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A relatively free algebra $F$ has a free generating set (basis) $X$ such that any map $f : X \to F$ can be extended to an endomorphism of $F$. It is known that, in general the notion of rank of $F$ (as the cardinality of a basis of $F$) is not well-defined, that is, there are examples of relatively free algebras whose bases are not necessarily of the same cardinality.

Question:

What about relatively free groups? It seems likely that the rank of a relatively free group $F$ must not be well-defined in general. Surely, in some nice cases when, for instance, the abelianization $F/[F,F]$ of $F$ is a free abelian group, the rank of $F$ is well-defined.

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The rank of relatively free group in a variety of groups is well defined.

See pages 11 to 12 of Hanna Neumann's Varieties of Groups. $F/[F,F]$ is either a free abelian group or a free abelian group of exponent m, and in either case it is a free module over a PID and has a well-defined rank, which is in fact the rank of the free generating set X.

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    What if F=[F,F]? It seems to be possible. In this case no information can be derived from the abelianization.2012-01-21
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    The only laws outside of the commutator subgroup are (equivalent mod the commutator subgroup to) exponent laws, so if F=[F,F], then the exponent of F is 1, and so indeed every relatively free group in that variety is the group of order 1, and so the rank is not well-defined. However, most people ignore the variety of groups of order 1.2012-01-21
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    Ah, clear. Thanks!2012-01-21