An abelian group as an $\mathbb F_2$-vector space

Question

Consider an abelian group $G$ in which every element $x$ satisfies $x+x=0$. Then this abelian group (or a $\mathbb Z$-module) becomes an $\mathbb F_2$-module ($\mathbb F_2$-vector space).

I'm a bit confused by the statement in this answer asserting that $G$ is isomorphic to a sum of copies of $C_2$. I would say it is isomorphic (as a $\mathbb F_2$-vector space) to a sum of copies of $\mathbb F_2$. Is the above statement supposed to mean that $C_2$ is the additive group underlying the field $\mathbb F_2$? And is it correct that it is isomorphic to a sum of copies of $C_2$ because the vector space isomorphism I talked about satisfies the condition on the group isomorphism? If there are any other worthwhile comments/remarks that I need to take into account to sort out this confusion, that would be great too.

Answer 1

Yes, this is what they mean. The vector space isomorphism $G\cong\oplus_i \Bbb F_2$, in particular, is an isomorphism of the underlying abelian groups. Since $\Bbb F_2$ is the same thing as $C_2$ as abelian group, you have the statement you want.

Answer 2

It might be worth looking at this more generally.

Let $p$ be a prime.

(A) Suppose we have a vector spaces $V_1$ and $V_2$ over the field of $p$ elements, $\mathbb{F}_p$, with a vector-space homorphism $\phi:V_1\to V_2$.

Then consider the structures $A_i$ ($i=1,2$) we get by taking the underlying set of $V_i$ equipped with the addition and zero of the vector space structure. Then these are clearly abelian groups since the abelian-group axioms are a subset of the vector-space axioms.

Since $p=0$ in $\mathbb{F}_p$ it's also the case that for each $x\in A_i$ we have that the sum of $p$ $x$'s, $x+x+\dots+x=p\cdot x=0\cdot x=0$.

It is also the case that the map $\phi:A_1\to A_2$ is an abelian-group homomorphism, since the requirements for that are a subset of the requirements for a vector-space homomorphism.

(B) In the other direction suppose that we have two abelian groups $A_1$ and $A_2$, each satisfying the condition that $x+x+\cdots +x=0$ ($p$ summands), and an abelian group homomorphism $\phi: A_1\to A_2$.

It is now the case that there is exactly one way to put an $\mathbb{F}_p$ vector space structure on $A_i$. This is because we must (for the vector space axioms to be true) define the scalar multiplication so that $0\cdot x=0$, $1\cdot x=x$. Then we must define $2\cdot x=(1+1)\cdot x=1\cdot x+ 1\cdot x=x+x$ and so on. We can then check that all the vector space aaxioms are satisfied. The fact that the sum of $p$ $x$s is zero ensures this doesn't lead to a contradiction.

So we can uniquely use the same underlying sets and additions, and this uniquely defined scalar multiplication, to make $\mathbb{F}_p$ vector spaces $V_i$.

Consider now the map $\phi: V_1 \to V_2$. This is actually a vector-space homomorphism. Why? Certainly the map preserves the additive structure. For linearity we have $\phi(s\cdot x)=\phi(x+x+\cdots x)=\phi(x)+\phi(x)+\cdots +\phi(x)=s\cdot \phi(x)$.