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

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)

• For $G$ to have the structure of a $\mathbb{F}_2$-vector space you need the condition $g^2 = 1$ (for all $g \in G$) such that everything works out with the scalar multiplication. Just try it for a small group like $\mathbb{Z}/3\mathbb{Z}$ and you will see.
– Con
Oct 18, 2019 at 9:20
• Alright. That does make sense ^^'. Thanks. So it does generalize to any prime number, doesn't it? If $\forall g : g^p = 1$, then we can find an isomorphism from $G$ to $F_p^n$ ? :)
– Azur
Oct 18, 2019 at 9:23
• Yes. To state it in a fancy way: There is an equivalence of categories $\mathbb{F}_p$-vector spaces to elementary abelian $p$-groups given by the forgetful functor. So one could say that these are the same if one just forgets about the vector space structure.
– Con
Oct 18, 2019 at 9:29
• That do be a fancy way. I guess I'll come back to look at this post when I know enough about this ;)
– Azur
Oct 18, 2019 at 9:30
• @Arthur It works for abelian groups, though. There are groups such that $g^3=1$ for all $g$, but which aren't isomorphich to any $\Bbb F_3^n$.
– user239203
Oct 18, 2019 at 9:36

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 $$x$$.

• That's the step I'd been missing, thanks mate :)
– Azur
Oct 18, 2019 at 9:28

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 $$2$$.

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 $$p=2$$.
• That makes sense, but to show it can be represented as an $F_p-$ vector space, we need it to be abelian, do we not? But it doesn't look like every $p$-group is abelian :/