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Consider the definition of free groups via the universal property:

Definition.We say that the group $F$ is the free group generated by the set $S$ if there’s a map $f:S\to F$ such that whenever there is another map $g:S\to G$ from $S$ to a group $G$, there exists a unique homomorpshism $\psi:F\to G$ such that $\psi\circ f = g$.

Using this definition, can we show that $f(S)$ generates $F)$?

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One way is by first observing that $F$ is unique upto isomorphim since it is defined by a universal property (the configuration $f:S\to F$ is an initial object in the category of all maps from the set $S$ to a group $G$ with $\psi\in \text{Hom}(f_1:S\to G_1,f_2:S\to G_2)$ if $\psi\circ f_1=f_2$ and hence is unique upto a unique isomorphsim).

Then, construct $F$ by “bare hands”, that is, define $F$ to the set of all the words formed by elements of $S$ and define multiplication in an appropriate way and finally show that such a construction satisfies the universal property. Clearly, this construction shows that $F$ is generated by $S$.

But it seems to me that there must exist an argument which doesn’t require us to explicitly construct a free group.

Can somebody help me with this?

Thanks.

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Let $F$ be free over $S$ and $G=\langle f(S)\rangle$ the subgroup generated by (the image of) the set $S$.

Let $X$ be any group and $g\colon S\to X$ a map. Then there exists a homomorpshism $\phi\colon G\to X$ such that $\phi\circ f=g$. With $F$ in place of $G$, this would immediately be unique. But also with $G$, uniqueness is clear because homomorphisms that agree on a generating set are equal. We conclude that $G$ (with $f\colon S\to G$) is free. Then the unique homomorphisms $F\to G$ and $G\to F$ show that $F,G$ are canonically isomorphic. Especially, since $G$ ids generated by $f(S)$, so is $F$.

In general, the “bare hands” construction of free groups (and other universal objects) is only required to show that free groups *exist* for any set $S$.

You already construction one free group for which the image of $S$ generates $F$. (Words etc…) Now, to show that the image of $S$ generates $F$ for any free group $F$ (that is, for any $F$ equipped with a map $f$ such that bla bla) you just have to show that any such $F$ is isomorphic to the one you constructed (not even up to unique isomorphism, even it is the case), and this is dued to the very definition of the free group by a universal property.

Important remark : the fact that the image of $S$ generates $F$ is not a consequence of the universal property. The situation is the same for tensor product of two modules over a ring $A$ : the tensor product is generated by simple tensors because one can construc at least one version of it for which it is true that simple tensors generate the tensor product.

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