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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.

José Carlos Santos
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Azur
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    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 '19 at 09:20
  • 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 '19 at 09:23
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    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 '19 at 09:29
  • 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 '19 at 09:30
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    @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 $\Bbb F_3^n$. –  Oct 18 '19 at 09:36

3 Answers3

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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 $x$.

slimmerikko
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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 $2$.

José Carlos Santos
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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 $p=2$.

J.-E. Pin
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  • 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 :/ – Azur Oct 18 '19 at 09:59