How does the following definiton of homomorphism preserve structure?

Question

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?

Answer 1

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.

Answer 2

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.

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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.

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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.