Intuition behind group theory problem

Question

I'm looking at the following problem (3.1:1 in invitation to general algebra and universal constructions by Bergman), and hoping for some more intuition/explanation.

Suppose $F$ and $G$ are groups, and $a,b,c\in F$ and $\alpha,\beta,\gamma\in G$, and consider the following statements.

a) every group theoretic equation $p=q$ satisfied by $(a,b,c)$ in $F$ is also satisfied by $(\alpha,\beta,\gamma) \in G$

b) There exists a homomorphism $h:F\to G$ under which $h(a)=\alpha$, $h(b)=\beta$, $h(c)= \gamma$.

Show by example it is possible for a) to hold but b) to fail.

I stumbled on the following counterexample. Take the tuples $(0,2)\in \mathbb{Z}_4\times \mathbb{Z}_4$ and $(0,1)\in\mathbb{Z}_2\times \mathbb{Z}_2$. The subgroup $\langle 0,2\rangle\subsetneq \mathbb{Z}_4$ is isomorphic to $\mathbb{Z}_2$, so evidently any equation in the former holds in the latter. But no hom $h:\mathbb{Z}_4\to\mathbb{Z}_2$ with $h(0)=0$ and $h(2)=1$ can exist, since if $h(1)=x$ say, we have $x+x=h(1)+h(1)=h(1+1)=h(2)=1$, which has no solutions for $x$ in $\mathbb{Z}_2$.

Apart from finding this example, I have no intuition for why this is happening. Could someone provide some explanation? All that I can see is that while condition a) guarantees us a hom from the generated subgroup $\langle a,b,c\rangle\to G$, it is not always possible to extend this to a hom out of the full group $F\to G$

Answer 1

Some comments:

1. Given that the problem asks for an "example", I would say that calling it a "counterexample" is a little confusing. Generally, we speak about a counterexample as an example the shows that a particular statement is false. Had the question been "Does (a) imply (b)?" then you might talk about a counterexample to the implication. But if the problem says "Given a example showing that (a) may hold even if (b) does not", then on first reading I would expect a "counterexample" to be something showing that what Bergman wrote is somehow false, or that a question has a negative answer, which is not the case.

2. You are being asked for an example satisfying certain properties. Specifically, you should name two groups, and three elements in each group, and show that (a) holds for those two groups and two triples of elements, but that (b) does not. You did not actually do that. Instead, you should specify that triples of elements you are choosing. You also confuse matters by talking about "tuple[] $(0,2)\in\mathbb{Z}_4\times\mathbb{Z}_2$" when the question asks about ordered triples of elements of the group. While technically $(0,2)$ is a "tuple", we are not thinking of it as a tuple here, but rather as simply an element of the group.

That said, your example can be whipped into proper form. We take $F=\mathbb{Z}_4\times Z_2$ and $G=\mathbb{Z}_2$ (not $\mathbb{Z}_4$ as you do...). Then we take elements $a=b=(0,0)\in F$ and $c=(2,0)\in F$, and corresponding elements $\alpha=\beta=0\in G$, $\gamma=1\in G$. Then you argue that (a) holds for $(a,b,c)$ and $(\alpha,\beta,\gamma)$ because there is an isomorphism $\phi\colon\langle a,b,c\rangle \to \langle \alpha,\beta,\gamma\rangle$ sending $a$ to $\alpha$, $b$ to $\beta$, and $c$ to $\gamma$, so any identity $p=q$ that is satisfied by $(a,b,c)$ will necessarily be satisfied by $(\alpha,\beta,\gamma)$. That is correct. Then you argue that (b) fails to hold because there is no group homomorphism $f\colon F\to G$ that sends $c$ to $\gamma$, which is also correct.

There are a couple of points to this problem. This problem is in the section that motivates the construction of free groups (in three different ways, in sections 3.2, 3.3, and 3.4). The free group is characterized by the fact that you can specify an identification of (certain) elements with elements of a different group, and that will necessarily yield a homomorphism.

But more importantly, this problem follows Lemma 3.1.1, which reads:

Lemma 3.1.1. Suppose that $F$ and $G$ are groups, such that $F$ is generated by three elements $a,b,c\in F$, while $\alpha,\beta,\gamma$ are any three elements of $G$. Then the following conditions are equivalent:

(a) Every group theoretic relation $p=q$ satisfied by $(a,b,c)$ in $F$ is also satisfied by $(\alpha,\beta,\gamma)$ in $G$.

(b) There exists a group homomorphism $h\colon F\to G$ under which $a\mapsto \alpha$, $b\mapsto\beta$, $c\mapsto\gamma$.

Further, when these conditions hold, the homomorphism $h$ of (b) is unique.

If the assumption that $a$, $b$, and $c$ generate $F$ is dropped, one still has (b)$\implies$(a).

So, what is the point of the problem? The point of the problem is to show that this statement is best possible. That if you drop the assumption that $a$, $b$, and $c$ generate, you no longer have that (a) implies (b), even though this assumption is not needed for the implication that (b) implies (a). What you "see", in the final paragraph, is precisely the reason this does not work. That was the other point.