avewright/proofwiki-math huggingface dataset:
Theorem: Group Direct Product of Infinite Cyclic Groups

The group direct product of two infinite cyclic groups is not cyclic.

Proof:
Let $G_1 = \struct {G_1, \circ_1}$ and $G_2 = \struct {G_2, \circ_2}$ be infinite cyclic groups.
Let $G = \struct {G, \circ} = G_1 \times G_2$.
Let $G_1 = \gen {g_1}, G_2 = \gen {g_2}$.
From Generators of Infinite Cyclic Group:
:$g_1$ and $g_1^{-1}$ are the only generators of $G_1$
:$g_2$ and $g_2^{-1}$ are the only generators of $G_2$.
So a generator of $G$ must be of the form $\tuple { g_1^{\pm 1}, g_2^{\pm 1} }$ to have any hope to generate all of $G$.
{{WLOG}}, suppose that $G = \gen {\tuple {g_1, g_2} }$.
Let $e_1$ be the identity element of $G_1$.
Let $x_2 \in G_2$.
From the definition of an infinite cyclic group, both $g_1$ and $g_2$ are of infinite order.
Suppose now $\tuple {e_1, x_2} \in \gen {\tuple {g_1, g_2} }$.
Then there is an $i \in \Z$ such that we have:
:$\tuple {g_1, g_2}^i = \tuple {g_1^i, g_2^i} = \tuple {e_1, x_2}$
However, as $g_1$ is of infinite order, it must be that $i = 0$. 
Hence $x_2 = g_2^0 = e_2$, where $e_2$ is the identity element of $G_2$.
It follows that $\tuple {e_1, g_2} \notin \gen {\tuple {g_1, g_2} }$.
Therefore:
:$G \ne \gen {\tuple {g_1, g_2} }$
It follows that $G$ cannot be generated by one element.
{{qed}}
Theorem: Order of Group Element in Group Direct Product

Let $G$ and $H$ be finite groups.
Let $g \in G: \order g = m, h \in H: \order h = n$.
Then $\order {\tuple {g, h} }$ in $G \times H$ is $\lcm \set {m, n}$.

Proof:
Let $G$ and $H$ be a groups whose identities are $e_G$ and $e_H$.
Let $l = \lcm \set {m, n}$.
From the definition of lowest common multiple:
:$\exists x, y \in \Z: l = m x = n y$
From the definition of order of an element:
:$g^m = e_G, h^n = e_H$
Thus:
{{begin-eqn}}
{{eqn | l = \tuple {g, h}^l
      | r = \tuple {g^l, h^l}
      | c = 
}}
{{eqn | r = \tuple {\paren {g^m}^x, \paren {h^n}^y}
      | c = 
}}
{{eqn | r = \tuple {e_G^x, e_H^y}
      | c = 
}}
{{eqn | r = \tuple {e_G, e_H}
      | c = 
}}
{{end-eqn}}
Now suppose $\exists k \in \Z: \tuple {g, h}^k = \tuple {e_G, e_H}$.
It follows that:
:$\tuple {g, h}^k = \tuple {g^k, h^k} = \tuple {e_G, e_H}$
Thus:
:$g^k = e_G, h^k = e_H$
It follows from $g^k = e_G$ and $\order g = m$, by Absolute Value of Integer is not less than Divisors, that:
:$m \divides k$
Similarly it follows that:
:$n \divides k$
So $k$ is a positive common multiple of both $m$ and $n$.
Since $l$ is the lowest common multiple:
:$l \le k$
Therefore $l$ is the smallest such that $\tuple {g, h}^l = \tuple {e_G, e_H}$.
The result follows by definition of the order of an element.
{{qed}}
Category:Group Direct Products
qn3x5oeiex39ck218cbpsj3tjzn6b74
Theorem: Subgroup Product is Internal Group Direct Product iff Surjective

Let $G$ be a group.
Let $\sequence {H_n}$ be a sequence of subgroups of $G$.
Let $\ds \phi: \prod_{k \mathop = 1}^n H_k \to G$ be a mapping defined by:
:$\ds \map \phi {h_1, h_2, \ldots, h_n} = \prod_{k \mathop = 1}^n h_k$
Then $\phi$ is surjective {{iff}}:
:$\ds G = \prod_{k \mathop = 1}^n H_k$
That is, {{iff}} $G$ is the internal group direct product of $H_1, H_2, \ldots, H_n$.

