My question is about finitely generated $p$-groups. In general, a subgroup of a finitely generated group is not necessarily finitely generated. But, my question is about finite and finitely generated $p$-groups. More specifically: if $G$ is a finitely generated $p$-group, say, $m$-generated and $U$ is a finitely generated subgroup of $G$, then is $U$ at most $m$-generated? If not, can $U$ be generated by a number of elements that depends only on $m$?
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$\begingroup$ Is G finite? That's not 100% clear from the question. $\endgroup$– yatima2975Mar 28, 2013 at 18:40
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$\begingroup$ I don't think this is true, take a maximal elementary abelian subgroup. $\endgroup$– Alexander Gruber ♦Mar 28, 2013 at 18:42
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$\begingroup$ Yes, yatima2975. $G$ is finite. $\endgroup$– user59969Mar 28, 2013 at 18:42
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1 Answer
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A wreath product of a cyclic group of order $p$ with a cyclic group of order $p^k$ is 2-generated, but the base group of the wreath product requires $p^k$ generators. So the answer to both questions is no.
answered Mar 28, 2013 at 18:49
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$\begingroup$ Derek, this theorem M. Hall presented below is true? Do you have a copy of this book for me to read? For me it seems that if the theorem is true then the answer to my question would be true too! thank you $\endgroup$– user59969Mar 28, 2013 at 19:19
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$\begingroup$ @AgenorAndrade DerekHolt's answer doesn't contradict Hall. If a group is $k$ generated it's also $m$ generated for $m> k$. $\endgroup$– Alexander Gruber ♦Mar 28, 2013 at 20:46
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2$\begingroup$ Proving that a group of order $p^k$ can be generated by at most $k$ elements is easy. Just keep choosing elements that are not in the subgroup generated by the elements chosen so far, and use Lagrange's Theorem. $\endgroup$ Mar 28, 2013 at 20:51