Let $(A,O)$ be an algebraic system ($A\ne \varnothing$ is the underlying set and $O$ is the set of operators on $A$ ).

Subalgebras of an Algebra

Let $B$ be a non-empty subset of $A$ . $B$ is closed under operators of $A$ if for each $n$ -ary operator $\omega_A$ on $A$ , and any $b_1,\ldots,b_n\in B$ , we have $\omega_A(b_1,\ldots,b_n)\in B$ .

Suppose $B$ is closed under operators of $A$ . For each $n$ -ary operator $\omega_A$ on $A$ , we define $\omega_B:B^n\to B$ by $\omega_B(b_1,\ldots,b_n):= \omega_A(b_1,\ldots,b_n)$ . Each of these operators is well-defined and is called a restriction (of the corresponding $\omega_A$ ). Furthermore, $(B,O)$ is a well-defined algebraic system, and is called the subalgebra of $(A,O)$ . When $(B,O)$ is a subalgebra of $(A,O)$ , we also say that $(A,O)$ is an extension of $(B,O)$ .

$(A,O)$ is clearly a subalgebra of itself. Any other subalgebra of $(A,O)$ is called a proper subalgebra.

Remark. If $(A,O)$ contains constants, then any subalgebra of $(A,O)$ must contain the exact same constants. For example, the ring $\mathbb{Z}$ of integers is an algebraic system with no proper subalgebras. Indeed, if $R$ is a subring of $\mathbb{Z}$ , $1\in R$ , so $R=\mathbb{Z}$ .

Since we are operating under the same operator set, we can, for convenience, drop $O$ and simply call $A$ an algebra, $B$ a subalgebra of $A$ , etc... If $B_1,B_2$ are subalgebras of $A$ , then $B_1\cap B_2$ is also a subalgebra. In fact, given any set of subalgebras $B_i$ of $A$ , their intersection $\bigcap B_i$ is also a subalgebra.

Generating Set of an Algebra

Let $C$ be any subset of an algebra $A$ . Consider the collection $[C]$ of all subalgebras of $A$ containing $C$ . This collection is non-empty because $A\in [C]$ . The intersection of all these subalgebras is again a subalgebra containing the set $C$ . Denote this subalgebra by $\langle C\rangle$ . $\langle C\rangle$ is called the subalgebra spanned by $C$ , and $C$ is called the spanning set of $\langle C\rangle$ . Conversely, any subalgebra $B$ of $A$ has a spanning set, namely itself: $B=\langle B\rangle$ .

Given a subalgebra $B$ of $A$ , a minimal spanning set $X$ of $B$ is called a generating set of $B$ . By minimal we mean that the set obtained by deleting any element from $X$ no longer spans $B$ . When $B$ has a generating set $X$ , we also say that $X$ generates $B$ . If $B$ can be generated by a finite set, we say that $B$ is finitely generated. If $B$ can be generated by a single element, we say that $B$ is cyclic.

Remark. $\langle \varnothing\rangle =$ the subalgebra generated by the constants of $A$ . If no such constants exist, $\langle \varnothing \rangle :=\varnothing$ .

From the discussion above, the set of subalgebras of an algebraic system forms a complete lattice. Given subalgebras $A_i$ , $\bigvee A_i$ is the intersection of all $A_i$ , and $\bigvee A_i$ is the subalgebra $\langle \bigcup A_i\rangle$ . The lattice of all subalgebras of $A$ is called the subalgebra latttice of $A$ , and is denoted by $\operatorname{Sub}(A)$ .