TITLE: Intuition behind the Frattini subgroup
QUESTION [32 upvotes]: I am trying to get a better feel for what the Frattini subgroup really is, intuitively.
Let $G$ be a group and denote its Frattini subgroup by $\Phi(G)$.  I know that $\Phi(G)$ is the intersection of the maximal subgroups of $G$, and I know that it is the set of 'non-generators' (Isaacs calls them 'useless' elements) of $G$, i.e. elements $u$ for which if $\langle X \cup \{u\} \rangle =G$, then $\langle X \rangle = G$, or equivalently, if $\langle X \rangle \ne G$, then $\langle X \cup \{u\} \rangle \ne G$, where $X \subseteq G$ is a subset of $G$, and $u \in \Phi(G)$.
Since $\Phi(G)$ is the set of these elements, it would help to better understand what exactly these elements are.  Is it true that such an element $u \in \Phi(G)$ is necessarily a product of elements in $X \subseteq G$ ($u$ and $X$ as above)?  If not, what is an example where it isn't?
Finally, where exactly does the connection lie between these 'non-generators' and (the intersection of) maximal subgroups?  How do we see that they must lie in a maximal subgroup, and conversely that if an element lies in all maximal subgroups then it must be a 'non-generator'?
Thanks for the help, as always.

REPLY [3 votes]: What helped me become friendly with the Frattini subgroup, in the context of groups of prime-power order, was a related problem:  
Let $p$ be a prime, and let $G$ be a $p$-group, with $|G|=p^{n}$. Show that $|Aut(G)| \mid \Pi_{k=0}^{n-1}{(p^{n}-p^{k})}$.  
This result is sharp in the sense that, if $G$ is elementary abelian, $Aut(G)$ will be a general linear group whose order is exactly that product. So the Frattini subgroup is a way of getting a handle on $p$-groups of an arbitrary nature using the way elementary abelian $p$-groups work.
As has been mentioned, $\dim_{\mathbb{Z}/(p)} G/\Phi(G)$ is the smallest number of elements needed to generate $G$. This is one measure of the complexity of $G$. If it is $1$, $G$ is cyclic. If it is low, obtaining a presentation for $G$ should not be too difficult. If it is high, $G$ needs many generators and there can be many ways they satisfy their relations, and $G$ can be hard to describe. (Of course, when this is $n$, $G$ is elementary abelian and that's not hard to describe.)
Solving the problem I alluded to also brings out something else, which is the order of a Sylow $p$-subgroup of $Aut(G)$. When the minimum size of a generating set of $G$ is $1$, this is $p^{n-1}$. (In fact, this group will be trivial or cyclic, except when $p=2$ and $n \geq 3$.) If $G$ needs $r$ elements to be generated, then the order of a Sylow $p$-subgroup of $Aut(G)$ will divide (equal? I don't know enough about $p$-groups to answer this question) $p^{n-1+ \ldots + n-r} = p^{nr - \frac{r^{2}+r}{2}}$.  
One more remark about being the group of non-generators: if and only if $H < G$ is a subgroup and the only subgroup $K \leq G$ such that $HK = G$ (so $K$ straddles all the cosets of $H$, with some unspecified multiplicity) is $K = G$ itself, then $H \leq \Phi(G)$. To see this, try choosing $K$ to be a maximal subgroup of $G$ and remember that since $G$ is a $p$-group, any maximal subgroup of $G$ is normal.