Proof:
=== Necessary Condition ===
Let $\phi$ be a surjection.
Then $\Img \phi$ consists of all the products $\ds \prod_{k \mathop = 1}^n h_k$ such that:
:$\forall k \in \closedint 1 n: h_k \in H_k$
Thus, as $\phi$ is surjective, every element of $G$ must be representable in this form.
Using the notation:
:$\ds \prod_{k \mathop = 1}^n H_k = \set {\prod_{k \mathop = 1}^n h_k: \forall k \in \closedint 1 n: h_k \in H_k}$
the result follows immediately.
Thus every element of $G$ is the product of an element of each of $H_k$.
{{qed|lemma}}
Theorem: Internal Group Direct Product is Injective

Let $G$ be a group whose identity is $e$.
Let $H_1, H_2$ be subgroups of $G$.
Let $\phi: H_1 \times H_2 \to G$ be a mapping defined by:
:$\map \phi {h_1, h_2} = h_1 h_2$
Then $\phi$ is injective {{iff}}:
:$H_1 \cap H_2 = \set e$

Proof:
=== Necessary Condition ===
Let $\phi$ be an injection.
Let $\map \phi {h_1, h_2} = \map \phi {k_1, k_2}$.
As $\phi$ is injective, this means that:
:$\tuple {h_1, h_2} = \tuple {k_1, k_2}$
and thus:
:$h_1 = k_1, h_2 = k_2$
From the definition of $\phi$, this means:
:$h_1 h_2 = k_1 k_2$
Thus, each element of $G$ that can be expressed as a product of the form $h_1 h_2$ can be thus expressed ''uniquely''.
Now, suppose $h \in H_1 \cap H_2$.
We have:
{{begin-eqn}}
{{eqn | l = h = h e
      | o = :
      | r = h \in H_1, e \in H_2
      | c = 
}}
{{eqn | l = h = e h
      | o = :
      | r = e \in H_1, h \in H_2
      | c = 
}}
{{end-eqn}}
Thus we see that:
{{begin-eqn}}
{{eqn | l = \map \phi {h, e}
      | r = \map \phi {e, h}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \tuple {h, e}
      | r = \tuple {e, h}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = h
      | r = e
      | c = 
}}
{{end-eqn}}
Thus:
:$H_1 \cap H_2 = \set e$
{{qed|lemma}}
Theorem: Internal Group Direct Product Isomorphism

Let $G$ be a group.
Let $H_1, H_2$ be subgroups of $G$.
Let $\phi: H_1 \times H_2 \to G$ be the mapping defined by $\map \phi {h_1, h_2} := h_1 h_2$.
If $\phi$ is a (group) isomorphism, then both $H_1$ and $H_2$ are normal subgroups of $G$.

