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,

<cite authors="Richter, Linda Jean">_Richter, Linda Jean_, [**Degrees of structures**](http://dx.doi.org/10.2307/2273222), J. Symb. Log. 46, 723-731 (1981). [ZBL0512.03024](https://zbmath.org/?q=an:0512.03024).</cite>

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][1] 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.

  [1]: https://math.berkeley.edu/~slaman/talks/sw.pdf