Detail in proof that groups of order $mp$ with $1<m < p$ are simple

Question

The following is from Aluffi's Algebra:

In the underlined equality, where $1$ and $p-1$ comes from? If it uses The Class Formula (the picture below) then $|Z(G)|=1$ and why there are $p-1$ elements in $A$.

Answer 1

You are assuming that there must exist $H_1, \dots, H_{p+1}$ subgroups of order $p$ of $G$ with $p-1$ elements of order $p$ each. Note that this accounts for a total of $(p+1)(p-1)$ elements of order $p$, becase no element of these can be in two of the subgroups at the same time (as $H_i \cap H_j = 1$ when $i \neq j$). We also have that $1 \in G$, so this gives a lower bound of $1 + (p-1)(p+1)$.

Answer 2

It comes from

Any two distinct subgroups prime order can only meet at the identity.

Therefore, if $P_1,\cdots, P_k$ are the subgroups of order $p$, then $$P_1\cup\cdots \cup P_k=(P_1\setminus\{e\})\cup\cdots\cup(P_k\setminus\{e\})\cup \{e\}$$ Where the union on the right is disjoint. Its cardinality is thus $k(p-1)+1$.

Answer 3

All Aluffi is saying is that since the number $N_p$ of subgroups of order $p$ is congruent to $1$ modulo $p$, if there is more than subgroup of order $p$, we must have $$ N_p = 1 + \underbrace{k}_\text{some positive integer}p \ge 1 + p, $$ and each such subgroup contributes $p-1$ distinct elements. Hence the number of elements in the group $G$ is at least $$ \underbrace{1}_{\text{from the identity element}} + \underbrace{(p-1)}_{\text{from the number of elements per subgroup of order $p$}} N_p \ge 1 + (p-1)(p+1) . $$