# How does the following definiton of homomorphism preserve structure?

Quoting from wikipedia:

A homomorphism is a map between two algebraic structures of the same type (that is of the same name), that preserves the operations of the structures. This means a map $${\displaystyle f:A\to B}$$ between two sets $${\displaystyle A}, {\displaystyle B}$$ equipped with the same structure such that, if $$\cdot$$ is an operation of the structure (supposed here, for simplification, to be a binary operation), then $${\displaystyle f(x\cdot y)=f(x)\cdot f(y)}$$ for every pair $${\displaystyle x}$$, $${\displaystyle y}$$ of elements of $${\displaystyle A}$$. One says often that $${\displaystyle f}$$ preserves the operation or is compatible with the operation.

And while dealing with groups we do have a binary operation '$$\cdot$$' such that the above definiton follows exactly.

I do not however see what 'it means' to preserve structure from the above definiton.

In particular if I had defined, for groups, homomorphism as follows:

A map $$H: G \to G_1$$ such that for all $$a,b\in G$$ $$H(a\cdot b)=(Ha)^2 \cdot (Hb)^2$$

Then in what sense would this definiton not preserve structure while the original one does?

• In case $f$ is also bijective, "preserve the strucure" can be given a very precise meaning: math.stackexchange.com/a/3929910/1007416
– user1007416
Commented Mar 3, 2022 at 11:57

## 2 Answers

Generally "to preserve structure" is a vague term (and I don't believe you should pay too much attention to it), and in fact in case of groups it is simply defined by $$f(xy)=f(x)f(y)$$. Intuition here is that in order to multiply $$f(x)$$ by $$f(y)$$ we only need to know how to multiply $$x$$ by $$y$$, and how $$f$$ acts on it. And so the structure of the image of $$f$$ is fully determined by its domain and $$f$$ itself (which is later formalized by the first isomorphism theorem).

But as I will show here, your $$H$$ is either trivial, or it doesn't actually tell us how to multiply elements in the image, even though at first glance it looks like it does.

First, let $$e$$ be the neutral element of $$G$$ and $$e'$$ the neutral element of $$G_1$$. Then $$H(e)=H(ee)=H(e)^2H(e)^2=H(e)^4$$

which means that $$H(e)^3=e'$$. This implies that either $$H(e)=e'$$ or $$H(e)$$ is an element of order $$3$$. Either way $$H(e)^2=H(e)^{-1}$$. And in that situation

$$H(x)=H(ex)=H(e)^2H(x)^2=H(e)^{-1}H(x)^2$$

which implies that $$H(e)=H(x)$$, i.e. $$H$$ is a constant function with the value being either $$e'$$ or an element of order $$3$$ (and in fact you can easily verify that any such choice is valid).

If $$H(x)=e'$$ then we end up coincidentally with a group homomorphism and it does preserve the structure, although in a very limited way.

If $$H(x)\neq e'$$ is an element of order $$3$$ then the situation is worse. The image of $$H$$ isn't even a subgroup of $$G_1$$ now. Moreover we have no idea how to calculate $$H(x)H(y)$$ given $$x,y\in G$$ and $$H$$ only. Our $$H$$ doesn't give us this information, it doesn't preserve that structure.

Of course the reasoning above is valid mainly because you've chosen the concrete form of $$H$$. In particular $$H(e)^2=H(e)^{-1}$$ was very helpful. Perhaps for other, more sophisticated formulas there will be many interesting $$H$$s to work with. But then the general rule of thumb is: the simpler the better.

let's look at an simple example why $$H(a \cdot b) = (H(a))^2 \cdot (H(b))^2$$ isn't a good definition. As you probably know, each group $$G$$ has an identity element $$e_G$$, which is an very important part of the structure, hence $$H$$ should preserve this element (i.e $$H(e_G) = e_{G_1}$$).

Let's imagine a map $$H : G \to G_1$$ with $$H(a \cdot b) = (H(a))^2 \cdot (H(b))^2$$. Then $$H(a) = e_{G_1}$$ holds for each element.

To see this, we use $$a \cdot e_G = a$$. We conclude $$H(a) = H(a \cdot e_G) = H(a)^2 \cdot H(e_G)^2 = H(a)^2 \cdot (e_{G_1})^2 = H(a)^2.$$ By taking the inverse on both side, we get $$e_{G_1} = H(a)$$.

Actually, $$H$$ is a group homomorphism, but a quite boring one.