The Frattini subgroup contains a power of a certain subgroup This is from an article by Hall from 1961; it's probably one of the most trivial observations in that article, but I can't get the reasoning.
Let G be some group and let A be a normal abelian p-subgroup of G for some prime p. We want to show that $A^p$, the subgroup of p-th powers of elements of A, is contained in the Frattini subgroup of G.
So let M be a maximal subgroup in G which doesn't contain A and denote $N=M\cap A$. I have managed to show that N is normal in G and that A/N is a minimal normal subgroup of G/N. So
A/N can't have any nontrivial proper characteristic subgroups. Therefore $A^pN=N$, in which case we are done, or $A^pN=A$. I don't know where to go from here. I suppose the fact that A is a p-group should come into play somewhere, since all of the above works for any exponent n instead of p. 
 A: Let $G$ be a group with a normal abelian p-subgroup $A$.  Let $M$ be any maximal subgroup of $G$ not containing $A$, then $N=M\cap A$ is normal in both $M$ and $A$ so is normal in $G=MA$.  If there is a normal subgroup $B$ of $G$ with $N < B\le A$, then $B=B(A\cap M)=BM\cap A = A$. So $A/N$ is a minimal normal subgroup of $G/N$. Thus $A/N$ is an abelian p-group which has no non-trivial characteristic subgroups.  Since $A/N$ certainly contains an element of order p, the subgroup of all elements of order p is $A/N$, and thus $(A/N)^p=A^pN/N$ is trivial, and so $A^p\le N\le M$.
A: The Frattini subgroup of $G$ is also the intersection of the maximal (proper) normal subgroups of $G$.
Let $M$ be such a subgroup.
If $M \cap A = A$, no problem.
So assume $M$ does not contain $A$.
Let $\{g_1,\ldots,g_k\}$ be a maximal subset of $A$ such that $M' = \langle M , g_1, \ldots, g_k \rangle$ does not contain $A$.
Then $M'$ is normalized by $A$ (the assumption that $A$ is abelian is used here), so for any $g \in A \setminus \left( M' \cap A\right)$, $\langle M',g \rangle = \left\{m'g^k\ |\ m' \in M',\ k \in \mathbb{Z} \right\}$.
By maximality of our subset of $A$, for any $a \in A$, $a = m'g^k$ for some $m' \in M'$ and some $k \in \mathbb{Z}$.
$m'=ag^{-k} \in A$, so we get that $M' \cap A$ is a maximal subgroup of $A$, so contains $A^p$ (the Frattini subgroup of a $p$-group $A$ is equal to $A^p[A,A]$, which is equal to $A^p$ in this case).
Since $M'$ does not contain $A$, $M'$ is a proper subgroup of $G$, containing $A^p$ and $M$.
Since $M = \cap_{g \in G} gM'g^{-1}$ by maximality, and $gA^pg^{-1}=A^p$ for any $g \in G$, $A^p \subset M$, and we win.
There must be an easier proof of this.
