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Hi: The definition I'll use is this: Let $F$ be an abelian group and $X$ a subset of $F$. Then $F$ is a free abelian group on $X$ if for every abelian group $G$ and every function $f$ from $X$ to $G$ there is a homomorphism $\phi$ from $F$ to $G$ that extends $f$.

Let $G$ be a finite group and $X$ a subset of $G$. Let $F$ be the free abelian group on $X$. Then $F=\langle X\rangle$ and so $F\subseteq G$. That is, every finite group has an infinite subgroup. What am I doing wrong?

EDIT: It will be easier to make myself clear working with free groups. I'll quote from Derek Robinson, A Course in the Theory of Groups, 2nd ed.

From this a free group is not only always free on a subset but additionally that subset generates it. If $G$ is a group and $X$ is a subset, however, indeed there will exist a free group on $X$ but I am unable to show it will be generated by $X$ based in the above quote. Which is very natural, of course. Thanks for the posts. Honestly none of the feedback, up to now, throws light in the paradox (paradox for me, of course).

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    $\begingroup$ How do you conclude that $F\subseteq G$? Even if $G$ is trivial, the free group on the identity $e$ is infinite, and not contained in $G$. Remember, the relations from $G$ are not relations in $F$. $\endgroup$
    – lulu
    Mar 14 '20 at 0:44
  • $\begingroup$ $X\subseteq G$. Then $\langle X \rangle \subseteq G$. But $F=\langle X\rangle$. $\endgroup$
    – stf91
    Mar 14 '20 at 0:56
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    $\begingroup$ Nowhere does the definition say that the homomorphism $\phi : F \to G$ has to be injective, which is what you would need to conclude that $F \subseteq G$. $\endgroup$ Mar 14 '20 at 0:59
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    $\begingroup$ $F$ is the free group on the letters represented by $X$. I thought you were writing that as $\langle X \rangle$. People sometimes use that notation to represent the subgroup of $G$ generated by the elements in $X$. Those two notions are not the same. $\endgroup$
    – lulu
    Mar 14 '20 at 1:09
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    $\begingroup$ For what it's worth, I'd say it was more typical to denote the Free group on a set $X$ by $F_X$. See, e.g., this But, in any case, don't let the notation confuse the meaning. $\endgroup$
    – lulu
    Mar 14 '20 at 1:10
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You are using $\langle X \rangle$ to mean two different things, and conflating them:

  1. You are using it to mean the free abelian group on $X$.
  2. You are using it to mean the subgroup of $G$ generated by $X$.

These are not the same thing, but you assume that they are.

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  • $\begingroup$ Thanks. I have edited the OP. See under "EDIT" if you please. Honestly none of the feedback throws light in the paradox (paradox only for me, of course). $\endgroup$
    – stf91
    Mar 14 '20 at 11:51
  • $\begingroup$ @stf91 Exactly the same problem exists: you are still using $\langle X \rangle$ to mean two entirely different things (or, if you prefer, you're using two different group operations to define it). $\endgroup$ Mar 14 '20 at 12:44
  • $\begingroup$ I'll quote myself (OP): "Let G be a finite group and X a subset of G. Let F be the free abelian group on X. Then F=⟨X⟩ and so F⊆G". I already realized this inference is entirely gratuitous. At least starting from the statements in the text I quoted in the OP. Thanks for your kindness. PS: Always defining $\langle S \rangle$, where $S$ is a subset of some group as the intersection of all the subgroups including $S$. $\endgroup$
    – stf91
    Mar 14 '20 at 15:13
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The $X$ in the free abelian group definition is a collection of formal symbols. They are assumed to have the minimum properties necessary for $\langle X\rangle$ to be an abelian group. That is what it means to be "free" in this context.

However, for any $\Xi\subseteq G$ for a finite group $G$, there are extra constraints on the subgroup $\langle \Xi\rangle_G$ of $G$; for instance, each element of $\langle \Xi\rangle_G $ is of finite order. Thus it is not free.

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Consider $G=\mathbb{Z}/5\mathbb{Z}$ and $[1]\in G$, let $X=\{[1]\}$ $F=<X>=G$, let $f:X\rightarrow\mathbb{Z}$ defined by $f([1])=1$, $f$ cannot be extended to $G$ in $g$, since we must have $g(5.[1])=5.f([1])=5$ and $g(5.[1])=g([0])=0$ since $g$ is a morphism of groups.

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