta {y K}$
Let $x, y \in G_1: x K = y K$.
Then:
{{begin-eqn}}
{{eqn | l = x K
      | r = y K
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = x^{-1} y
      | o = \in
      | r = K = \map \ker \phi
      | c = Left Cosets are Equal iff Product with Inverse in Subgroup
}}
{{eqn | ll= \leadstoandfrom
      | l = \map \phi {x^{-1} y}
      | r = e_{G_2}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {\map \phi x}^{-1} \map \phi y
      | r = e_{G_2}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \map \phi x
      | r = \map \phi y
      | c = 
}}
{{end-eqn}}
Thus we see that $\theta$ is well-defined.
Since we also have that:
:$\map \phi x = \map \phi y \implies x K = y K$
it follows that:
:$\map \theta {x K} = \map \theta {y K} \implies x K = y K$
So $\theta$ is injective.
We also note that:
:$\Img \theta = \set {\map \theta {x K}: x \in G}$
So:
{{begin-eqn}}
{{eqn | l = \Img \theta
      | r = \set {\map \theta {x K}: x \in G}
      | c = 
}}
{{eqn | r = \set {\map \phi x: x \in G}
      | c = 
}}
{{eqn | r = \Img \phi
      | c = 
}}
{{end-eqn}}
We also note that $\theta$ is a homomorphism:
{{begin-eqn}}
{{eqn | q = \forall x, y \in G
      | o = 
      | r = \map \theta {x K y K}
      | c = 
}}
{{eqn | r = \map \theta {x y K}
      | c = 
}}
{{eqn | r = \map \phi {x y}
      | c = 
}}
{{eqn | r = \map \phi x \map \phi y
      | c = 
}}
{{eqn | r = \map \theta {x K} \map \theta {y K}
      | c = 
}}
{{end-eqn}}
Thus $\theta$ is a monomorphism whose image equals $\Img \phi$.
The result follows.
{{Qed}}
Theorem: First Isomorphism Theorem/Rings

Let $\phi: R \to S$ be a ring homomorphism.
Let $\map \ker \phi$ be the kernel of $\phi$.
Then:
:$\Img \phi \cong R / \map \ker \phi$
where $\cong$ denotes ring isomorphism.

Proof:
From Ring Homomorphism whose Kernel contains Ideal, let $J = \map \ker \phi$.
This gives the ring homomorphism $\mu: R / \map \ker \phi \to S$ as follows:
::<nowiki>$\begin {xy} \xymatrix@L + 2mu@ + 1em {
 R \ar[rr]^*{\nu}
   \ar[rdrd]_*{\phi}
& & R / \map \ker \phi \ar[dd]_*{\mu} \\
\\
& & S 
} \end {xy}$</nowiki>
That is:
:$\phi = \mu \circ \nu$
Then we have:
:$\map \ker \mu = \map \ker \phi / \map \ker \phi$
This is the null subring of $R / \map \ker \phi$ by Quotient Ring Defined by Ring Itself is Null Ring.
Then from Kernel is Trivial iff Monomorphism it follows that $\mu$ is a monomorphism.
From $\phi = \mu \circ \nu$, we have:
:$\Img \mu = \Img \phi$
It follows that $\mu$ is an isomorphism.
{{qed}}
Theorem: Second Isomorphism Theorem/Groups

Let $G$ be a group, and let:
:$(1): \quad H$ be a subgroup of $G$
:$(2): \quad N$ be a normal subgroup of $G$.
Then:
:$\dfrac H {H \cap N} \cong \dfrac {H N} N$
where $\cong$ denotes group isomorphism.

Proof:
The fact that $N$ is normal, together with Intersection with Normal Subgroup is Normal, gives us that $N \cap H \lhd H$.
Also, $N \lhd N H = \gen {H, N}$ follows from Subset Product with Normal Subgroup as Generator.
Now we define a mapping $\phi: H \to H N / N$ by the rule:
:$\map \phi h = h N$
Note that $N$ need not be a subset of $H$.
Therefore, the coset $h N$ is an element of $H N / N$ rather than of $H / N$.
Then $\phi$ is a homomorphism, as:
:$\map \phi {x y} = x y N = \paren {x N} \paren {y N} = \map \phi x \map \phi y$
Then:
{{begin-eqn}}
{{eqn | l = \map \ker \phi
      | r = \set {h \in H: \map \phi h = e_{H N / N} }
      | c = 
}}
{{eqn | r = \set {h \in H: h N = N}
      | c = {{LinkWanted}}
}}
{{eqn | r = \set {h \in H: h \in N}
      | c = 
}}
{{eqn | r = H \cap N
      | c = 
}}
{{end-eqn}}
Then we see that $\phi$ is a surjection because $h n N = h N \in H N / N$ is $\map \phi h$.
The result follows from the First Isomorphism Theorem.
{{qed}}
Theorem: Second Isomorphism Theorem/Rings

Let $R$ be a ring, and let:
:$S$ be a subring of $R$
:$J$ be an ideal of $R$.
Then:
:$(1): \quad S + J$ is a subring of $R$
:$(2): \quad J$ is an ideal of $S + J$
:$(3): \quad S \cap J$ is an ideal of $S$
:$(4): \quad \dfrac S {S \cap J} \cong \dfrac {S + J} J$
where $\cong$ denotes group isomorphism.

