Minimum size of generating set is bounded by sum of exponents of prime divisors of order

Statement
Suppose $$G$$ is a finite group of order $$n$$, which has prime factorization:

$$\! n = p_1^{k_1}p_2^{k_2} \dots p_r^{k_r}$$

Then, the minimum size of generating set for $$G$$ is at most equal to:

$$\sum_{i=1}^r k_i$$

Moreover, equality holds if and only if $$G$$ is a direct product of elementary abelian groups of order $$p_i^{k_i}$$, i.e.:

$$G \cong (\mathbb{Z}/p_1\mathbb{Z})^{k_1} \oplus (\mathbb{Z}/p_2\mathbb{Z})^{k_2} \oplus \dots \oplus (\mathbb{Z}/p_r\mathbb{Z})^{k_r}$$

Facts used

 * 1) Any generating set of minimum size must be an minimal generating set (in the sense of being irredundant).
 * 2) uses::Size of minimal generating set is bounded by max-length
 * 3) uses::Max-length is bounded by sum of exponents of prime divisors of order

Proof
The proof follows directly by combining Facts (1)-(3).