Prove that $G$ is a vector space over $\mathbb Z_2$ Suppose $G$ is an abelian group such that all non-identity elements in $G$ has order $2$.
Prove that $G$ is a vector space over $\mathbb Z_2$
Since $G$ is an abelian group only thing to show is to find an external composition .define $\circ :G\times \mathbb Z_2\rightarrow G$ by $g\circ 0=0$ and $g \circ 1=g$
Is this the solution
?Where $g^2=e$ forall g is used I cant identify?
 A: "Is this the solution?" - well if I were your instructor I would want lots more detail.  But I'm not, so you should ask your instructor.
"Where is $g^2=e$ used?" - first of all, the vector "addition" will be the group operation.  So it would be much less confusing to write the group operation as addition: write $g+h$ instead of $gh$, and $g+g$ instead of $g^2$, and $0$ instead of $e$.  Then it is not too hard to see that you need the property $g+g=0$ to prove the scalar distributive law
$$(\alpha+\beta)g=(\alpha g)+(\beta g)\ .$$
Proof: for the case $\alpha=\beta=1$ we have
$$\eqalign{
  LHS&=(1+1)g=0g=0\cr
  RHS&=(1g)+(1g)=g+g=0\cr}$$
so $LHS=RHS$; the other three cases are easy.
A: Perhaps using $+$ as the group operation might help see things clearly. Need also that $(a+b) \circ x = a\circ x+ b \circ x$ and the distributivity for vector addition across scalar multiplication. I think if not for order 2, the equation $x+x=2x=0$ would give problems as generally $x+x$ is not zero in most abelian groups, but, the interlinking of $\mathbb{Z}_2$ through distributive axioms demands this.
