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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!

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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.

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  • $\begingroup$ 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! $\endgroup$ – JacobsonRadical Nov 14 at 17:30
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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 $G$.

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    $\begingroup$ 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.. $\endgroup$ – JacobsonRadical Nov 14 at 17:32
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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 $G$.

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    $\begingroup$ Thank you for your example! $\endgroup$ – JacobsonRadical Nov 14 at 17:31

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