# Question about the definition of a homomorphism

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.

• No, there is no implication that the operation is the same. This is a consequence of abuse of notation in which we do not generally explicitly name the operation of a group, and instead of use juxtaposition for 'whatever it is the operation of the group is". Formally speaking you should say that you have a group $G$ with operation $\cdot$, and a group $K$ with operation $\odot$, and a function $f\colon G\to K$, and $f$ is a homomorphism if and only if for all $x,y\in G$, $f(x\cdot y) = f(x)\odot f(y)$. But by absue of notation we just say $f(xy)=f(x)f(y)$. Sep 26, 2022 at 3:11
• Thank you. Fraleigh unfortunately uses examples immediately after where the operation between groups is the same, such as a homomorphism where $f(x+y) = f(x)+f(y)$. Because he used examples where it's ordinary addition in both groups, I felt that I had to perform a "guessing game" to try and interpret what he was saying. I obviously also take notational shortcuts when doing my own work and I understand that professionals can understand this notation, but I feel casually leaking it into definitions might cause issues if the nukes drop and humanity has to rebuild knowledge using only books.
– Nate
Sep 26, 2022 at 3:37

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").

No, the distinction between homomorphism and isomorphism is NOT about whether or not the operations are the same on both sides of the equation. Rather, an isomorphism means a map $$f$$ which is a homomorphism, is bijective, and $$f^{-1}$$ is a homomorphism. For groups, the bijectivity is the key thing, and I find it hard to believe that the textbook doesn't mention bijectivity... perhaps you overlooked this?

More explicitly:

Definition.

Let $$(S_1,*_1)$$ and $$(S_2,*_2)$$ be sets equipped with respective binary operations. A homomorphism of $$(S_1,*_1)$$ into $$(S_2,*_2)$$ is a function $$f:S_1\to S_2$$ such that for all $$x,y\in S_1$$, we have that $$f(x*_1y)=f(x)*_2f(y).$$

An isomorphism of $$(S_1,*_1)$$ onto $$(S_2,*_2)$$ means a function $$f:S_1\to S_2$$ such that $$f$$ is bijective, $$f$$ is a homomorphism, and $$f^{-1}$$ is a homomorphism.

Now, in the definition of isomorphism, the third condition that $$f^{-1}$$ be a homomorphism is redundant; but the fact that it is redundant needs to be proven.

In general, homomorphisms/morphisms are meant to 'respect the structure'. So, if you're just dealing with sets with binary operations, you only have one condition to check (for all $$x,y\in S_1$$, $$f(x*_1y)=f(x)*_2f(y)$$). The more structure you add, the more that is required to be considered a homomorphism.

For example, if you consider a group $$(G,\cdot_G)$$, then it has 3 pieces of structure. The first is the binary map $$\cdot_G:G\times G\to G$$, the next is the existence of a special identity element $$e_G\in G$$, and the last is the existence of an 'inversion map' $$\text{inv}_G:G\to G$$, which sends $$g\mapsto g^{-1}$$. So, the definition of a homomorphism from a group $$(G,\cdot_G)$$ into a group $$(H,\cdot_H)$$ is a function $$f:G\to H$$ which satisfies three conditions:

• for all $$x,y\in G$$, $$f(x\cdot_Gy)=f(x)\cdot_Hf(y)$$ (respecting binary operations)
• $$f(e_G)=e_H$$ (respecting identity element)
• for all $$x\in G$$, $$f(\text{inv}_G(x))=\text{inv}_H(f(x))$$, i.e $$f(x^{-1})=(f(x))^{-1}$$ (respecting inversion).

It turns out that for groups, the second two conditions follow as consequences of the first, which is why some textbooks merely use the first bullet point as a definition, and prove the remaining two as theorems (while other textbooks use all three as a definition to show the recurring idea of 'respecting structure' and show afterwards the last two are 'redundant').

I'm sure you can guess the definition of an isomorphism of a group: it is a function $$f:G\to H$$ such that it is bijective, and $$f$$ is a group homomorphism, and $$f^{-1}$$ is a group homomorphism.

At this point I should make the comment that it is only in the first few pages of an algebra textbook that we explicitly write out the operations as $$*_1,*_2,\cdot_G,\cdot_H$$. After some time, we get lazy and just write everything as $$\cdot$$ (or $$\circ$$ or $$+$$ or whatever is context-appropriate). It is the duty of the student to keep in mind, if the confusion arises, where each operation is precisely defined.

Hopefully the above is clear, but to hammer the point home:

• if we talk about other algebraic structures, for example fields, then these consist of the data $$(F,+,\cdot, 0,1)$$, so there are two operations, there are two special elements, and there is a compatibility conditions. A homomorphism is supposed to 'respect' all of these (but again, it will turn out some of these conditions are redundant). An isomorphism means a bijective map such that both the map and its inverse are homomorphisms.
• You can also come to the situation of linear algebra, where you have a vector space over a field $$(V,F,+,\cdot, 0_V)$$, where now this $$+,\cdot$$ are the addition on the vector space, and scalar multiplication (not those of $$F$$), and $$0_V$$ is the special zero element of $$V$$. A 'vector space homomorphism', $$T$$, from $$(V,F,+_V,\cdot_V, 0_V)$$ to $$(W,F,+_W,\cdot_W, 0_W)$$ is more commonly simply called a linear map from $$V$$ into $$W$$, and it is a function which has to 'respect' all this structure (but turns out a lot of it follows easily, so the definition simplifies to: for all $$x,y\in V$$ and all $$c\in F$$, $$T(cx+y)=cT(x)+T(y)$$). An isomorphism of vector spaces means as usual a bijective linear map with linear inverse (though the inverse is automatically linear). An extra piece of terminology: if you look at linear maps $$T:V\to V$$ having the same vector space as domain and target, then these are sometimes called endomorphisms.
• If you study topology, then the data you have is $$(X,\tau_X)$$, where $$X$$ is a given set, $$\tau_X\subset \text{power set}(X)$$ is a collection known as a topology on $$X$$. Here the 'homomorphisms' are called simply 'continuous maps'. An 'isomorphism', more commonly called a homeomorphism (note the $$e$$) means a bijective map such that the map and its inverse are both continuous. Here, it is good to note that $$f$$ being continuous doesn't automatically imply $$f^{-1}$$ is continuous. So, the condition on the inverse is no longer superfluous!

The point is that isomorphisms in whatever category of spaces you're working on tell us that things are 'same' up to a 'relabelling' (i.e moving things around by a bijective map).

• "For groups, the bijectivity is the key thing, and I find it hard to believe that the textbook doesn't mention bijectivity... perhaps you overlooked this?" – he stumbled over the fourth criterion with a different notation than for homomorphisms. I assume one of the others is the bijectivity (or something equivalent to it). Sep 26, 2022 at 22:21
• @PaŭloEbermann lol I should have read more closely then Sep 26, 2022 at 23:34