# On the confusing definition of free group in Munkres

This question arose from a statement from Munkres Section 69 that is seeming contradictory to his definition of free group.

He defines the free group in the following way:

Let $$\{a_{\alpha}\}$$ be a family of elements of a group $$G$$. Suppose each $$a_{\alpha}$$ generates an infinite cyclic subgroup $$G_{\alpha}$$ of $$G$$. If $$G$$ is the free product of the groups $$\{G_{\alpha}\}$$, then $$G$$ is said to be a free group, and the family $$\{a_{\alpha}\}$$ is called a system of free generators for $$G$$.

Then on the page 424, when he talked about generators and relations, he said

Given $$G$$, suppose we are given a family $$\{a_{\alpha}\}_{\alpha\in J}$$ of generators of $$G$$. Let $$F$$ be the free group on the element $$\{a_{\alpha}\}$$. Then the obvious map $$h(a_{\alpha})=a_{\alpha}$$ of these elements into $$G$$ extends to a homomorphism $$h:F\longrightarrow G$$ that is surjective.

I have no problem with the third sentence, it is the extension theorem, or universal mapping property.

However, in the second sentence, he directly put $$F$$ to be free group on $$\{a_{\alpha}\}$$. Can he do that? I mean, in the definition, he requires $$a_{\alpha}$$ generates an infinite cyclic group, but here perhaps some $$a_{\alpha}$$ has finite order, which cannot generate infinite cyclic group.

I tried generate such free group on a subset of $$\{a_{\alpha}\}$$ which consists of all $$a_{\alpha}$$ with infinite order. But then $$h$$ cannot be surjective.

What am I missing here? Thank you!

When we say "the free group on $$\{a_\alpha\}$$", it means that you have to treat the $$a_\alpha$$ as abstract symbols, forgetting for a second that they are elements of the group $$G$$.

So the group $$F$$ is a free group with generators $$a_\alpha$$, but the product is completely different from the product in $$G$$. It does not matter if $$a_\alpha$$ has finite order in $$G$$, it always has infinite order in $$F$$ since $$F$$ is a free group. Likewise, any relation between the $$a_\alpha$$ that holds in $$G$$ disappears in $$F$$.

Maybe it would have been clearer if the book said: let $$\{b_\alpha\}$$ be symbols indexed by the same set as $$\{a_\alpha\}$$, and let $$F$$ be the free group on $$\{b_\alpha\}$$, then we define $$h:F\to G$$ by $$h(b_\alpha)=a_\alpha$$.

On the other hand it is very convenient to use the same symbols, it makes the map intuitive to write.

• Thanks! Now I understand the whole logic of this section. In lemma 69.1, he actually gives a extension lemma (universal mapping property) that makes sure you can always have such map. Therefore we can always define a free group on a set of generators which actually belongs to other group. Thanks! – JacobsonRadical Nov 14 at 17:30

That's not a great definition. I'd define the free group on a set $$S$$ to be the group $${\cal F}(S)$$ such that any function $$f:S\to X$$ with $$X$$ a group extends to a unique homomorphism $$f:{\cal F}(S) \to X$$. (It's not clear a priori that such a group exists or is unique. The latter is easy from the definition; the former requires a bit of work.) Munkres is then saying that $$F$$ is the free group on a set $$S\subset F$$ iff the obvious map $$*_{s\in S}{\cal F}(s) \to F$$ is an isomorphism. (Implicit in that definition is the notion of free product and the fact that the free product of free groups is itself free. I don't have a copy of the book offhand (and I have to admit that I think Munkres' algebraic topology book is terrible), so I don't know what exactly he's doing there.)

In the latter part, Munkres is talking about choosing a set of generators $$S = \{s_\alpha\}_{\alpha\in A}$$ of $$G$$, then defining $${\cal F}(A) \to G$$ by $$\alpha \to s_\alpha$$. (Like you said, the existence and uniqueness of this map is part of the definition of a free group.) The issue is just notation; he's using $$s_\alpha$$ for both the element of $${\cal F}(A)$$ and the element of $$G$$.

• Yes I am now kind of regret of following Munkres for free group. He spends a long time on the direct sum of abelian, and free product and then use free product to define free group. This gives us a lot of conveniences but seems not really standard now.. – JacobsonRadical Nov 14 at 17:32

Munkres is not saying that $$F$$ is a subgroup of $$G$$. Nor is he saying that the group operation for $$F$$ behaves the same way as the group operation for $$G$$. Consider the example where $$G$$ is the integers modulo $$7$$. The element $$\bar1$$ is a generator of $$G$$, and we can let $$F$$ be the free group generated by $$\bar1$$, which is isomorphic to $$\mathbb Z$$. The map that takes $$\bar1\in F$$ to $$\bar1\in G$$ can be extended to a surjective homomorphism from $$F$$ to $$G$$.

• Thank you for your example! – JacobsonRadical Nov 14 at 17:31