# Proof

### Uniqueness

• Assume, two groups $$(G_1,\ast)$$ and $$(G_2,\circ)$$ fulfil both properties (1) and (2).
• Because of the property (2), $$(G_1,\ast)$$ is a subgroup of $$(G_2,\ast)$$ and vice versa.
• Therefore, both groups are, in fact, the same groups.

### Existence

Our proof will be comprised of 4 steps:

• Step 1: Construct $$(G,\ast)$$ as a non-empty set with a binary operation "$$\ast$$"
• By assumption, $$(H,\circ)$$ is a commutative and cancellative semigroup.
• As a first step, we define an equivalence relation on the cartesian product $$H\times H=\{(a,b)~|~a,b\in H\}$$ as follows: $(a,b)\sim (c,d)\Leftrightarrow a\circ d=b\circ c~~~(a,b,c,d\in H).$
• Please note that this is indeed an equivalence relation, since it is reflexive, symmetric and transitive: * The reflexivity of "$$\sim$$" follows from the commutativity of $$H$$: $(a,b)\sim(a,b)\Leftrightarrow a\circ b=b\circ a.$ * Also the symmetry of "$$\sim$$" follows from the commutativity of $$H$$:$\begin{array}{ccl} (a,b)\sim (c,d)&\Leftrightarrow&a\circ d=b\circ c\\ &\Leftrightarrow&c\circ b=d\circ a\\ &\Leftrightarrow&(c,d)\sim (a,b). \end{array}$ * The transitivity of "$$\sim$$" follows from $$H$$ being cancellative and commutative: * Assume for $$a,b,c,d,e,f\in H$$: $\begin{array}{ccc} (a,b)\sim (c,d)&\Leftrightarrow&a\circ d=b\circ c\\ (c,d)\sim (e,f)&\Leftrightarrow&c\circ f=d\circ e.\\ \end{array}$ * After multiplying both sides of both equations, rearranging the variables and cancelling we get $\begin{array}{ccc} (a\circ d)\circ (c\circ f)&=&(b\circ c)\circ (d\circ e)\\ a\circ d\circ c\circ f&=&b\circ c\circ d\circ e\\ (a\circ f)\circ \cancel{(d\circ c)}&=&(b\circ e)\circ \cancel{(d\circ c)}\\ &(a,b)&\sim&(e,f).& \end{array}$
• We have just shown that $$\sim$$ is an equivalence relation.
• Therefore, each ordered pair $$(a,b)$$ represents its own equivalence class $$[a,b]\subseteq H\times H$$ and we can set $G:=( H\times H)/\sim$ as the set of all these equivalence classes.
• However, because the relation $$\sim$$ is symmetric and transitive, each equivalence class $$[a,b]$$ does not(!) depend on the special choice of its representative $$(a,b)$$.
• Therefore, we can define the operation $$\ast$$ on $$G$$ using any representatives, e.g. the representatives $$(a,b)$$ and $$(c,d)$$ as we do in the following $[a,b]\ast[c,d]:=[a\circ c,b\circ d].$
• Please note that $$G$$ does contain equivalence classes suitable to define the above binary operation $\ast,$ since $$H$$ is not empty and so $$G$$ is.
• Therefore, we have succeeded to define a non-empty set with a binary operation $$(G,\ast)$$.
• Step 2: Demonstrate that $$(G,\ast)$$ is a commutative group.
• We will show that the operation $$\ast$$ is associative and commutative and that $$G$$ contains a unique identity as well as each of its elements has a unique inverse.
• By construction, the operation $$\ast$$ is associative: $\begin{array}{rcl}([a,b]\ast[c,d])\ast[e,f]&=&([a\circ c,b\circ d])\ast[e,f]\\ &=&[a\circ c\circ e,b\circ d\circ f]\\ &=&[a,b]\ast([c\circ e,d\circ f])\\ &=&[a,b]\ast([c,d]\ast [e,f]).\end{array}$
