I am reading "Algebra I Groups and Rings" written by Toshiyuki KATSURA, and the following proposition is used in the book without a proof.
Please show me a proof of the proposition below:
Suppose that $G$ is a finite group with $n = p_1^{m_1} \cdot p_2^{m_2} \cdots p_k^{m_k}$ elements. Then $G$ has a set of generators $\{a_1, a_2, \dots, a_n\}$, where $n$ is less than or equal to $m_1 + m_2 + \cdots + m_k$.
I assume the $p_{i}$ are distinct primes, although this does not really matter.
For a matter of orders (see Lagrange's theorem) $G$ is generated by its $p_{i}$-Sylow subgroups.
Let $P$ be one of this Sylow subgroups, of order $p^{m}$. I claim that there are $m$ elements of $P$ which generate $P$.
One way of proving this requires the Frattini subgroup $\Phi(G)$: if $G/\Phi(G)$ has order $p^{k}$, then $G$ is generated by $k \le m$ elements.
Another way of proving this uses the fact that there is a series of normal subgroups $P_{i}$ $$ 1 = P_{0} < P_{1} < P_{2} < \dots < P_{n} = P $$ with $P_{i}$ of order $p^{i}$. If you choose $g_{i} \in G_{i} \setminus G_{i-1}$, for $1 \le i \le n$, the the $m$ elements $g_{i}$ generate $P$.