$p$-Group as union of subgroups

Question

It is well known that a group can not be union of two proper subgroups. For finite $p$-groups, we can say more:

A finite $p$-group can not be union of $p$ proper subgroups.

Moreover,

Theorem. If $G$ is a non-cyclic finite $p$-group, which is union of $p+1$ proper subgroups $A_1,A_2,\dots, A_{p+1}$, then $A_i$'s are maximal and $\cap_{i=1}^{p+1} {A_i}$ has index $p^2$ in $G$.

(See Berkovich, Groups of Prime power order, vol. 3, Chapter on "Groups covered by few proper subgroups".)

The problem, I want to post is to understand its proof; the main obstacle in my understanding is a set theoretic statement; it may be simple, but I couldn't understand. The proof is as follows:

Proof: Let $|G|=p^{n}$. Then $G=\cup_{i=1}^{p+1} A_i= A_{p+1}\cup (A_1 - A_{p+1})\cup (A_2-A_{p+1})\cup \cdots \cup (A_p-A_{p+1})$. Comparing sizes,

$|G|\leq p^{n-1} + [(p^{n-1}-p^{n-2})+ (p^{n-1}-p^{n-2})+\cdots (p^{n-1}-p^{n-2})]_{p-\text{times}}=p^n=|G|$,

hence the above union (before comparing sizes) is disjoint, and $A_i$'s should be maximal.

(Next, the author of the book says:) It follows that $A_i\cap A_{p+1}=A_j\cap A_{p+1}$ for $i,j<p+1$.

Question: Why $ A_i\cap A_{p+1}=A_j\cap A_{p+1}$ for $i,j<p+1$?

Answer 1

The argument is a little tricky. From the disjoint union we obtain the following fact:

$$A_i\cap A_j\cap A_{p+1}^c=\varnothing$$

for all $i\ne j$, with $1\le i,j< p+1$. However, there is nothing special about $A_{p+1}$, so we may as well allow ourselves to use the more general relationship $$A_i\cap A_j\cap A_k^c=\varnothing$$ for all distinct $i,j$, and $k$. Applying the above relationship twice gives us the following:

$$\begin{align}A_i\cap A_{p+1}&=(A_i\cap A_{p+1}\cap A_j)\cup(A_i\cap A_{p+1}\cap A_j^c)\\&=A_i\cap A_{p+1}\cap A_j\\&=(A_i\cap A_{p+1}\cap A_j)\cup(A_i^c\cap A_{p+1}\cap A_j)\\&=A_j\cap A_{p+1}\end{align}$$

for all $1\le i,j< p+1$

Answer 2

$A_i\cap A_j\cap A_{p+1}^c=\phi$, i.e. $A_i\cap A_j\subseteq A_{p+1}$ for all $i,j<p+1$. Fixing $i,j$, and working for all such $A_{p+1}$'s, we've $A_i\cap A_j\subseteq \cap_{k=1}^{p+1} A_k$. q.e.d.