Proof:
$\phi$ is an isomorphism, so in particular a (group) homomorphism.
Thus by Induced Group Product is Homomorphism iff Commutative, every element of $H_1$ commutes with every element of $H_2$.
Now suppose $a \in G$.
As $\phi$ is an isomorphism, it follows that $\phi$ is surjective.
Thus by Subgroup Product is Internal Group Direct Product iff Surjective:
:$\exists h_1 \in H_1, h_2 \in H_2: a = h_1 h_2$
Now any element of $a H_1 a^{-1}$ is in the form $a h a^{-1}$ for some $h \in H_1$.
Thus:
{{begin-eqn}}
{{eqn | l = a h a^{-1}
      | r = \paren {h_1 h_2} h \paren {h_1 h_2}^{-1}
      | c = 
}}
{{eqn | r = h_1 h_2 h h_2^{-1} h_1^{-1}
      | c = 
}}
{{eqn | r = h_1 h h_2 h_2^{-1} h_1^{-1}
      | c = $h \in H_1$ and $h_2 \in H_2$ commute
}}
{{eqn | r = h_1 h h_1^{-1} \in H_1
      | c = 
}}
{{end-eqn}}
Thus $a H_1 a^{-1} \subseteq H_1$, and $H_1$ is therefore a normal subgroup of $G$.
Similarly, any element of $a H_2 a^{-1}$ is in the form $a h a^{-1}$ for some $h \in H_2$.
Thus:
{{begin-eqn}}
{{eqn | l = a h a^{-1}
      | r = \paren {h_1 h_2} h \paren {h_1 h_2}^{-1}
      | c = 
}}
{{eqn | r = h_1 h_2 h h_2^{-1} h_1^{-1}
      | c = 
}}
{{eqn | r = h_1 h' h_1^{-1}
      | c = as $h' = h_2 h h_2^{-1} \in H_2$
}}
{{eqn | r = h' h_1 h_1^{-1}
      | c = $h_1 \in H_1$ and $h' \in H_2$ commute
}}
{{eqn | r = h' \in H_2
      | c = 
}}
{{end-eqn}}
Thus $a H_2 a^{-1} \subseteq H_2$ and $H_2$ is normal as well.
{{qed}}
Category:Internal Group Direct Products
Category:Group Isomorphisms
Category:Normal Subgroups
tal7o18xi8em9ebt61g1ghj7qd3spdp
Theorem: Internal Group Direct Product of Normal Subgroups

Let $G$ be a group whose identity is $e$.
Let $H_1$ and $H_2$ be normal subgroups of $G$ such that $H_1 \cap H_2 = \set e$.
Let $\phi: H_1 \times H_2 \to G$ be a mapping defined by $\map \phi {h_1, h_2} = h_1 h_2$.
Then $\phi$ is a (group) homomorphism.

Proof:
Let $H_1$ and $H_2$ be normal subgroups of $G$.
Let $h_1 \in H_1, h_2 \in H_2$.
Consider $x \in G: x = h_1 h_2 h_1^{-1} h_2^{-1}$.
{{begin-eqn}}
{{eqn | l = x
      | r = h_1 h_2 h_1^{-1} h_2^{-1}
      | c = 
}}
{{eqn | r = \paren {h_1 h_2 h_1^{-1} } h_2^{-1}
      | c = 
}}
{{end-eqn}}
As $H_2$ is normal, we have $h_1 h_2 h_1^{-1} \in H_2$ and thus $x \in H_2$.
Similarly, we can show that $x \in H_1$ and so $x \in H_1 \cap H_2$ and thus $x = e$.
From Product of Commuting Elements with Inverses, $h_1 h_2 h_1^{-1} h_2^{-1} = e$ {{iff}} $h_1$ and $h_2$ commute.
Thus $h_1 h_2 = h_2 h_1$. 
As $h_1$ and $h_2$ are arbitrary elements of $H_1$ and $H_2$, it follows that every element of $H_1$ commutes with every element of $H_2$.
Thus from Induced Group Product is Homomorphism iff Commutative, $\phi$ is a homomorphism.
{{qed}}
Category:Internal Group Direct Products
Category:Normal Subgroups
Category:Group Homomorphisms
b0d6rb2df8xwikh17apons5oa4d87vz
Theorem: Inclusion Mapping is Surjection iff Identity

Let $T$ be a set.
Let $S\subseteq T$ be a subset.
Let $i_S: S \to T$ be the inclusion mapping.
Then:
:$i_S: S \to T$ is surjective {{iff}} $i_S: S \to T = I_S: S \to S$
where $I_S: S \to S$ denotes the identity mapping on $S$.
Alternatively, this theorem can be worded as:
:$i_S: S \to S = I_S: S \to S$
It follows directly that from Surjection by Restriction of Codomain, the surjective restriction of $i_S: S \to T$ to $i_S: S \to \Img {i_S}$ is itself the identity mapping.

