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Concatenation joins two strings end to end, retaining the order of their symbols. For example, joining $ab$ and $ba$ gives $abba$. It has the associative property: joining $(uv)w$ or $u(vw)$ gives the same word over an alphabet. The empty word is its identity element. All finite words over an alphabet with this binary operation form a free monoid.

The matricesgenerate a free monoid in ${\mathrm{SL}}_2(\mathbb{Z})$. Hence the $2^n$ positive words of length $n$ are distinct and ${\mathrm{SL}}_2(\mathbb{Z})$ has exponential growth of a group.

The free monoid ${\Sigma }^{\ast }$ on an alphabet $\Sigma$ consists of all finite words over an alphabet, including the empty word $\epsilon$, with concatenation as its binary operation. Distinct strings represent distinct elements: there are no imposed equalities between different sequences of letters. The empty word is its identity element, and concatenation has the associative property.^[ref]

For $\Sigma =\{a,b\}$, the elements begin $\epsilon ,a,b,aa,ab,ba,bb,\dots$. Both $(ab)a$ and $a(ba)$ give the same word $aba$, but $ab\ne ba$, so this monoid is not commutative. The length of a word over an alphabet obeys $\ell (uv)=\ell (u)+\ell (v)$; it is a monoid homomorphism to the nonnegative integers with addition. A nonempty word has no inverse element, since joining it to any other word cannot decrease its length to zero.

The defining benefit of being free is that letters can stand for arbitrary elements of another monoid. Choose a value $h(a)$ for each letter $a\in \Sigma$; then $h(a_1\cdots a_n)=h(a_1)\ast \cdots \ast h(a_n)$, with $h(\epsilon )=e$, is the unique monoid homomorphism extending those choices. Associativity and the identity element make this evaluation consistent for every finite word. For example, interpreting the letters as self-maps evaluates a word as a composed transformation; interpreting them as nonnegative counts evaluates it by addition.

A monoid is a semigroup with a two-sided identity element $e$: $e\ast a=a\ast e=a$ for every element $a$. Associativity handles the grouping of a nonempty sequence; the identity element also supplies the result for an empty sequence. This makes the same rules apply to combining zero, one, or many contributions. ^[ref].

The nonnegative integers $\{0,1,2,\dots \}$ with addition form a commutative monoid, with identity element zero. They do not form a group: a positive integer has no additive inverse in that set. All words over an alphabet with concatenation form a free monoid, whose identity element is the empty word. The full transformation monoid consists of all self-maps of a set, with function composition and the identity function.

These examples explain why monoids are useful: accumulation needs an operation and a starting value, but it need not be reversible. Adding counts, joining strings, and composing transformations all share the same grouping and identity laws. Requiring every element to have an inverse element gives a group, a stronger structure suitable for reversible operations.

For the free monoid on $x,y$, use the functionThe strict inequality is important. For any prefix $v$, $|vw|>|v|+n$ is equivalent to $|w|>n$. Whenever it holds, $w$ is nonempty and prefixing $v$ does not change its final letter. When it fails, both values are zero. Thuswhich proves equivariance for the diagonal action on $M\times \mathbb{N}$ and the trivial action on $B$. Hence $f$ is an element of the exponential described above. Let $g$ be the constant-zero equivariant function. They differ at $(x,0)$.

But every word $wy$ ends in $y$, sofor every $w,n$. The action of $y$ on $B^A$ is not injective, and $B^A$ is not decidable, even though $B$ is decidable. This is a nondecidable exponential of decidable monoid sets. The monoid condition in part (a) also fails here: $|pmq|=|p|+|m|+|q|$ cannot equal $|p|$ for a nonempty word $m$.

Consider the matricesTheir projective action on the positive real line isThese disjoint images give the monoid version of the ping-pong lemma: after removing a common initial letter, two different positive words act differently. Thus $A,B$ generate a free monoid. There are $2^n$ distinct positive words of length $n$, all lying in the radius-$n$ word ball for any generating set enlarged to contain $A,B$. Therefore the Exponential growth of SL2 of Z gives

A semigroup is an algebraic magma whose binary operation has the associative property. Its products can be regrouped while preserving the order of their elements. An identity element and inverse elements are not required.

For example, the positive integers with addition form a semigroup, since $(a+b)+c=a+(b+c)$ and a sum of positive integers is positive. They have no identity element: it would have to be zero, which is outside that set. Nonempty words over an alphabet with concatenation give a noncommutative example without an identity element. Including the empty word instead gives a free monoid.

The usefulness of a semigroup is that an ordered list has one product regardless of its parentheses. This permits replacing a consecutive block by its already computed value. Here the empty set is also allowed as a semigroup; a monoid is necessarily nonempty because it contains an identity element.