# Does the general linear group equals the automorphism group of the additive group of a K-vector space?

I am wondering given a $K$-vector space $E$ of dimension $n$ if $GL_n(K)=Aut(E,+)$. For a finite field it seems to be true. For example $Aut(\mathbb Z/2\mathbb Z^3,+)=GL_3(\mathbb Z/2\mathbb Z)$.
If $K=\mathbb Q$ it also seems to be true. Is this correct in general?

I have a another question. The axioms of the external law of a $K$-vector space E are equivalent to the given of a ring morphism from $(K,+,\times)$ to $(Aut(E),+,\circ)$. For instance on a $\mathbb R^3$, does it exist another morphism than the standard one $\lambda \mapsto (x \mapsto\lambda x)$? Do you know some intersting non trivial examples of external laws?

• To answer your first question, the automorphism group of the additive group of $\mathbb R$ is horrific because it is a vector space of uncountable dimension over $\mathbb Q$, so in particular the automorphism group is not the same as the multiplicative group of nonzero real numbers. Feb 4, 2015 at 8:43
No. For $K=\mathbb C$, we have $GL_1(K)$ consists of multiplications with nonzero constantsd, but $(\mathbb C,+)$ also has $z\mapsto \bar z$ as automorphism (plus an infinitude of weird non-contiunuous ones).
Any automorphism of $(E,+)$ of course leads to a new action of $K$ on $E$. With $K=E=\mathbb C$ and the automorphism $z\mapsto \bar z$, this makes no difference; but the noncontinuous automorphisms of $E$ would really create a different vector space structure. Admittedly, these extreme examples are maybe not interesting. But I suggest you interest yourself in Galois theory.