TITLE: On the minimal number of generators of a finite group
QUESTION [5 upvotes]: Let $G$ be  finite group and d(G) be the minimal  number of generators of $G$ and $H$ be subgroup of $G$.
Then prove or disprove $$d(H)\leq d(G)?$$ 
What about abelian groups?
For example is true for cyclic groups.

REPLY [4 votes]: There is a nice family of groups where $d(H)\leq d(G)$. 
In the finite case we have the following definition: a finite $p$-group $G$ is powerful if $p$ is odd and $G^p\leq [G,G]$, or if $p=2$ and $G^4\leq [G,G]$. Examples of these groups are the elementary abelian groups. It is a theorem that if $G$ is a powerful $p$-group then $d(H)\leq d(G)$ for any subgroup $H$ of $G$. You can look up the proof of this result in Analytic Pro-p Groups.