We know that every group is isomorphic to a subgroup of a symmetric group. So, the question arises, is every monoid a submonoid of a full transformation monoid, where a full transformation monoid is the set of all functions from a set $X$ to $X$, the operation is composition of functions, and the identity element is the identity function.
Is every monoid isomorphic to a submonoid of a full transformation monoid
monoid
Best Answer
Elaborating on Eric's comment, the exact same argument works for monoids.
Specifically, given a monoid $\mathcal{M}=(M,*)$, we have a natural action of $\mathcal{M}$ on the set $M$ given by left multiplication: each $a\in M$ corresponds to the function $f_a: b\mapsto a*b$.
The map $act: a\mapsto f_a$ is easily checked$^1$ to be an injective monoid homomorphism from $\mathcal{M}$ to $Sym(M)$, so $\mathcal{M}$ is isomorphic to the submonoid of $Sym(M)$ with domain $im(act)$.
$^1$Note that the presence of an identity element is crucial here for showing injectivity (if $a\not=b$ then $f_a(e)=a\not=b=f_b(e)$). Consider the two-element semigroup $\mathcal{S}$ with underlying set $\{0,1\}$ and multiplication given by the constant map $(x,y)\mapsto 1$. If we run the construction above we get a non-injective homomorphism from $\mathcal{S}$ to $Sym(\{0,1\})=S_2$. It's a good exercise to check whether $\mathcal{S}$ does in fact embed into any $Sym(X)$.
Associativity is also necessary, and we know it has to be: no non-associative magma can embed into any $Sym(X)$. So what goes wrong if we don't assume associativity, in the above argument?