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!
Best Answer
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.