• By construction, the operation $$\ast$$ is commutative: $\begin{array}{rcl}[a,b]\ast[c,d]&=&[a\circ c,b\circ d]\\ &=&[c\circ a,d\circ b]\\ &=&[c,d]\ast[a,b].\end{array}$
• Because $$H$$ is cancellative, it follows for any $$(h,h)\in H\times H$$ that the equivalence class $$[h,h]$$ is an identity element of $$(G,\ast$$): $\begin{array}{rcl}[a,b]\ast[h,h]&=&[a\circ \cancel h,b\circ \cancel h]\\ &=&[\cancel h\circ a,\cancel h\circ b]\\ &=&[h,h]\ast[a,b]\\ &=&[a,b].\end{array}$ Please note that the identity $$[h,h]$$ is unique because, as an equivalence class, it does not dependent on the special choice of the representative $$(h,h)$$.
• Because $$\circ$$ is commutative by assumption, for any $$[a,b]\in (G,\ast)$$, the equivalence class $$[a,b]$$ is the unique inverse: $\begin{array}{rcl}[a,b] \ast[b,a]&=&[a\circ b, b\circ a]\\ &=&[a\circ b, a\circ b]\\ &=&[h,h].\end{array}$
• Step 3: Demonstrate that $$(G,\ast)$$ fulfills property (1)
• We shall show that there is a subset $$S\subseteq G$$, which is a semigroup isomorphic to $$H$$, i.e. where $$(S,\ast)\simeq (H,\circ)$$.
• We will do so by constructing the subset $$S\subseteq G$$ and finding a bijective homomorphism $$f:H\mapsto S$$.
• For the identity element $$[h,h]\in G$$, consider the subset $S:=\{[a\circ h,h]~|~a\in H\},$ and set $f:\begin{cases}H&\mapsto S\\a&\mapsto [a\circ h,h].\end{cases}$
• Then $$f$$ surjective by definition and it is injective since we can conclude that $\begin{array}{ccl} f(a)=f(b)&\Leftrightarrow&[a\circ h,h]=[b\circ h,h]\\ &\Rightarrow&(a\circ h,h)\sim(b\circ h,h)\\ &\Rightarrow&(a\circ h)\circ h=h\circ (b\circ h)\\ &\Rightarrow&a\circ \cancel{h^2}=b\circ \cancel{h^2}\\ &\Rightarrow&a=b\\ \end{array}.$ Since $$f$$ is both, surjective and injective, it is bijective.
• $$f$$ is also a homomorphism, since we have $\begin{array}{ccl} f(a\circ b)&=&[(a\circ b)\circ h,h]\\ &=&[(a\circ b)\circ h,h]\ast[h,h]\\ &=&[a\circ b\circ h\circ h,h\circ h]\\ &=&[(a\circ h)\circ (b\circ h),h\circ h]\\ &=&[a\circ h,h]\ast[b\circ h,h]\\ &=&f(a)\ast f(b). \end{array}$
• Altogether, in steps 1 to 3, we have shown the isomorphism $$(S,\ast)\simeq (H,\circ)$$.
• Step 4: Demonstrate that $$(G,\ast)$$ fulfills property (2)
• We shall show that $$(G,\ast)$$ is minimal with the property (1).
• By construction of $$S\subseteq G$$, $$G$$ contains all elements of the form $$[a\circ h,h]$$ for all $$a\in H$$.
• Because $$G$$ is a group, it must also contain the inverse elements $$[h,a\circ h]$$ for all $$a\in H$$.
• Because $$(G,\ast)$$ is closed under $$\ast$$, we also have for all $a,b\in H$: $\begin{array}{rcl}[a\circ h,h]\ast[h,b\circ h]&=&[a\circ \cancel{h\circ h},b\circ \cancel{h\circ h}]\\ &=&[a,b]\end{array}.$
• This demonstrates that we cannot reduce $$G$$ any further, e.g. by taking away any single equivalence class $$[a,b]$$.
• Therefore $$G$$ is minimal with the property (1).

Github: ### References

#### Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013