Why doesn't every group have order $2^n$?

Question

sorry for the clickbait title... but it does come from a genuine misunderstanding in the following proof:

Let G be a group such that $\forall x \in G: x^2 = 1$. G is abelian. If G is finite, then G can be seen as a vector space on the field $F_2$, and is of finite dimension $n$. Hence, $G \cong F_2^n \implies |G| = 2^n$.

The proof is pretty straightforward, but I can't see why it would not hold if we replaced $G$ by any arbitrary finite abelian group... though it obviously doesn't (not every finite group has order $2^n$).

I assume there has to be a subtlety when seeing $G$ as an $F_2-$, but I cannot see it (it is the first time I have seen this trick). I assume every finite abelian group can be seen as an $F_2-$vector space? (I'm not sure about that)

I read this post but cannot find an answer :/

Thanks for your help.

Answer 1

Not every finite group can be seen as a vector space over $F_2$. Let $G$ be a group.

By the axioms of a vector space it must hold for all $x \in G$ that $(1 + 1)x = x \circ x$, where $\circ$ is the group operation on $G$. Since $1 + 1 = 0$ in $F_2$, this means that $x^2$ must be the identity element of $G$ for every $x$.

Answer 2

If $V$ is a $\mathbb F_2$-vector space and if $v\in V$, then$$v+v=1.v+1.v=(1+1).v=0.v=0.$$Therefore, the order of every element of $V$ is $1$ or $2$.

Answer 3

More generally, let $p$ be a prime number. A group in which the order of each element is a power of $p$ is a $p$-group. A well-known result states that a finite group is $p$-group if and only if its order is a power of $p$.

In particular, a $p$-group is a $2$-group if and only if $p=2$.