On the minimal number of generators of a finite group

Question

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.

Answer 1

False.

Since every finite group is a subgroup of some symmetric group $S_n$ (group of all n-element permutations under composition) and each symmetric group can be generated from two elements (rotate and swap_first), it suffices to find a finite group that cannot be generated from two elements.

Consider the group of 3-bit numbers under bitwise addition (XOR) (isomorphic to $Z_2^3$). Then for any two generators $A$ and $B$, the four elements $\{0, A, B, A+B\}$ are closed under addition (and self-inverse), meaning that $A, B$ don't generate the whole group.

Thus,

$S_8$ has a pair of generators
$Z_2^3$ doesn't have a pair of generators
$d(Z_2^3) \leq d(S_8)$ is not true
$Z_2^3$ is a subgroup of $S_8$
$H$ being a subgroup of $G$ doesn't imply $d(H)\leq d(G)$

Answer 2

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.