Groups and vectorspaces Let $(G, +)$ be a commutative Group with the neutral Element $e$ and $\mathbb{Z}_2$ be the field of remainders mod 2. The multiplication is declared as  $\star : \mathbb{Z}_2 \times G \rightarrow G : \overline{0} \star a := e, \overline{1} \star a := a$. What necessary and sufficient condition must $G$ satisfy, so that $G$ is a vectorspace over $\mathbb{Z}_2$? 
 A: These axioms are satisfied because $G$ is an Abelian group:


*

*Associativity of addition:  $u + (v + w) = (u + v) + w$

*Commutativity of addition:  $u + v = v + u$

*Identity element of addition:   $\exists e \in V: \forall v \in V: v + e = v$.

*Inverse elements of addition:   $\forall v \in V: \exists −v ∈ V: v + (−v) = e$


These still need to be checked to see if they impose any conditions on $G$:


*

*Identity element of scalar multiplication: $\bar{1}v = v$

*Distributivity of scalar multiplication with respect to field addition: $(a + b)v = av + bv$

*Distributivity of scalar multiplication with respect to vector addition: $a(u + v) = au + av$

*Compatibility of scalar multiplication with field multiplication:   $a(bv) = (ab)v$


The first of these doesn't impose any restrictions on $G$. Distributivity of scalar multiplication with respect to field addition implies $\bar{0}v = e$. This can be seen as follows:
$(\bar{1} + \bar{0})v = \bar{1}v + \bar{0}v \Rightarrow v = \bar{0}v + v \Rightarrow \bar{0}v = e$
This allows us to conclude that every element of $G$ needs to be it's own inverse:
$(\bar{1} + \bar{1})v = \bar{1}v + \bar{1} v \Rightarrow e = v + v \Rightarrow v = (-v)$
Given that we already know that $\bar{0}v = e$ and $\bar{1}v = v$, it is easy to check that distributivity of scalar multiplication with respect to vector addition and compatibility of scalar multiplication with field multiplication impose no new conditions on $G$.
Hence, the only requirement on $G$ is that each element is its own inverse: $\forall v \in G: v = (-v)$
