Linear Independence & Group Theory Given an elementary abelian p-group $G$ . 
Can someone please explain me why it can be seen as a vector space over $\mathbb{Z}_p $ ? 
It should be elementary but I can't figure it out.
As a consequence, can  someone please give me some advice for possible papers / theorem that prove some kind of linear independence in the group context when regarding a group as a vector space? (something like proving that some elements in a group $G$ are linearly independent mod $\phi(G)$ when $ \phi(G) $ is the Frattini -group of $G$ ) .
Can someone help me?
Thanks in advance ! 
 A: Any abelian group is a $\mathbb Z$-module. Since $p\mathbb Z$ annihilates any $p$-group $G$, we see that $G$ is a $\mathbb Z/p\mathbb Z$ module as well (as the action of the subgroup $p\mathbb Z$ is trivial on $G$, we can quotient out by it). A vector space over $\mathbb Z/p\mathbb Z$ is the same as a module over $\mathbb Z/p\mathbb Z$.
A: In an elementary abelian $p$-group every nontrivial element has order $p$, so by the classification of finitely generated abelian groups it is isomorphic to $(\mathbb Z/p\mathbb Z)^n = \mathbb F_p^n$ for some $n$. An element of $\mathbb F_p^n$ is just a vector with entries in $\mathbb F_p$ and these obviously form a vector space over $\mathbb F_p$.
For another way to see this note that every abelian group $G$ is a $\mathbb Z$-module by $(n,g) \mapsto g^n$, i.e. there is a canonical homomorphism $\mathbb Z \to \operatorname{End}(G), n \mapsto [g \mapsto g^n]$. If $G$ is an elementary $p$-group then $g^p = 1$ for all $g \in G$, so $p$ is in the kernel and the homomorphism induces $\mathbb Z/p\mathbb Z \to \operatorname{End}(G)$ which equips $G$ with the structure of a $\mathbb Z/p\mathbb Z$ vector space.
Yet another way (the most direct one) is to define $\mathbb F_p \times G \to G, (\overline a, g) \mapsto g^a$ and to verify that this satisfies the axioms of a vector space (where the addition is the group operation). This is well-defined since if $a \equiv b \pmod p$ then $g^a = g^b$ where we are using that $g^p = 1$ for all $g \in G$.

For an application involving linear independece let $G$ be a finite $p$-group (not necessarily abelian) and $\Phi(G)$ its Frattini subgroup, i.e. the intersection of all maximal subgroups. Then $\Phi(G)$ is normal in $G$ and the Frattini quotient $G/\Phi(G)$ is an elementary abelian $p$-group, hence may be viewed as a vector space over $\mathbb F_p$ by the above considerations.
Claim: The size of a minimal set of generators of $G$ equals $\dim_{\mathbb F_p}(G/\Phi(G))$.
Proof: Let $\{x_1,\ldots,x_n$} be a minimal set of generators of $G$. Then obviously the residue classes $\{\overline{x_1},\ldots,\overline{x_n} \}$ generate $G/\Phi(G)$ as an $\mathbb F_p$-vector space, so it suffices to show that they are linearly independent over $\mathbb F_p$. Assume for contradiction that
$$\overline{x_1}^{a_1}\cdots \overline{x_n}^{a_n} = \overline {1}$$
is a linear relation in $G/\Phi(G)$ where at least one $a_i$ is not divisible by $p$, say $a_1 \not\equiv 0 \mod p$. By exponentiating both sides with the mod $p$ inverse of $a_1$ we find
$$\overline{x_1} \cdot \overline{x_2}^{b_2}\cdots \overline{x_n}^{b_n} = \overline{1}$$
for certain $b_i$, hence
$$x_1 = g \cdot x_n^{-b_n}\cdots x_2^{-b_2}$$
for some $g \in \Phi(G)$. This shows that $\{g, x_2,\ldots,x_n\}$ is also a generating set of $G$. We show that we can remove $g$ and still generate all of $G$, contradicting the minimality of our initial generating set. Assume $\{x_2,\ldots,x_n\}$ generates a proper subgroup of $G$, say $H$. Then $H$ is contained in a maximal subgroup $M$ of $G$. By definition of $\Phi(G)$, we have $g \in M$, hence the subgroup generated by $\{g,x_2,\ldots,x_n\}$ is also contained in $M$, contradicting the fact that it generates all of $G$.
