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This partial solution by Stuart Margolis: The

The answer to question 2$2$ is definitely yes, the number of elements required to generate G$G$ can't be more than D=2^(d-1)$D=2^{d-1}$. Suppose that G$G$ requires more than D$D$ generators, then every D$D$-tuple of elements of G$G$ must belong to a proper subgroup, which therefor must be solvable of derived length at most d-1$d-1$. So every D$D$-tuple of elements satisfies the identity (in D$D$ variables) that defines being solvable of derived length at most d-1$d-1$. So G$G$ itself must be solvable of derived length at most d-1$d-1$. This

This is just a special example of the fact that if a variety (in the Hanna Neumann sense) is defined by an n$n$-variable law then: a group which is not in the variety, but all its subgroups do belong to the variety, must be generated by at most n$n$ elements.

This partial solution by Stuart Margolis: The answer to question 2 is definitely yes, the number of elements required to generate G can't be more than D=2^(d-1). Suppose that G requires more than D generators, then every D-tuple of elements of G must belong to a proper subgroup, which therefor must be solvable of derived length at most d-1. So every D-tuple of elements satisfies the identity (in D variables) that defines being solvable of derived length at most d-1. So G itself must be solvable of derived length at most d-1. This is just a special example of the fact that if a variety (in the Hanna Neumann sense) is defined by an n-variable law then: a group which is not in the variety, but all its subgroups do belong to the variety, must be generated by at most n elements.

This partial solution by Stuart Margolis:

The answer to question $2$ is definitely yes, the number of elements required to generate $G$ can't be more than $D=2^{d-1}$. Suppose that $G$ requires more than $D$ generators, then every $D$-tuple of elements of $G$ must belong to a proper subgroup, which therefor must be solvable of derived length at most $d-1$. So every $D$-tuple of elements satisfies the identity (in $D$ variables) that defines being solvable of derived length at most $d-1$. So $G$ itself must be solvable of derived length at most $d-1$.

This is just a special example of the fact that if a variety (in the Hanna Neumann sense) is defined by an $n$-variable law then: a group which is not in the variety, but all its subgroups do belong to the variety, must be generated by at most $n$ elements.

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This partial solution by Moshe NewmanStuart Margolis: The answer to question 2 is definitely yes, the number of elements required to generate G can't be more than D=2^(d-1). Suppose that G requires more than D generators, then every D-tuple of elements of G must belong to a proper subgroup, which therefor must be solvable of derived length at most d-1. So every D-tuple of elements satisfies the identity (in D variables) that defines being solvable of derived length at most d-1. So G itself must be solvable of derived length at most d-1. This is just a special example of the fact that if a variety (in the Hanna Neumann sense) is defined by an n-variable law then: a group which is not in the variety, but all its subgroups do belong to the variety, must be generated by at most n elements.

This partial solution by Moshe Newman: The answer to question 2 is definitely yes, the number of elements required to generate G can't be more than D=2^(d-1). Suppose that G requires more than D generators, then every D-tuple of elements of G must belong to a proper subgroup, which therefor must be solvable of derived length at most d-1. So every D-tuple of elements satisfies the identity (in D variables) that defines being solvable of derived length at most d-1. So G itself must be solvable of derived length at most d-1. This is just a special example of the fact that if a variety (in the Hanna Neumann sense) is defined by an n-variable law then: a group which is not in the variety, but all its subgroups do belong to the variety, must be generated by at most n elements.

This partial solution by Stuart Margolis: The answer to question 2 is definitely yes, the number of elements required to generate G can't be more than D=2^(d-1). Suppose that G requires more than D generators, then every D-tuple of elements of G must belong to a proper subgroup, which therefor must be solvable of derived length at most d-1. So every D-tuple of elements satisfies the identity (in D variables) that defines being solvable of derived length at most d-1. So G itself must be solvable of derived length at most d-1. This is just a special example of the fact that if a variety (in the Hanna Neumann sense) is defined by an n-variable law then: a group which is not in the variety, but all its subgroups do belong to the variety, must be generated by at most n elements.

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This partial solution by Moshe Newman: The answer to question 2 is definitely yes, the number of elements required to generate G can't be more than D=2^(d-1). Suppose that G requires more than D generators, then every D-tuple of elements of G must belong to a proper subgroup, which therefor must be solvable of derived length at most d-1. So every D-tuple of elements satisfies the identity (in D variables) that defines being solvable of derived length at most d-1. So G itself must be solvable of derived length at most d-1. This is just a special example of the fact that if a variety (in the Hanna Neumann sense) is defined by an n-variable law then: a group which is not in the variety, but all its subgroups do belong to the variety, must be generated by at most n elements.