On the Frattini Subgroup For a prime $p$, let $H=\{x\in \mathbb{C}\colon x^{p^n}=1 \mbox{ for some } n\geq 1\}$ be the Prufer $p$-group, $C_2=\langle y\colon y^2=1\rangle$, and $G=H\oplus C_2$. Then $H$ is the unique maximal subgroup of $G$, hence it is Frattini subgroup of $G$. Let $S=\{e^{2\pi i/p^n} \colon n\geq 1\}$. Then $S$ generates the subgroup $H\leq G$, and $S\cup \{y\}$ generate whole $G$. Now, as $H$ is Frattini subgroup of $G$, it is the set of non-generators of $G$. But as $S\cup \{y\}$ generate $G$, we must have $\{y\}$ generates $G$, I reached to a wrong conclusion. Can one clarify mistake in the arguments, if any?

Your mistake is in your interpretation of "non-generators". A non-generator is an element that can be removed from any generating set and still leave a generating set. It is true that $S\cup\{y\}$ generates $G$, but given any $z\in S$, we may put $S':=S\setminus\{z\},$ and $S'\cup\{y\}$ will still generate $G$. We cannot remove all elements of $S$ from $S\cup\{y\}$ and still have a generating set for $G$.
