TITLE: Basic examples of monoids?
QUESTION [8 upvotes]: What are some (simple/elementary) examples of noncommutative monoids with no additional structure? I'm having a hard time thinking of examples of "pure" monoids that aren't monoids simply because they are groups...
I've read this and this and some of this, but would like more examples that presuppose little to no algebra.

REPLY [4 votes]: Let $\mathfrak{M}_n(\mathbb{R})$ be the set of real square matrices of order $n$ and $\circ$ the operation of matrix product, then $(\mathfrak{M}_n(\mathbb{R}),\circ)$ is a monoid for all $n\ge 2$.
It is not commutative: for $n=2$ it is
$$\begin{bmatrix}
0 & 1\\0 & 0
\end{bmatrix}\circ\begin{bmatrix}
1 & 0\\0 & 0
\end{bmatrix}\ne\begin{bmatrix}
1 & 0\\0 & 0
\end{bmatrix}\circ\begin{bmatrix}
0 & 1\\0 & 0
\end{bmatrix}.$$
It is not a group, since all matrices with determinant $0$ have no inverse. For example, for $n=2$, the identity element is
$$\begin{bmatrix}
1 & 0\\0 & 1
\end{bmatrix}$$
and there's no $A\in\mathfrak{M}_2(\mathbb{R})$ such that
$$A\circ\begin{bmatrix}
1 & 0\\0 & 0
\end{bmatrix}=\begin{bmatrix}
1 & 0\\0 & 1
\end{bmatrix}$$
(SLM, pp. 37-38).

Let $A$ be a non empty set, $A^A$ denote the set of all functions $f\colon A\to A$ and $\circ$ denote the composition of functions defined, for all $f,g\in A^A$, as
$$(f\circ g)(x)=f(g(x)),\ \forall x\in A.$$
Then $(A^A,\circ)$ is a monoid with identity element $Id$, defined
$$Id(x)=x,\ \forall x\in A$$
(SVTL, pp. 164-165). Again, such monoid is not commutative and is not a group: can you prove it?