A question on finite $p$-groups.

Question

Is true that if $G$ is a $p$-group finite, say, $\mid G \mid = p^d$, then $G$ is $d$-generated?

Answer 1

If $d$ generated means "generated by d elements", much more is true: If the Frattini quotient $G/\phi(G)$ has order $p^n$, then $G$ is generated by n elements.

Answer 2

Yes, because it is $k$-generated, where $p^k$ is the order of $G/\phi(G)$, where $\phi(G)$ is the Frattini subgroup of $G$.

See: https://en.wikipedia.org/wiki/Frattini_subgroup and https://www.math.colostate.edu/~clayton/courses/602/602_4.pdf

Answer 3

Something stronger is true: if $|G| = p_1^{e_1} p_2^{e_2} \ldots p_n^{e_n}$, then there is a generating set for $G$ of size $\sum_{i=1}^n{e_n}$. The bound is sharp (at least) for products of elementary $p$-groups.

I don't have a reference at hand, but it's easy to prove. The idea is that if you have a set $X$ such that $\langle X \rangle \neq G$, you can find an element $y \in G - \langle X \rangle$. Since $\langle X \cup \{ y\} \rangle \neq \langle X \rangle$, the size will increase by at least one prime factor: $| \langle X \cup \{ y\} \rangle : \langle X \rangle | \gt 1$. Therefore, starting with the empty set, you'll find a generating set for $G$ in a number of steps that's at most equal to the sum of the exponents in the prime factorisation of $|G|$.

This latter number is called the multiprimality of $|G|$, see https://mathworld.wolfram.com/PrimeFactor.html or the delightful paper "Counting groups: gnus, moas and other exotica" by Conway et al.