Automorphism group of the Turing degrees It is conjectured that the automorphism group of the Turing degrees, $Aut(\mathcal{D})$, is trivial. However, to the best of my knowledge, the current state-of-the-art is that $Aut(\mathcal{D})$ is countable. 
My question is twofold: first, is my understanding correct? Is there any countable group $G$ which we know can't be isomorphic to $Aut(\mathcal{D})$?
Second, and more interesting to me: what if anything could we conclude from some group-theoretic information about $Aut(\mathcal{D})$? I.e., what new computability theory would we know if we knew that $Aut(\mathcal{D})$ is abelian, or is simple, or is not finitely generated? (This is admittedly a slightly awkward question, given that we think $ZFC$ proves "$Aut(\mathcal{D})=\lbrace e\rbrace$.")
My reason for asking is that as a rule I am interested in interaction between computability theory and other subjects (for instance, the proof via Higman that there is a universal finitely presented group is dear to my heart), but I am especially interested in examples of other mathematics being applied to computability theory, and I'd be very interested in what the group-theoretic nature of $Aut(\mathcal{D})$ (assuming it's nontrivial) has to say about other parts of computability theory.
 A: Let $p_i$ denote the $i$th prime number, and let $\oplus$ be the recursive join on $\omega$. Let $\mathcal O$ be Kleene's $\Pi^1_1$-complete set and $\mathcal O'$ its Turing jump.
For any $B$, let $G_B$ be the direct sum of $\mathbb Z/p_i\mathbb Z$ over all $i\in B\oplus\overline B$. So $G_B$ is a countably infinite abelian group.
I claim that

Aut($\mathcal D$) is not isomorphic to $G_B$ with $B=\mathcal O'$.

I'll show this by showing that Aut($\mathcal D$) has a presentation which is recursive in $\mathcal O$, hence not $\ge_T B$. This will suffice because Richter, in her famous paper,
Richter, Linda Jean, Degrees of structures, J. Symb. Log. 46, 723-731 (1981). ZBL0512.03024.
showed that for all $B$, $G_B$ has isomorphism type of degree $[B]_T$, i.e., all presentations of $G_B$ have degree $\ge_T B$.
Note that if Aut($\mathcal D$) is finite then it is not isomorphic to $G_B$ for any $B$, since the latter is countably infinite. So assume Aut($\mathcal D$) is infinite.
Slaman and Woodin showed that each automorphism $\pi$ of $\mathcal D$ is represented by an arithmetic function in the sense that there is an $n_0$ such that for all $\pi$ and all $X$, $\pi([X]_T)= [P(X)]_T$ where $P(X)=\{e\}({X^{(n_0)}})$.
Now we consider the necessary questions about numbers $e$ in order to present Aut($\mathcal D$).
We let $E=\{e_0<e_1<e_2<\dots\}$ be the set of those $e$ for which $P_e$ given by $X\mapsto \{e\}^{X^{(n_0)}}$ is an arithmetic representation of some automorphism.
We claim that the set $E$ is $\Pi^1_1$:
First, let $F$ be the $\Pi^1_1$ set of all $e$ for which
\begin{equation}
 \forall A(P_e(A)\text{ is total}),
\end{equation}
\begin{equation}
 \forall A\forall B(A\le _T B\to P_e(A)\le_T P_e(B)),\text{ and }
\end{equation}
\begin{equation}
 \forall A\forall B(P(A)\equiv_T P(B)\to A\equiv_T B).
\end{equation}
Then
$$E=\{e: e\in F\text{ and }(\exists d\in F) \forall A(P_d(P_e(A))\equiv_T A\text{ and }P_d(P_e(A))\equiv_T A)\}.$$
The multiplication is given by defining $*$ by
$$
 P_{e_1* e_2} = P_{e_1}\circ P_{e_2}
$$
which is equivalent to
$$\forall A\forall B\forall C(B=P_{e_2}(A)\text{ and }C=P_{e_1}(B)\to C=P_{e_1*e_2}(A))$$
We also have to mod out by equality of the automorphisms induced by $e_1$ and $e_2$, which we check by:
$$\forall A (P_{e_1}(A)\equiv_T P_{e_2}(A))$$
Overall, we get a subset of $\omega$ recursive in the $\Pi^1_1$-complete set Kleene's $\mathcal O$, with an $\mathcal O$-recursive group operation. This is then isomorphic to all of $\omega$ with an $\mathcal O$-recursive group operation, as desired.
