ABSTRACT ALGEBRAIC STRUCTURES OPERATIONS AND ...

Proof : The first thing to show is that the substructure is a semigroup. This
follows from T62. Next we show that there is a neutral element. A substructure
is per definition closed wrt each operation in the structure, and the neutral
element is (considered to be ) an operation. The neutral element will therefore
be a member of the substructure.
A somewhat more operational way is
72. Theorem: Criterion for submonoid
A subset B of en monoid (A, ♢ , e) is a submonoid if and kun if B is closed
wrt the operation (B ♢ B ⊆ B) and e ∈ B.
Even though the following result may seem somewhat meager it is useful to
have it coined for later reference (T77). But logically it belongs right here:
73. Theorem: The trivial submonoid
The singleton {e} is a submonoid
Proof : It is obviously a substructure and therefore a submonoid.
74. Theorem: Structures induced by a monoid.
If (A, ♢ , e) is a monoid and M is a set, then F(M, A) is a monoid with the
induced operations. The neutral element is constant mapping M ∋ x ↦→ e.
Product structures of monoids (Ai, ♢ i
e = (e1, . . . , en).
, ei) is a monoid. The neutral element is
A quotient structure (A, ♢ , e) modulo ∼ of a monoid is a monoid. The neutral
element is [e]∼.
Proof : We only need to check the axioms: In all three cases we know that the
underlying structure is a semigroup, which is the first axiom. Which remains
is to check the neutral element. This is straightforward in all the cases. Lets
do it.
For F(M, A) the induced operation is the constant mapping defined on M with
the constant value e. This is seen to be neutral by simple inspection..
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