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In this question it is asked whether subgroups of a finite, 2-generated group are also 2-generated. The answer is no, with a nice counterexample. However, when the group is Abelian, this is true. My question, albeit a bit open ended, is what are some other nice conditions under which this is true? I have a particular example where it turns out to be true, and I'm trying to understand why. Two properties that seem to be relevant in this example is that the orders of the two generators of the large group are coprime, and that although the group is non-abelian, every element is of the form

$$a^ib^j$$

Where $G=\langle a,b\rangle$. Are either of these conditions sufficient to guarantee the claim? If not, what other conditions can be added to guarantee it?

Carlyle
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    Coprime generators are not enough. Take something like $C_2^3\rtimes C_7$, where the Sylow 7-group is permuting non-identity elements in $C_2^3$. This is 2-generated by elements of order $2$ and $7$, but $C_2^3$ is a rank-3 Abelian subgroup. – Steve D Jul 05 '24 at 15:43
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    Even "nilpotent of class $2$" is not sufficient. The group $\langle x,y\mid x^{p^2} = y^{p^2}=[x,y]^p=[x,y,x]=[x,y,y]=1\rangle$ is a $2$-generated nilpotent group of class $2$, but the subgroup $\langle x^p,y^p,[x,y]\rangle$ is an abelian subgroup of order $p^3$ and exponent $p$ which requires $3$ generators. – Arturo Magidin Jul 05 '24 at 17:08

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