Proof:
It is apparent from the definitions of both the inclusion mapping and the identity mapping that:
:$(1): \quad \Dom {i_S} = S = \Dom {I_S}$
:$(2): \quad \forall s \in S: \map {i_S} s = s = \map {I_S} s$
Theorem: Cyclic Group of Order 6

Let $C_n$ be the cyclic group of order $n$.
Then:
: $C_2 \times C_3 \cong C_6$
: $C_6$ is the internal group direct product of $C_2$ and $C_3$.

Proof:
From Group Direct Product of Cyclic Groups noting that $2 \perp 3$:
:$C_2 \times C_3 \cong C_6$
* $C_6$ is the internal group direct product of $C_2$ and $C_3$:
Let $\left({C_6, \circ}\right) = \left \langle {x} \right \rangle$, let $\left({C_2, \circ}\right) = \left \langle {x^3} \right \rangle$, and let $\left({C_3, \circ}\right) = \left \langle {x^2} \right \rangle$.
From Subgroup of Abelian Group is Normal, $C_2 \lhd C_6$ and $C_3 \lhd C_6$.
We can factorise the elements of $C_6$ thus:
:$e = e \circ e, x = x^3 \circ \left({x^2}\right)^2, x^2 = e \circ x^2, x^3 = x^3 \circ e, x^4 = e \circ \left({x^2}\right)^2, x^5 = x^2 \circ x^3$ 
Hence the result.
{{qed}}
Category:Examples of Cyclic Groups
Category:Groups of Order 6
400xedj6n0gkkj3rx3sgk7htrg1f6bj
Theorem: Internal Direct Product Generated by Subgroups

Let $G$ be a group whose identity is $e$.
Let $\sequence {H_n}$ be a sequence of subgroups of $G$.
Then:
:the subgroup generated by $\ds \bigcup_{k \mathop = 1}^n H_k$ is the internal group direct product of $\sequence {H_n}$
{{iff}}:
:$\sequence {H_n}$ is an independent sequence of subgroups such that every element of $H_i$ commutes with every element of $H_j$ whenever $1 \le i < j \le n$.

Proof:
In the following, the notation $\closedint m n$ is to be understood to mean a (closed) integer interval:
:$\closedint m n := \set {x \in \Z: m \le x \le n}$
for $m, n \in \Z$.
For each $k \in \closedint 1 n$, let $\ds L_k = \prod_{j \mathop = 1}^k H_j$ be the cartesian product of the subgroups $H_1, H_2, \ldots, H_k$ of $G$.
Let $\ds C_k: L_k \to G: \map {C_k} {x_1, x_2, \ldots, x_k} = \prod_{j \mathop = 1}^k x_j$.
Theorem: Internal Group Direct Product Commutativity

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H$ and $K$ be subgroups of $G$.
Let $\struct {G, \circ}$ be the internal group direct product of $H$ and $K$.
Then:
:$\forall h \in H, k \in K: h \circ k = k \circ h$

