In Fraleigh's abstract algebra book, he gives definitions on how structure carries over between two isomorphic binary structures. The first definition is given relatively early in the book (Section 3) as the fourth criteria for defining an isomorphism:
$$ \phi(x\space *\space y) = \phi(x) \space \bar* \space \phi(y)$$
Where $*$ and $\bar *$ are potentially two different operations.
Later on, in Section 13 he gives the definition of a homomorphism as such:
A map $\phi$ of a group G into a group $\bar G$ is a homomorphism if the homomorphism property
$$ \phi(ab) = \phi(a)\phi(b)$$
Holds for all $a,b \in G$
Unlike for his property of an isomorphism, where $*$ and $\bar *$ are implied to be different operations, his definition of a homomorphism implies that
$$ \phi(a \cdot b) = \phi(a) \space \cdot \space \phi(b)$$
Where the multiplicative notation implies that the operation $\cdot$ must be the same in both groups. Does this mean then for example, that defining an isomorphic relation between two binary structures allows something of the form
$$ \phi(a \space + \space b) = \phi(a) \space \times \space \phi(b) $$
Where $+$ and $\times$ are different operations, but I must define a homomorphism with the form
$$\phi(a \space \times \space b) = \phi(a) \space \times \space \phi(b) $$
Where $\times$ is the same operation for both groups? If that is the case, why can an isomorphism use two different operations, but a homomorphism must use the exact same operation? Thank you for your time.
Best Answer
It is a common convention to have the multiplication/operation implicit when using homomorphisms, or in general in topics in abstract algebra.
More explicitly: take $(G,\ast),(H,\circ)$ as groups. Then $\phi : G \to H$ is a homomorphism if and only if, $\forall a,b \in G$,
$$\phi(a \ast b) = \phi(a) \circ \phi(b)$$
But this notation can be a bit cumbersome sometimes. And notice something else: $G,H$ only have one operation defined on each. So if we were to define $ab$, sans the $\ast$, there's only really one way to interpret it that makes mathematical sense. Similarly, $\phi(a)\phi(b) \equiv \phi(a) \circ \phi(b)$, because there is no other sensible way to interpret that multiplication.
Be it isomorphisms or homomorphisms, this inherently stays the same: the multiplication implied is that of whatever group you lie in. Indeed, many times $\ast$ in one group will be a lot different from $\circ$ in another, even if the two are isomorphic (at least on a surface level, because, remember, "isomorphic" just means that "for all intents and purposes in this theory, the two items are functionally identical").