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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?

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    $\begingroup$ Are you definitely sure such $G$ does not exists or it is your private conjecture? $\endgroup$ – Bonbon May 1 at 21:31
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    $\begingroup$ 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 $\varphi_q\circ\varphi_r = \varphi_{q+r}$. $\endgroup$ – Arturo Magidin May 1 at 21:47
  • $\begingroup$ Relevant article; academic.oup.com/qjmath/article-abstract/41/2/179/… $\endgroup$ – Servaes May 1 at 22:04
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    $\begingroup$ Well, the automorphism group of $\mathbb Q^+$ is $\mathbb Q^\times$ -- so near and yet so far ... $\endgroup$ – Henning Makholm May 1 at 22:32
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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.

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    $\begingroup$ $f=1 \rightarrow G$ 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 $gcd(\alpha(x),\alpha(y))?$ $\endgroup$ – Mathematical Mushroom May 6 at 12:33
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    $\begingroup$ 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. $\endgroup$ – Eric Wofsey May 6 at 15:52
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    $\begingroup$ Thanks man, I appreciate these great answers. Hypothetically speaking, if $dim_{Z_2}(G)=2$, then would the automorphism group of $G$ be $GL(2,\mathbb{Z}_2)$? $\endgroup$ – Mathematical Mushroom May 6 at 20:41
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    $\begingroup$ Yes, that's right. $\endgroup$ – Eric Wofsey May 6 at 20:48
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    $\begingroup$ 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 $a \in \frac{G}{Z(G)}$? $\endgroup$ – Mathematical Mushroom May 8 at 15:39

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