Generating Prüfer 2-group without its Frattini Subgroup I must be reading this wrong; please help me!
It seems to say we can always remove the non-generating elements from any of a group's Generating Sets and still have a Generating Set.  But it also seems to say the Frattini Subgroup of any Prufer p-group is the group itself.
So by my reckoning this implies we can take the entire Prufer p-group away from itself and therefore the empty set is a generating set of the Prufer p-group, which must surely be a contradiction.
Where am I going wrong?
One possibility is that I'm wrong to assume any Prufer P-Group is its own generating set.   Where a Generating Set is defined here, perhaps it should say Proper Subset?
The other possibility is that where it says "$\ldots\Phi(g)$ is the set of all non-generating elements...", it should say "Provided $\Phi(G)\neq G,$ then $\Phi(g)$ is the set of all non-generating elements..."
Or perhaps I am just being daft again?
 A: I'm basically stealing Tobias' should've-been-an-answer-comment:
Here's what you are thinking:

If $S$ is a generating set of $G$ then so is $S\setminus\Phi(G)$.

But what is actually true is:

If $S$ is a generating set for a group $G$ and $s\in S\cap\Phi(G)$, then $S-\{s\}$ is a generating set for $G$.

You can do this once, twice, three times, or any finite number of times.  It's a one-at-a-time process.  However, this does not permit you to eliminate infinitely many generators at once in any arbitrary group and generating set, as your example shows.
Possibly you are confused because the "eliminating redundant generators" idea was introduced to you in the case of finite groups, or with finite generating sets.  Here you can can drop all redundant generators at once without issue because you are by assumption only ever removing finitely many of them. If a concerted effort to point out the importance of the finiteness is not made, it then becomes easy to erroneously think that if $S$ is a generating set then so is $S\setminus\Phi(G)$.
