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.


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 $i,j<p+1$?

  • 2
    $\begingroup$ What did you intend to say with your first sentence "It is well known that a group can not be union of proper subgroups."? As every group is the union of its cyclic subgroups your statement holds only for cyclic groups. Did you want to write "of two proper subgroups"? $\endgroup$ – j.p. Apr 13 '13 at 13:54
  • $\begingroup$ Do you happen to know the Frattini subgroup (the intersection of all maximal subgroups)? If you look at the group modulo its Frattini subgroup everything becomes more easy. $\endgroup$ – j.p. Apr 13 '13 at 14:03
  • $\begingroup$ @jug: Then the question you are solving is for elementary abelian $p$-group; again, the solution (or the following answer) is not obvious!! $\endgroup$ – Beginner Apr 16 '13 at 6:04
  • $\begingroup$ @MarshalKurosh: You're right that the Frattini subgroup doesn't help directly for the question. One can maybe get a better picture of what's going on, but it wouldn't shorten a proof. $\endgroup$ – j.p. Apr 16 '13 at 10:14

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


$A_i\cap A_j\cap A_{p+1}^c=\phi$, 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.


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