# Prove that there is no group $G$ s.t. $\operatorname{Aut}(G)=\mathbb{Q}$

Prove that there is no group $$G$$ s.t. $$\operatorname{Aut}(G)=\mathbb{Q}$$

I get the feeling that we should proceed by contradiction.

So let $$G$$ be a group s.t. $$\operatorname{Aut}(G)=\mathbb{Q}$$. Then we can identify elements of $$\mathbb{Q}$$ with automorphisms of $$G$$... and identities such as $$\frac{1}{2}*2(g)=1(g)=g$$

Can somebody help me find a contradiction?

• Are you definitely sure such $G$ does not exists or it is your private conjecture? – Bonbon May 1 at 21:31
• Actually, $1g$ couldn't be always $g$: the neutral element of $\mathrm{Aut}(G)$ is the identity, but the operation you have on $\mathbb{Q}$ to make it a group must be addition, not multiplication. So $0$ is the identity map, and the automorphism corresponding to $2$ composed with the automorphism corresponding to $\frac{1}{2}$ is not the automorphism corresponding to $1$, but the one corresponding to $\frac{5}{2}$. You'd be better served thinking of the maps as $\varphi_q\colon G\to G$ with $q\in\mathbb{Q}$ and $\varphi_q\circ\varphi_r = \varphi_{q+r}$. – Arturo Magidin May 1 at 21:47
• Relevant article; academic.oup.com/qjmath/article-abstract/41/2/179/… – Servaes May 1 at 22:04
• Well, the automorphism group of $\mathbb Q^+$ is $\mathbb Q^\times$ -- so near and yet so far ... – Henning Makholm May 1 at 22:32

Suppose $$G$$ is a group with $$\operatorname{Aut}(G)\cong\mathbb{Q}$$. If $$G$$ is abelian, then $$f(x)=-x$$ is an automorphism of $$G$$ which satisfies $$f^2=1$$. But $$\mathbb{Q}$$ is torsion-free, so this implies $$f=1$$. But then $$G$$ is a vector space over $$\mathbb{Z}/(2)$$, and so its automorphism group is nonabelian if its dimension is greater than $$1$$ and finite otherwise.
So, $$G$$ must be nonabelian; say $$x,y\in G$$ do not commute. Now note that $$G$$ acts on itself by conjugation, and this gives a homomorphism $$\varphi:G\to\operatorname{Aut}(G)\cong\mathbb{Q}$$ whose kernel is $$Z(G)$$, the center of $$G$$. Note that the subgroup of $$\mathbb{Q}$$ generated by $$\varphi(x)$$ and $$\varphi(y)$$ is cyclic (since every finitely generated subgroup of $$\mathbb{Q}$$ is cyclic); say it is generated by $$\varphi(a)$$ for some $$a\in G$$. Then there are $$m,n\in\mathbb{Z}$$ and $$z,z'\in Z(G)$$ such that $$x=a^nz$$ and $$y=a^mz'$$. But now we see that $$x$$ and $$y$$ actually do commute (since $$z$$ and $$z'$$ commute with everything), so we have a contradiction.
• $f=1 \rightarrow G$ is a vector space over $\frac{\mathbb{Z}}{(2)}$? Can you explain a little? Also, the subgroup generated by $\alpha(x)$ and $\alpha(y)$ is cyclic because all finitely generated subgroups of $\mathbb{Q}$ is cyclic. Is the generator for this subgroup $gcd(\alpha(x),\alpha(y))?$ – Mathematical Mushroom May 6 at 12:33
• If $f=1$ then $2x=0$ for all $x\in G$. As for your second question, I suppose that depends on what you mean by GCD, but if you take a common denominator for $\alpha(x)$ and $\alpha(y)$ then you can get a generator by taking the GCD of their numerators over the same denominator. – Eric Wofsey May 6 at 15:52
• Thanks man, I appreciate these great answers. Hypothetically speaking, if $dim_{Z_2}(G)=2$, then would the automorphism group of $G$ be $GL(2,\mathbb{Z}_2)$? – Mathematical Mushroom May 6 at 20:41
• Also, since the subgroup of $\mathbb{Q}$ generated by $\alpha(x),\alpha(y)$ is cyclic and generated by $\alpha(a)$, this implies that $x,y$ are both generated by $a \in \frac{G}{Z(G)}$. How do we know $a \in \frac{G}{Z(G)}$? – Mathematical Mushroom May 8 at 15:39