Show that a subset $X \subseteq G$ generates $G$ if and only if $X+pG$ generates $G$. Let $p$ be a prime, and let $G$ be a finite abelian $p$-group. Define a subgroup $H<G$ to be maximal if there is no other subgroup $K<G$ such that $H<K<G$. Define the set $pG = \left\{ pg: g\in G \right\}$. Show that a subset $X \subseteq G$ generates $G$ if and only if $X+pG$ generates $G$.
I think I am close to figuring this question out, but cannot quite put the last few pieces together. So far I have worked out in a previous question that $$H \text{ is maximal if and only if } G/H \cong \mathbb{Z}_p \tag{1}$$ I worked this out using the correspondence theorem, and the fundamental theorem of finitely generated abelian groups. I then worked out in another previous question that
$$
X \subseteq G \text{ generates }G \text{ if and only if } X \text{ is not contained in a maximal subgroup of } G \tag{2}
$$
Now, back to the original statement I am trying to prove. It is quite trivial proving the forwards implication of this statement. I am now trying to prove the backwards implication of the statement by showing that if $X$ does not generate $G$, then $X+pG$ does not generate $G$. To do this, I have shown that if $X$ does not generate $G$, then $X$ is contained in a maximal subgroup $H$ of $G$, by $(2)$. We then have by $(1)$ that $G/H \cong \mathbb{Z}_p$. I now want to somehow show that $pG$ must now also be contained by H, which would prove the statement, again by using $(2)$. Is this the right approach to this question? Can anybody help me prove that $pG\subseteq H$, or is this simply the wrong approach to the question? Thanks for any help!
 A: Since $G/H$ is isomorphic to $\mathbb Z_p$, every element of the quotient group (which recall are just the cosets of $H$) has order $p$. What this means formally is that
$$ p(gH) = (pg) H = H $$
for all $g \in G$. This holds only if $pg \in H$, so $pG \subseteq H$.
As a side remark, to show (1) you don’t really need the full power of the fundamental theorem (using a sledgehammer to crack a nut, so to speak), something weaker like the Sylow theorems will suffice.
A: $\DeclareMathOperator{\Frat}{Frat}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$What follows is possibly too general a setting, but I believe it clarifies a bit what's  going on here.
An element $g$ of a finite group $G \ne \Set{1}$ is called a non-generator if for any $X \subseteq G$ one has
$$
G = \Span{X, g} \quad\text{implies}\quad G = \Span{X},
$$
or equivalently
$$
G \ne \Span{X} \quad\text{implies}\quad G \ne \Span{X, g},
$$
It is not difficult to see that the set $\Frat(G)$ of non-generators is a subgroup (the so-called Frattini subgroup), as
$$
\Frat(G) = \bigcap \Set{ M \le G : \text{$M$ is a maximal subgroup of $G$}}.
$$
This is because if $g$ is a non-generator, and $M$ is a maximal subgroup of $G$, then $\Span{M, g} = M$, as $\Span{M} = M \ne G$, so that $g \in M$. And conversely, if $g$ lies in every maximal subgroup, and $X \subseteq G$ is such that $\Span{X} \ne G$, then $\Span{X}$ is contained in a maximal subgroup $M$ of $G$, so that $\Span{X, g} \le M$, and $\Span{X, g} \ne G$.
Now if $G$ is a finite, abelian $p$-group, then for every maximal subgroup $M$ you have $p G \le M$, as you have noted that $G / M$ has order $p$. Hence $p G \le \Frat(G)$, so that $p G$ is all made of non-generators.
