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5 stated second sentence more precisely

Note that $n$ is the sum over prime divisors $p$ of $|G|$ of the minimal number of generators of the distinct Sylow $p$-subgroups of $G.$ The minimal number sizes of generators all minimal generating sets of a finite $p$-group is well defined are the same by properties of the Frattini subgroup. Use of the Frattini subgroup helps to prove the leftmost inequality: take a prime $p$ which divides $d_1 .$ Then a Sylow $p$-subgroup of $G$ can't be generated by fewer than $k$ elements, so $G$ itself certainly can't be generated by fewer than $k$ elements, as each Sylow $p$-subgroup of $G$ is a homomorphic image of $G.$ On the other hand, take a minimal generating set $S$ for $G$ of maximal cardinality, and minimize the sum of the orders of elements of $S$ subject to that. Then each element of $S$ must have prime power order, for if $s \in S$ has order divisible by more than one prime, then we may write $s = t + u$ where $t$ and $u$ have coprime orders (each greater than one) whose product is the order of $s$. Then $(S \backslash \{ s \}) \cup \{t,u\}$ is still a minimal generating set for $G,$ contradicting the maximality of the cardinality of $S.$ The fact that $S$ is a minimal generating set means that if we now collect the elements of $S$ whose orders are powers of a fixed prime $p$, we must obtain a generating set for a Sylow $p$-subgroup of $G,$ and this must be minimal by the choice of $S$. Hence the cardinality of $S$ is at most $n,$ as defined above.

4 minor typo

Note that $n$ is the sum over prime divisors $p$ of $|G|$ of the minimal number of generators of the distinct Sylow $p$-subgroups of $G.$ The minimal number of generators of a finite $p$-group is well defined by properties of the Frattini subgroup. Use of the Frattini subgroup helps to prove the leftmost inequality: take a prime $p$ which divides $d_1 .$ Then a Sylow $p$-subgroup of $G$ can't be generated by fewer than $k$ elements, so $G$ itself certainly can't be generated by fewer than $k$ elements, as each Sylow $p$-subgroup of $G$ is a homomorphic image of $G.$ On the other hand, take a minimal generating set $S$ for $G$ of maximal cardinality, and minimize the sum of the orders of elements of $S$ subject to that. Then each element of $S$ must have prime power order, for if $s \in S$ has order divisible by more than one prime, then we may write $s = t + u$ where $t$ and $u$ have coprime orders (each greater than one) whose product is the order of $s$. Then $(S \backslash \{ s \}) \cup \{t,u\}$ is still a minimal generating set for $G$, G,$contradicting the maximality of the cardinality of$S.$The fact that$S$is a minimal generating set means that if we now collect the elements of$S$whose orders are powers of a fixed prime$p$, we must obtain a generating set for a Sylow$p$-subgroup of$G,$and this must be minimal by the choice of$S$. Hence the cardinality of$S$is at most$n,$as defined above. 3 clarified that$t$and$u$are nonidentity elements Note that$n$is the sum over prime divisors$p$of$|G|$of the minimal number of generators of the distinct Sylow$p$-subgroups of$G.$The minimal number of generators of a finite$p$-group is well defined by properties of the Frattini subgroup. Use of the Frattini subgroup helps to prove the leftmost inequality: take a prime$p$which divides$d_1 .$Then a Sylow$p$-subgroup of$G$can't be generated by fewer than$k$elements, so$G$itself certainly can't be generated by fewer than$k$elements, as each Sylow$p$-subgroup of$G$is a homomorphic image of$G.$On the other hand, take a minimal generating set$S$for$G$of maximal cardinality. Then each element of$S$must have prime power order, for if$s \in S$has order divisible by more than one prime, then we may write$s = t + u $where$t$and$u$have coprime orders (each greater than one) whose product is the order of$s$. Then $(S \backslash \{ s \}) \cup \{t,u\}$ is still a minimal generating set for$G$, contradicting the maximality of the cardinality of$S.$The fact that$S$is a minimal generating set means that if we now collect the elements of$S$whose orders are powers of a fixed prime$p$, we must obtain a generating set for a Sylow$p$-subgroup of$G,$and this must be minimal by the choice of$S$. Hence the cardinality of$S$is at most$n,\$ as defined above.

2 Minor typo
1