algebra

Binary operation

$A\ne \varnothing$, binary operation on the set a: $\circ :A\times A\to A$

$$(x,y)\mapsto x\circ y$$

$(A,\circ )$ … algebraic structure or “grupoid” or “algebra” or “magma” or …

Example: $(\mathbb{N},+),(\mathbb{N},\cdot ),(\mathbb{Z},\cdot ),\dots$

closure

Closure

$(A^A,\circ )$ … set of all functions from $A$ to $A$ with composition as operation
$B^A:=\{f:A\to B\}$ … set of all functions from $A$ to $B$ (intuition for this notation)
$B\ne \varnothing ,B\subseteq A,\text{closure}:=\forall x,y\in B:x\circ y\in B$
$B$ is closed under $\circ$, which means any operation on elements of $B$ results in an element of $B$.
$B\subseteq A$ is called a substructure of $(A,\circ )$.

In simple words: A set is closed under an operation if the result of the operation on any points in the set is within the same set.

EXAMPLE

$(\mathbb{N},+)\le (\mathbb{Z},+)$
$\mathbb{N}\subseteq \mathbb{Z}$ is closed under addition.
$\mathbb{Q}\setminus \{0\},\cdot \le (\mathbb{Q},\cdot )$

The set of all bijective functions $S_A$ is a substructure of $(A^A,\circ )$.

$S_A:=\{f|f:A\mapsto A,f\text{ bijective}\}$, $(S_A,\circ )\le (A^A,\circ )$
This means that the composition of two bijective functions is also bijective.
I.e. the set of all bijective functions is closed under composition.

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Transclude of semigroup

Transclude of monoid#^0293a3

Transclude of group#^a092fc

Transclude of abelian-group#^cc4dda

Subgroup

A subgroup is a substructure which also has group properties:
$(G,\circ )$ group, $U\subseteq G$ subgroup $⟺(U,\circ )$ group.
We denote it as $U\le G$.

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Ordnung/order = cardinality of a group