Proof:
The relations being defined can be illustrated by this commutative diagram:
:620px
$(1): \quad S + J$ is a subring of $R$
From Sum of All Ring Products is Additive Subgroup, $S + J$ is an additive subgroup of $R$.
Suppose $s, s' \in S, j, j' \in J$.
Then:
:$\paren {s + j} \paren {s' + j'}$
{{begin-eqn}}
{{eqn | l = \paren {s + j} \paren {s' + j'}
      | r = s s' + b{s j' + s' j + j j'}
      | c = 
}}
{{eqn | o = \in
      | r = S + J
      | c = as $J$ is an ideal of $R$
}}
{{end-eqn}}
so by the Subring Test $S + J$ is a subring of $R$.
{{qed|lemma}}
$(2): \quad J$ is an ideal of $S + J$
Let $s + j \in S + J$ and let $j \in J$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {s + j} j'
      | r = s j + s j'
      | c = 
}}
{{eqn | o = \in
      | r = J
      | c = as $s j, s j' \in J$ as $J$ is an ideal of $R$
}}
{{end-eqn}}
So $J$ is an ideal of $S + J$.
{{qed|lemma}}
$(3): \quad S \cap J$ is an ideal of $S$
Let $\nu: R \to R / J$ be the quotient epimorphism.
Let $\nu'$ be the restriction of $\nu$ to $S$.
Then $\nu': S \to R / J$ is a homomorphism.
The image of $\nu'$ is the set of all cosets $s + J$ for $s \in S$:
:$\image {\nu'} = \dfrac {S + J} J$
Now, the kernel of $\nu'$ is the set of all elements of $S$ which are sent to $0_{S/J}$ by $\nu$.
That is, all the elements of $S$ which are also in $J$ itself, which is how the quotient ring behaves.
That is:
:$\ker \paren {\nu'} = S \cap J$
and so from Kernel of Ring Homomorphism is Ideal, $S \cap J$ is an ideal of $S$.
$(4): \quad \dfrac S {S \cap J} \cong \dfrac {S + J} J$
This follows directly from the First Ring Isomorphism Theorem.
{{qed}}
Theorem: Third Isomorphism Theorem/Groups

Let $G$ be a group, and let:
:$H, N$ be normal subgroups of $G$
:$N$ be a subset of $H$.
Then:
:$(1): \quad N$ is a normal subgroup of $H$
:$(2): \quad H / N$ is a normal subgroup of $G / N$
:::where $H / N$ denotes the quotient group of $H$ by $N$
:$(3): \quad \dfrac {G / N} {H / N} \cong \dfrac G H$
:::where $\cong$ denotes group isomorphism.

Proof:
From Normal Subgroup which is Subset of Normal Subgroup is Normal in Subgroup, $N$ is a normal subgroup of $H$.
We define a mapping:
:$\phi: G / N \to G / H$ by $\map \phi {g N} = g H$
Since $\phi$ is defined on cosets, we need to check that $\phi$ is well-defined.
Suppose $x N = y N$.
Then:
:$y^{-1} x \in N$
Then:
:$N \le H \implies y^{-1} x \in H$
and so:
:$x H = y H$
So:
:$\map \phi {x N} = \map \phi {y N}$
and $\phi$ is indeed well-defined.
Now $\phi$ is a homomorphism, from:
{{begin-eqn}}
{{eqn | l = \map \phi {x N} \map \phi {y N}
      | r = \paren {x H} \paren {y H}
      | c = 
}}
{{eqn | r = x y H
      | c = 
}}
{{eqn | r = \map \phi {x y N}
      | c = 
}}
{{eqn | r = \map \phi {x N y N}
      | c = 
}}
{{end-eqn}}
Also, since $N \subseteq H$, it follows that:
:$\order N \le \order H$
So:
:$\order {G / N} \ge \order {G / H}$, indicating $\phi$ is surjective.
So:
{{begin-eqn}}
{{eqn | l = \map \ker \phi
      | r = \set {g N \in G / N: \map \phi {g N} = e_{G / H} }
      | c = 
}}
{{eqn | r = \set {g N \in G / N: g H = H}
      | c = 
}}
{{eqn | r = \set {g N \in G / N: g \in H}
      | c = 
}}
{{eqn | r = H / N
      | c = 
}}
{{end-eqn}}
The result follows from the First Isomorphism Theorem.
{{qed}}
Theorem: Third Isomorphism Theorem/Groups

Let $G$ be a group, and let:
:$H, N$ be normal subgroups of $G$
:$N$ be a subset of $H$.
Then:
:$(1): \quad N$ is a normal subgroup of $H$
:$(2): \quad H / N$ is a normal subgroup of $G / N$
:::where $H / N$ denotes the quotient group of $H$ by $N$
:$(3): \quad \dfrac {G / N} {H / N} \cong \dfrac G H$
:::where $\cong$ denotes group isomorphism.

Proof:
From Normal Subgroup which is Subset of Normal Subgroup is Normal in Subgroup, $N$ is a normal subgroup of $H$.
Let $q_H$ and $q_N$ be the quotient mappings from $G$ to $\dfrac G H$ and $G$ to $\dfrac G N$ respectively.
Hence:
:$N \subseteq \map \ker {q_H}$
{{explain|This is confusing me: having difficulty identifying exactly what $\map \ker {q_H}$ is becausew I've lost track of what maps to what}}
From Quotient Theorem for Group Homomorphisms: Corollary 2, it therefore follows that:
:there exists a group epimorphism $\psi: \dfrac G N \to \dfrac G H$ such that $\psi \circ q_N = q_H$
Then we have that :
:there exists a group epimorphism $\phi: \dfrac {G / N} {H / N} \to \dfrac G N$ such that $\phi \circ q_{H / N} = \psi$
{{iff}}:
:$H / N \subseteq \map \ker \psi$
{{finish|Just need to prove that $H / N \subseteq \map \ker \psi$ and the job is done.}}
Thus we form the composite:
:$\phi \circ q_{H / N} \circ q_N = q_H$
and it remains to be shown that $\phi$ is an isomorphism.
{{finish}}