I am very new at the topic of algebras in functional analysis and I am currently self-studying through the text "Complete Normed Algebras" by Bonsall and Duncan. In Chapter 1 section 9, page 49, the definition of a left $ A $-Module is given as follows:
"Let $ A $ be an algebra over $ \mathbb{F} $ (= $ \mathbb{R} $ or $ \mathbb{C} $), and $ M $ a linear space over $ \mathbb{F} $. $ M $ is said to be a left $ A $-module if a mapping $ (a, m) \to am $ of $ A \times M $ into $ M $ is specified which satisfies the following axioms:
LM1: For each fixed $ a \in A $, the mapping $ m \to am $ is linear on $ M $.
LM2: For each fixed $ m \in M $, the mapping $ a \to am $ is linear on $ M $.
LM3: $ a_1(a_2m) = (a_1a_2)m $ for all $ a_1, a_2 \in A $ and for all $ m \in M $.
The specified mapping $ (a, m) \to am $ is called the module multiplication."
A similar definition can be given to define what it means for a linear space to be a right $ A $-module (but is not explicitly given in the text)
If $ A $ is an algebra with product $ \cdot $, it was defined earlier in the chapter that corresponding reversed algebra denoted $ \mathrm{Rev}(A) $ is the set $ A $ with the same operations of addition and scalar multiplication on $ A $, but with a new product $ * $ called the reversed product, defined for all $ x, y \in A $ by:
$$ x * y = y \cdot x $$
The author then states that if $ A $ is an algebra and $ M $ is a right $ A $-module then $ M $ is also a left $ \mathrm{Rev}(A) $-module.
This seems intuitively obvious but I am still having a bit of difficulty in explicitly showing this because I'm not sure how the product in $ A $ or in $ \mathrm{Rev}(A) $ comes into play.
For example, if $ M $ is a right $ A $-module, then for every fixed $ a \in A $, $ m \to ma $ is linear. So for all $ m_1, m_2 \in M $ we have $ (m_1 + m_2)a = m_1a + m_2a $. I'm not sure how to use the reversed product here, since it is only defined between two elements of $ A $ (and $ m_1, m_2 $ may not be in $ A $?) I think I am probably missing something quite obvious and getting confused with there being three products here: the product on $ A $, the reversed product, and then the module product.
Any help would be greatly appreciated!
You don't need the product on $A$ or on $\mathrm{Rev}(A)$ in the definition of the induced left module, until the point when it turns to prove the axiom
LM3.If $M$ is a right $A$-module, then, what a surprise, $am:=ma$ defines the left $\mathrm{Rev}(A)$ action.