Contents
Context
Algebra
Monoid theory
monoid theory in algebra:

monoid, infinitymonoid

monoid object, monoid object in an (infinity,1)category

Mon, CMon

monoid homomorphism

submonoid, quotient monoid?

divisor, multiple?, quotient element?

inverse element, unit, irreducible element

ideal in a monoid

principal ideal in a monoid

commutative monoid

cancellative monoid

GCD monoid

unique factorization monoid

Bézout monoid

principal ideal monoid

group, abelian group

absorption monoid

free monoid, free commutative monoid

graphic monoid

monoid action

module over a monoid

localization of a monoid

group completion

endomorphism monoid

super commutative monoid
Contents
Idea
A ternary function which simultaneously exhibits an action on a set from both the left and the right side.
Sets with biactions are the bimodule objects internal to Set.
Definition
Given a set $S$ and monoids $(M, e_M, \mu_M)$ and $(N, e_N, \mu_N)$, a $M$$N$biaction or twosided action is a ternary function $\alpha:M \times S \times N \to S$ such that

for all $s \in S$, $\alpha(e_M, s, e_N) = s$

for all $s \in S$, $a \in M$, $b \in M$, $c \in N$, and $d \in N$, $\alpha\big(a, \alpha(b, s, c), d\big) = \alpha\big(\mu_M(a, b), s, \mu_N(c, d)\big)$
Left and right actions
The left $M$action is defined as
$\alpha_M(a, s) \coloneqq \alpha(a, s, e_N)$
for all $a \in M$ and $s \in S$. It is a left action because
$\alpha_M(e_M, s) = \alpha(e_M, s, e_N) = s$
$\alpha_M\big(a, \alpha_L(b, s)\big) = \alpha\big(a, \alpha(b, s, e_N), e_N\big) = \alpha\big(\mu_M(a, b), s, \mu_N(e_N, e_N)\big) = \alpha\big(\mu_M(a, b), s, e_N\big) = \alpha_M\big(\mu_M(a, b), s\big)$
The right $N$action is defined as
$\alpha_N(s, c) \coloneqq \alpha(e_M, s, c)$
for all $c \in N$ and $s \in S$. It is a right action because
$\alpha_N(s, e_N) = \alpha(e_M, s, e_N) = s$
$\alpha_N\big(\alpha_N(s, c), d\big) = \alpha\big(e_M, \alpha(e_M, s, c), d\big) = \alpha\big(\mu_M(e_M, e_M), s, \mu_N(c, d)\big) = \alpha\big(e_M, s, \mu_N(c, d)\big) = \alpha_N\big(s, \mu_N(c, d)\big)$
The left $M$action and right $N$action satisfy the following identity:
 for all $s \in S$, $a \in M$ and $c \in N$, $\alpha_M\big(a, \alpha_N(s, c)\big) = \alpha_N\big(\alpha_M(a, s), c\big)$.
This is because when expanded out, the identity becomes:
$\alpha\big(a, \alpha(e_M, s, c), e_N\big) = \alpha\big(e_M, \alpha(a, s, e_N), c\big)$
$\alpha\big(\mu_M(a, e_M), s, \mu_N(c, e_N)\big) = \alpha\big(\mu_M(e_M, a), s, \mu_N(e_N, c)\big)$
$\alpha(a, s, c) = \alpha(a, s, c)$
See also