Proof:
Let $G$ be the internal group direct product of $H$ and $K$.
Then by definition the mapping:
:$\phi: H \times K \to G: \map \phi {h, k} = h \circ k$
is a (group) isomorphism from the (external) direct product $\struct {H, \circ \restriction_H} \times \struct {K, \circ \restriction_K}$ onto $\struct {G, \circ}$.
Let the symbol $\circ$ also be used for the operation induced on $H \times K$ by $\circ \restriction_H$ and $\circ \restriction_K$.
Let $h \in H, k \in K$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {e, k} \circ \tuple {h, e}
      | r = \tuple {e \circ h, k \circ e}
      | c = {{Defof|External Direct Product}}
}}
{{eqn | r = \tuple {h, k}
      | c = {{Defof|Identity Element}}
}}
{{eqn | ll= \leadsto
      | l = h \circ k
      | r = \map \phi {h, k}
      | c = {{Defof|Internal Direct Product}}
}}
{{eqn | r = \map \phi {\tuple {e, k} \circ \tuple {h, e} }
      | c = {{apriori}}
}}
{{eqn | r = \map \phi {e, k} \circ \map \phi {h, e}
      | c = as $\phi$ is a homomorphism
}}
{{eqn | r = \paren {e \circ k} \circ \paren {h \circ e}
      | c = {{Defof|Internal Direct Product}}
}}
{{eqn | r = k \circ h
      | c = {{Defof|Identity Element}}
}}
{{end-eqn}}
{{qed}}
Theorem: Internal Group Direct Product Commutativity

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H$ and $K$ be subgroups of $G$.
Let $\struct {G, \circ}$ be the internal group direct product of $H$ and $K$.
Then:
:$\forall h \in H, k \in K: h \circ k = k \circ h$

Proof:
Let $\sqbrk {x, y}$ denote the commutator of $x, y \in G$:
:$\sqbrk {x, y} := x^{-1} y^{-1} x y$
We have that:
{{begin-eqn}}
{{eqn | n = 1
      | l = y x \sqbrk {x, y}
      | r = y x x^{-1} y^{-1} x y
      | c = {{Defof|Commutator of Group Elements}}
}}
{{eqn | r = y y^{-1} x y
      | c = {{Group-axiom|3}}
}}
{{eqn | r = x y
      | c = {{Group-axiom|3}}
}}
{{end-eqn}}
Let $h \in H$, $k \in K$.
We have:
{{begin-eqn}}
{{eqn | l = \sqbrk {h, k}
      | r = h^{-1} k^{-1} h k
      | c = 
}}
{{eqn | r = h^{-1} \paren {k^{-1} h k}
      | c = {{Group-axiom|1}}
}}
{{eqn | o = \in
      | r = h^{-1} H
      | c = {{Defof|Normal Subgroup}}
}}
{{eqn | ll= \leadsto
      | l = \sqbrk {h, k}
      | o = \in
      | r = H
      | c = as $h^{-1} H = H$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \sqbrk {h, k}
      | r = h^{-1} k^{-1} h k
      | c = 
}}
{{eqn | r = \paren {h^{-1} k^{-1} h} k
      | c = {{Group-axiom|1}}
}}
{{eqn | o = \in
      | r = K k
      | c = {{Defof|Normal Subgroup}}
}}
{{eqn | ll= \leadsto
      | l = \sqbrk {h, k}
      | o = \in
      | r = K
      | c = as $K k = K$
}}
{{end-eqn}}
Thus:
:$\sqbrk {h, k} \in H \cap K$
But as $H \cap K = \set e$, it follows that:
:$\sqbrk {h, k} = e$
It follows from Commutator is Identity iff Elements Commute that:
:$h k = k h$
and the result follows.
{{qed}}
Theorem: Internal and External Group Direct Products are Isomorphic

Let $G$ be a group whose identity is $e$.
Then $G$ is the (external) group direct product of $G_1, G_2, \ldots, G_n$ {{iff}} $G$ is the internal group direct product of $N_1, N_2, \ldots, N_n$ such that:
:$\forall i \in \N_n: N_i \cong G_i$
where:
:$\cong$ denotes (group) isomorphism
:$\N_n$ denotes $\set {1, 2, \ldots, n}$

