# 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?

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Did you try to verify the axioms of vector space ? since there is an element of order $p$ in the group it creates a subgroup of order $p$ hence isomorphic to $\mathbb{Z}_p$. you can add two elements in the group and it acts on an isomorphic copy of $\mathbb{Z}_p$ using the addition of the group (someone correct me if I got this wrong) – Belgi Aug 22 '12 at 21:26

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 $G$.

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Great!!! Thanks a lot !!! – joshua Aug 23 '12 at 15:22

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 $\mathbb Z/p\mathbb Z$.

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Still can't figure it out... Why the fact that $p\mathbb{Z}$ annihilates any p-group implies that it is a vector space over $\mathbb{Z} _p$ ? Thanks – joshua Aug 23 '12 at 6:44
@joshua I added some more explanation, and fixed the word "group" which should have been "module". – Alex Becker Aug 24 '12 at 4:38
@AlexBecker: I think it should be "$p\mathbb Z$ annihilates every elementary $p$-group". $\mathbb Z/p^2\mathbb Z$ is a $p$-group, but not annihilated by $p\mathbb Z$. – marlu Aug 24 '12 at 16:57