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Let $V$ be a vector space of dimension $n$ over finite field $K=\mathbb{F}_q$ and let $A$ be an abelian group such that $V$ is simple, faithful $KA$-module. Then $A$ is cyclic.

Moreover, for every $a\in A$, we can define an $A$-homomorphism $\varphi_a : V\to V$ by $v\cdot\varphi_a = v\cdot a$. Since $V$ is faithful, then $A$ can be embedded into $Hom_A(V,V)$ via $a\mapsto \varphi_a$. Therefore $A\leq Hom_A(V,V)$.

Since $V$ is simple, $Hom_A(V,V)$ is a skew-field.

Is $|Hom_A(V,V)|=q^n-1$? If so then $|A|$ divides $q^n-1$. How to show that $n$ is the smallest integer satisfying this?

Thanks a lot in advance.

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1 Answer 1

Yes, $\lvert \mathrm{Hom}_{A}(V,V) \rvert = q^{n}$.

Let us say that $A$ has order $k$.

By the theory reported here, $V$ is isomorphic as a $K A$-module to $W = \mathbf{F}_{q}[x] / (f(x))$, where $f(x)$ is a factor of $x^{k} - 1$ of degree $n$, such that $f(x)$ is irreducible in $\mathbf{F}_{q}[x]$ (so that the module is simple) and $f(x)$ does not divide any $x^{h} - 1$ for $h < k$ (so that the module is faithful). Here a generator $a$ of the cyclic group $A$ acts as multiplication by $x$.

Now we see that $\lvert \mathrm{Hom}_{A}(W,W) \rvert$ consists of all maps $W \to W$ that send $w \mapsto w b$, for $b \in W$, so it has the same order as $W$, that is, $q^{n}$.

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Thank Andreas Caranti. But I have not seen the conclusion of my problem? –  9999 Jan 29 '13 at 21:59

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