Proof:
=== Necessary Condition ===
Let $G$ be the external direct product of groups $G_1, G_2, \ldots, G_n$.
For all $i \in \N_n$, let $N_i$ be defined as the set:
:$N_i = \set e \times \cdots \times \set e \times G_i \times \set e \times \cdots \times \set e$
of elements which have entry $e$ everywhere except possibly in the $i$th co-ordinate.
It remains to be checked that:
:$(1): \quad N_i$ is isomorphic to $G_i$
:$(2): \quad N_i$ is a normal subgroup of $G$
:$(3): \quad$ Every element of $G$ has a unique expression:
:$\tuple {g_1, \ldots, g_n} = \tuple {g_1, e, \ldots, e} \tuple {e, g_2, e \ldots e} \ldots \tuple {e, \ldots, g_n}$
as a product of elements of $N_1, \ldots, N_n$.
Theorem: Pullback of Quotient Group Isomorphism is Subgroup

Let $\struct {G, \circ}$ be a group whose identity element is $e_G$.
Let $\struct {H, *}$ be a group whose identity element is $e_H$.
Let $N \lhd G, K \lhd H$ be normal subgroups of $G$ and $H$ respectively.
Let:
:$G / N \cong H / K$
where:
:$G / N$ denotes the quotient of $G$ by $N$
:$\cong$ denotes group isomorphism.
Let $\theta: G / N \to H / K$ be such a group isomorphism.
Let $G \times^\theta H$ be the pullback of $G$ and $H$ via $\theta$.
Then $G \times^\theta H$ is a subgroup of $G \times H$.

Proof:
This result is proved by an application of the Two-Step Subgroup Test:
Theorem: Group/Examples/x+y over 1+xy

Let $G = \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.
Let $\circ: G \times G \to \R$ be the binary operation defined as:
:$\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$
The algebraic structure $\struct {G, \circ}$ is a group.

Proof:
Let $-1 < x, y, z < 1$.
We check the group axioms in turn:
Theorem: Group/Examples/x+y over 1+xy

Let $G = \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.
Let $\circ: G \times G \to \R$ be the binary operation defined as:
:$\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$
The algebraic structure $\struct {G, \circ}$ is a group.

Proof:
To prove $G$ is isomorphic to $\struct {\R, +}$, we need to find a bijective homorphism $\phi: \openint {-1} 1 \to \R$:
:$\forall x, y \in G: \map \phi {x \circ y} = \map \phi x + \map \phi y$
From Group Examples: $\dfrac {x + y} {1 + x y}$:
:the identity element of $G$ is $0$
:the inverse of $x$ in $G$ is $-x$.
This also holds for $\struct {\R, +}$.
This hints at the structure of a possible isomorphism, namely:
:$(1): \quad$ that it is an odd function
:$(2): \quad$ that it passes through $0$
:$(3): \quad$ that it is defined on the open interval $\openint {-1} 1$.
One such function is the inverse hyperbolic arctangent function $\tanh^{-1}$, which is indeed defined on $\openint {-1} 1$ and has the above properties.
Hence:
{{begin-eqn}}
{{eqn | l = \map {\tanh^{-1} } x
      | r = \frac 1 2 \map \ln {\frac {x + 1} {x - 1} }
      | c = {{Defof|Inverse Hyperbolic Tangent|subdef = Real|index = 2}}
}}
{{eqn | ll= \leadsto
      | l = \map {\tanh^{-1} } {x \circ y}
      | r = \frac 1 2 \map \ln {\frac {\frac {x + y} {1 + x y} + 1} {\frac {x + y} {1 + x y} - 1} }
      | c = Definition of $\struct {G, \circ}$
}}
{{eqn | r = \frac 1 2 \map \ln {\frac {x + y + 1 + x y} {x + y - 1 - x y} }
      | c = 
}}
{{eqn | r = \frac 1 2 \map \ln {\frac {\paren {x + 1} \paren {y + 1} } {\paren {x - 1} \paren {y - 1} } }
      | c = 
}}
{{eqn | r = \frac 1 2 \map \ln {\frac {x + 1} {x - 1} } + \frac 1 2 \map \ln {\frac {y + 1} {y - 1} }
      | c = 
}}
{{eqn | r = \tanh^{-1} x + \tanh^{-1} y
      | c = 
}}
{{end-eqn}}
This demonstrates the homorphism between $\struct {G, \circ}$ and $\struct {\R,  +}$.
We have that Real Inverse Hyperbolic Tangent Function is Strictly Increasing over $\openint {-1} 1$.
Hence from Strictly Monotone Real Function is Bijective, $\tanh^{-1}: \openint {-1} 1 \to \R$ is a bijection.
Hence the result that $\struct {G, \circ} \cong \struct {\R,  +}$.
{{Qed}}
Theorem: Group/Examples/x+y over 1+xy

