Definition 1. Let $\Omega$ and $E$ be two sets. A mapping of $\Omega$
into the set $E^E$ of mappings of into itself is called an action of
$\Omega$ on $E$.Let $\alpha\mapsto\ f_{\alpha}$ be an action of $\Omega$ on $E$. The
mapping $(\alpha,x)\mapsto f_{\alpha}(x)$ (resp. $(x,\alpha)\mapsto
f_{\alpha}(x)$) is called the law of left (resp. right) action of
$\Omega$ on E associated with the given action of $\Omega$ on E.The element $f_{a}(x)$ of $E$ is denoted by left (resp. right)
multiplicative notation $\alpha.x$ (resp. $x.\alpha$).
Question 1: Since both $\alpha.x$ and $x.\alpha$ denote the same element of $E$, namely $f_{\alpha}(x)$, does that mean $\alpha.x=x.\alpha$ (as elements that is; I know they are products of different operations) ?
Definition 2. Let $M$ be a monoid, its law written multiplicatively and its identity element denoted by $e$, and $E$ a set. An action
$\alpha\mapsto\ f_{\alpha}$ of $M$ on $E$ is called a left (resp.
right) operation of $M$ on $E$ if $f_{e}=id_{E}$ and
$f_{\alpha\beta}=f_{\alpha}\circ f_{\beta}$ (resp.
$f_{\alpha\beta}=f_{\beta}\circ f_{\alpha}$) for all $\alpha, \beta\in
M$.If the law of action corresponding to the action of $M$ is denoted by
left (resp. right) multiplication, the fact that this action is a left
(resp. right) operation may be expressed by the formulae$$e.x=x;\ \alpha.(\beta.x)=(\alpha\beta).x$$ $$x.e=x;\
(x.\alpha).\beta=x.(\alpha\beta)$$Let $M$ be a monoid that operates on a set $E$ on the left and let
$M^{op}$ denote the opposite monoid to $M$. Under the same action, the
monoid $M^{op}$ operates on the set $E$ on the right.
Question 2: What is meant by "under the same action, the monoid $M^{op}$ operates on the set $E$ on the right"? —Normally, a right operation of a monoid $M$ on a set $E$ can be understood as a monoid homomorphism $M\rightarrow (E^{E})^{op}$; the proof with the function notation is easy:
$$f_{\alpha\cdot_{op}\beta}=f_{\beta\alpha}=f_{\beta}\circ f_{\alpha}=f_{\alpha}\circ_{op} f_{\beta}.$$
On the other hand, how can one prove it using the multiplicative notation? I think you need the equality referred to in Question 1, no?
Thank you in advance.
Best Answer
Question 1. Yes, $x\cdot \alpha$ and $\alpha \cdot x$ represent the same element, but the point is that you have to choose between the left and the right multiplicative notation. Using both of them at the same time is confusing.
Question 2. If $M$ acts on the left on $\Omega$, one has $1\cdot x = x$ and $a \cdot(b\cdot x) = (ab) \cdot x$. Since the product $*$ on $M^{op}$ is defined by $a * b= ba$, writing the second equation as a right action yields $(x \cdot b) \cdot a = x \cdot (ab) = x \cdot (b * a)$. Thus $M^{op}$ acts on the right on $\Omega$.