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Is it possible for a finitely generated, recursively presented group to have a non-recursively presented outer automorphism group? Or is the following true,

$G$ finitely generated, recursively presented $\Rightarrow$ $\operatorname{Out}(G)$ recursively presented.

It has been known for some time (1989, I believe) that every group $Q$ can be realised as the outer automorphism group of some group $G_Q$. Indeed, $G_Q$ can be taken to have certain properties, such as simple or torsion-free metabelian with trivial centre. If $Q$ is assumed to be countable then $G_Q$ can be taken to be a locally finite $p$-group, or finitely generated. If $Q$ is assumed to be finitely presentable then $G_Q$ can be taken to be finitely generated and residually finite. If $Q$ is assumed to be finite then $G_Q$ can be taken to be the fundamental group of a closed hyperbolic $3$-manifold.

Hand-wavingly: As $Q$ acquires more finiteness properties then so can $G_Q$.

I cannot find anything which talks about recursive presentability though. I have a proof that if $Q$ is recursively presented then $G_Q$ can be taken to be finitely generated and recursively presented (and some other nice properties which are what I am really interested in but they are irrelevant to this question). I am (essentially) wondering if it is reasonable to try and take this further: will I ever be able to prove that $G_Q$ can be taken to be recursively presented even though $Q$ is not recursively presented?

My initial instinct was that this is silly - obviously a recursively presented group cannot have anything to do with a non-recursively presented group. I mean, they cannot be their subgroups, so, you know, wave your hands a bit and you have a proof...On the other hand, outer automorphism groups are very different from subgroups. I cannot think of any valid reason why finitely generated, recursively presented $G$ must have $\operatorname{Out}(G)$ recursively presented.

So, again, (just to reiterate) does the following hold,

$G$ finitely generated, recursively presented $\Rightarrow$ $\operatorname{Out}(G)$ recursively presented.

Is it possible for a recursively presented group to have a non-recursively presented outer automorphism group? Or is the following true,

$G$ recursively presented $\Rightarrow$ $\operatorname{Out}(G)$ recursively presented.

It has been known for some time (1989, I believe) that every group $Q$ can be realised as the outer automorphism group of some group $G_Q$. Indeed, $G_Q$ can be taken to have certain properties, such as simple or torsion-free metabelian with trivial centre. If $Q$ is assumed to be countable then $G_Q$ can be taken to be a locally finite $p$-group, or finitely generated. If $Q$ is assumed to be finitely presentable then $G_Q$ can be taken to be finitely generated and residually finite. If $Q$ is assumed to be finite then $G_Q$ can be taken to be the fundamental group of a closed hyperbolic $3$-manifold.

Hand-wavingly: As $Q$ acquires more finiteness properties then so can $G_Q$.

I cannot find anything which talks about recursive presentability though. I have a proof that if $Q$ is recursively presented then $G_Q$ can be taken to be recursively presented (and some other nice properties which are what I am really interested in but they are irrelevant to this question). I am (essentially) wondering if it is reasonable to try and take this further: will I ever be able to prove that $G_Q$ can be taken to be recursively presented even though $Q$ is not recursively presented?

My initial instinct was that this is silly - obviously a recursively presented group cannot have anything to do with a non-recursively presented group. I mean, they cannot be their subgroups, so, you know, wave your hands a bit and you have a proof...On the other hand, outer automorphism groups are very different from subgroups. I cannot think of any valid reason why recursively presented $G$ must have $\operatorname{Out}(G)$ recursively presented.

So, again, (just to reiterate) does the following hold,

$G$ recursively presented $\Rightarrow$ $\operatorname{Out}(G)$ recursively presented.

Is it possible for a finitely generated, recursively presented group to have a non-recursively presented outer automorphism group? Or is the following true,

$G$ finitely generated, recursively presented $\Rightarrow$ $\operatorname{Out}(G)$ recursively presented.

It has been known for some time (1989, I believe) that every group $Q$ can be realised as the outer automorphism group of some group $G_Q$. Indeed, $G_Q$ can be taken to have certain properties, such as simple or torsion-free metabelian with trivial centre. If $Q$ is assumed to be countable then $G_Q$ can be taken to be a locally finite $p$-group, or finitely generated. If $Q$ is assumed to be finitely presentable then $G_Q$ can be taken to be finitely generated and residually finite. If $Q$ is assumed to be finite then $G_Q$ can be taken to be the fundamental group of a closed hyperbolic $3$-manifold.

Hand-wavingly: As $Q$ acquires more finiteness properties then so can $G_Q$.

I cannot find anything which talks about recursive presentability though. I have a proof that if $Q$ is recursively presented then $G_Q$ can be taken to be finitely generated and recursively presented (and some other nice properties which are what I am really interested in but they are irrelevant to this question). I am (essentially) wondering if it is reasonable to try and take this further: will I ever be able to prove that $G_Q$ can be taken to be recursively presented even though $Q$ is not recursively presented?

My initial instinct was that this is silly - obviously a recursively presented group cannot have anything to do with a non-recursively presented group. I mean, they cannot be their subgroups, so, you know, wave your hands a bit and you have a proof...On the other hand, outer automorphism groups are very different from subgroups. I cannot think of any valid reason why finitely generated, recursively presented $G$ must have $\operatorname{Out}(G)$ recursively presented.

So, again, (just to reiterate) does the following hold,

$G$ finitely generated, recursively presented $\Rightarrow$ $\operatorname{Out}(G)$ recursively presented.

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ADL
ADL
  • 2.8k
  • 1
  • 24
  • 32

Recursive presentability of outer automorphism group

Is it possible for a recursively presented group to have a non-recursively presented outer automorphism group? Or is the following true,

$G$ recursively presented $\Rightarrow$ $\operatorname{Out}(G)$ recursively presented.

It has been known for some time (1989, I believe) that every group $Q$ can be realised as the outer automorphism group of some group $G_Q$. Indeed, $G_Q$ can be taken to have certain properties, such as simple or torsion-free metabelian with trivial centre. If $Q$ is assumed to be countable then $G_Q$ can be taken to be a locally finite $p$-group, or finitely generated. If $Q$ is assumed to be finitely presentable then $G_Q$ can be taken to be finitely generated and residually finite. If $Q$ is assumed to be finite then $G_Q$ can be taken to be the fundamental group of a closed hyperbolic $3$-manifold.

Hand-wavingly: As $Q$ acquires more finiteness properties then so can $G_Q$.

I cannot find anything which talks about recursive presentability though. I have a proof that if $Q$ is recursively presented then $G_Q$ can be taken to be recursively presented (and some other nice properties which are what I am really interested in but they are irrelevant to this question). I am (essentially) wondering if it is reasonable to try and take this further: will I ever be able to prove that $G_Q$ can be taken to be recursively presented even though $Q$ is not recursively presented?

My initial instinct was that this is silly - obviously a recursively presented group cannot have anything to do with a non-recursively presented group. I mean, they cannot be their subgroups, so, you know, wave your hands a bit and you have a proof...On the other hand, outer automorphism groups are very different from subgroups. I cannot think of any valid reason why recursively presented $G$ must have $\operatorname{Out}(G)$ recursively presented.

So, again, (just to reiterate) does the following hold,

$G$ recursively presented $\Rightarrow$ $\operatorname{Out}(G)$ recursively presented.