(related to Theorem: Construction of Groups from Commutative and Cancellative Semigroups)

- 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.

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).∎

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