Let $G = \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.
Let $\circ: G \times G \to \R$ be the binary operation defined as:
:$\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$
The algebraic structure $\struct {G, \circ}$ is a group.

Proof:
To prove $G$ is isomorphic to $\struct {\R, +}$, it is sufficient to find a bijective homorphism $\phi: \to \R \to G$:
:$\forall x, y \in G: \map \phi {x + y} = \map \phi x \circ \map \phi y$
From Group Examples: $\dfrac {x + y} {1 + x y}$:
:the identity element of $G$ is $0$
:the inverse of $x$ in $G$ is $-x$.
This also holds for $\struct {\R, +}$.
This hints at the structure of a possible such $\phi$:
:$(1): \quad$ that it is an odd function
:$(2): \quad$ that it passes through $0$
:$(3): \quad$ that its image is the open interval $\openint {-1} 1$.
One such function is the hyperbolic tangent function $\tanh$, which indeed has the above properties on $\R$.
Then we have that:
{{begin-eqn}}
{{eqn | l = \map {\tanh } {x + y}
      | r = \dfrac {\tanh x + \tanh y} {1 + \tanh x \tanh y}
      | c = Hyperbolic Tangent of Sum
}}
{{eqn | r = \tanh x \circ \tanh y
      | c = 
}}
{{end-eqn}}
This demonstrates the homorphism between $\struct {\R,  +}$ and $\struct {G, \circ}$.
We have that Real Hyperbolic Tangent Function is Strictly Increasing over $\R$.
Hence from Strictly Monotone Real Function is Bijective, $\tanh: \R \to \openint {-1} 1$ is a bijection.
Hence the result that $\struct {\R,  +} \cong \struct {G, \circ}$.
{{Qed}}
Theorem: Group/Examples/inv x = 1 - x

Let $S = \set {x \in \R: 0 < x < 1}$.
Then an operation $\circ$ can be found such that $\struct {S, \circ}$ is a group such that the inverse of $x \in S$ is $1 - x$.

Proof:
Define $f: \openint 0 1 \to \R$ by:
:$\map f x := \map \ln {\dfrac {1 - x} x}$
Let us show that $f$ is a bijection by constructing an inverse mapping $g: \R \to \openint 0 1$:
:$\map g z := \dfrac 1 {1 + \exp z}$
Theorem: Group/Examples/Self-Inverse and Cancellable Elements

Let $S$ be a set with an operation which assigns to each $\tuple {a, b} \in S \times S$ an element $a \ast b \in S$ such that:
:$(1): \quad \exists e \in S: a \ast b = e \iff a = b$
:$(2): \quad \forall a, b, c \in S: \paren {a \ast c} \ast \paren {b \ast c} = a \ast b$
Then $\struct {S, \circ}$ is a group, where $\circ$ is defined as $a \circ b = a \ast \paren {e \ast b}$.

Proof:
We verify the group axioms, in the following order (for convenience):
Theorem: Complex Numbers under Addition form Infinite Abelian Group

Let $\C$ be the set of complex numbers.
The structure $\struct {\C, +}$ is an infinite abelian group.

