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What is the difference between "a free group generated by A" and "a group generated by A"?

I saw the definition, but I don't know. However, since $\mathbb{Z}\times \mathbb{Z}$ is generated by two elements and is abelian, it is not a free group, so I understand that the two definitions are different.

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  • A "free group" has no relations between the generators. A "group generated by $A$" could potentially have many relations between the generators. – Joshua P. Swanson May 28 '21 at 07:28
  • Continuing Joshua P. Swanson's comment, in your example of $\mathbb{Z}^2$ you have already subdued your generators, say $a$ and $b$, to the the relation $ab=ba$, enforcing the commutativity. So $\mathbb{Z}^2$ is already a quotient of the free group. – nicrot000 May 28 '21 at 08:00

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Let me add a few examples. If $A=\{a\}$, then the free group generated by $A$ has the presentation $\langle a\rangle$ and is given by $$ \langle a \rangle =\{a^k\mid k\in \Bbb Z\}\cong \Bbb Z, $$ whereas the group generated by $A$ also can have a relation $a^n=1$ for some $n$, so that the group is given by $$ \langle a \mid a^n=1\rangle \cong C_n, $$ which is a finite cyclic group.

For $A=\{a,b\}$, the free group generated by $A$ is $$ F_2=\langle a,b\rangle. $$ This is an infinite, non-solvable group, which is quite different from "the free abelian group" $$ \langle a,b\mid ab=ba\rangle \cong \Bbb Z\times \Bbb Z. $$ Examples of special properties of $F_2$ include the fact that $F_2$ contains every free subgroup $F_k$ of rank $k\ge 2$.

The free group $F_2$ contains $F_k$

Dietrich Burde
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