Proof:
From Complex Numbers under Addition form Group, $\struct {\C, +}$ is a group.
Then:
:Complex Addition is Commutative
and:
:Complex Numbers are Uncountable.
{{qed}}
Theorem: Non-Zero Complex Numbers under Multiplication form Infinite Abelian Group

Let $\C_{\ne 0}$ be the set of complex numbers without zero, that is:
:$\C_{\ne 0} = \C \setminus \set 0$
The structure $\struct {\C_{\ne 0}, \times}$ is an infinite abelian group.

Proof:
From Non-Zero Complex Numbers under Multiplication form Group, $\struct {\C_{\ne 0}, \times}$ is a group.
Then we have:
:Complex Multiplication is Commutative
and:
:Complex Numbers are Uncountable.
{{qed}}
Theorem: Real Numbers under Addition form Infinite Abelian Group

Let $\R$ be the set of real numbers.
The structure $\struct {\R, +}$ is an infinite abelian group.

Proof:
From Real Numbers under Addition form Group, $\struct {\R, +}$ is a group.
Then:
:Real Addition is Commutative
and:
:Real Numbers are Uncountably Infinite.
{{qed}}
Theorem: Non-Zero Real Numbers under Multiplication form Abelian Group

Let $\R_{\ne 0}$ be the set of real numbers without zero:
:$\R_{\ne 0} = \R \setminus \set 0$
The structure $\struct {\R_{\ne 0}, \times}$ is an uncountable abelian group.

Proof:
Taking the group axioms in turn:
=== {{Group-axiom|0|nolink}} ===
From Non-Zero Real Numbers Closed under Multiplication: Proof 2, $\R_{\ne 0}$ is closed under multiplication.
Note that proof 2 needs to be used specifically here, as proof 1 rests on this result.
{{qed|lemma}}
=== {{Group-axiom|1|nolink}} ===
Real Multiplication is Associative.
{{qed|lemma}}
=== {{Group-axiom|2|nolink}}: Real Multiplication Identity is One ===
{{:Real Multiplication Identity is One}}
{{qed|lemma}}
=== {{Group-axiom|3|nolink}}: Inverse for Real Multiplication ===
{{:Inverse for Real Multiplication}}
{{qed|lemma}}
=== $\text C$: Commutativity ===
Real Multiplication is Commutative.
{{qed|lemma}}
=== Infinite ===
Real Numbers are Uncountably Infinite.
{{qed}}
Theorem: Non-Zero Real Numbers under Multiplication form Abelian Group

Let $\R_{\ne 0}$ be the set of real numbers without zero:
:$\R_{\ne 0} = \R \setminus \set 0$
The structure $\struct {\R_{\ne 0}, \times}$ is an uncountable abelian group.

Proof:
We have Real Numbers under Multiplication form Monoid.
From Inverse for Real Multiplication, the non-zero numbers are exactly the invertible elements of real multiplication.
Thus from Invertible Elements of Monoid form Subgroup of Cancellable Elements, the non-zero real numbers under multiplication form a group.
From:
:Real Multiplication is Commutative
:Subset Product within Commutative Structure is Commutative
it follows that this group is also Abelian.
{{qed}}
Theorem: Non-Zero Real Numbers under Multiplication form Abelian Group

Let $\R_{\ne 0}$ be the set of real numbers without zero:
:$\R_{\ne 0} = \R \setminus \set 0$
The structure $\struct {\R_{\ne 0}, \times}$ is an uncountable abelian group.

Proof:
From Non-Zero Real Numbers under Multiplication form Group, $\struct {\R_{\ne 0}, \times}$ forms a group.
{{qed|lemma}}
From Real Multiplication is Commutative it follows that $\struct {\R_{\ne 0}, \times}$ is abelian.
{{qed|lemma}}
From Real Numbers are Uncountably Infinite it follows that $\struct {\R_{\ne 0}, \times}$ is an uncountable abelian group.
{{qed}}