Return to Answer

EDIT: Apologies for the delay. I was too tired yesterday for anything more than a few comments. So let me turn this into a proper answer for everyone's sake, with a few more details added, and recap the discussion from yesterday.

1. Case: $G$ finite

It follows from Artin's theorem that every finite group $G$ is the Galois group of some finite Galois extension $L/L^G$, but we have no control over $L^G$ (this would be the inverse Galois problem). As Keith Conrad points out in the comments, however, this is a non-problem in the context of OP's question, because every finite group is the automorphism group of some finite, not necessarily Galois extension $L/\mathbb{Q}$ as shown in [Fr80]. A somewhat simpler proof of this is also given in [Ge83] (in German). This settles OP's question in the finite case.

2. Case: $G$ infinite

An infinite Galois group in the sense of infinite Galois theory is necessarily profinite. Conversely, every profinite group is the Galois group (in the sense of infinite Galois theory) of some extension $L/L^G$ as shown in [Wat74], alas once again we have no control over $L^G$.

But once again this is a non-problem in the context of OP's question. Even more generally, it is proved in [DugGöb87] that for every prescribed infinite group $G$ and every prescribed base field $K$ there exists an extension $L = L(G,K) / K$ (in their notation $R(K,G)$) such that $\operatorname{Aut}_K(L) \cong G$. Taking $K:=\mathbb{Q}$, we get $G \cong \operatorname{Aut}(L)$ (the full automorphism group of $L$), which settles OP's question in the infinite case.

Note that, as pointed out by Keith Conrad in the comments, the class of infinite groups is much wider than that of profinite groups, for example because profinite groups are either finite or uncountably infinite. I implicitly (and incorrectly) assumed from the paper's title that in the profinite case their construction would in fact produce a Galois extension of $K$. But firstly, this was overly optimistic, given that we don't even know the full answer for a general finite $G$ and $K:=\mathbb{Q}$, and secondly, we can simply take $K \cong \bar{K}$ algebraically closed (which in turn of course does not admit proper algebraic extensions). So, beware that their usage of the term "Galois group" is very non-standard as far as infinite Galois theory is concerned, as pointed out by Keith Conrad in the comments.

I hope I haven't forgotten anything.

References:

[DugGöb87] Manfred Dugas and Rüdiger Göbel. “All Infinite Groups Are Galois Groups Over Any Field.” Transactions of the American Mathematical Society, vol. 304, no. 1, 1987, pp. 355–84

[Fr80] M. Fried."A note on automorphism groups of algebraic number fields".Proc. Amer. Math. Soc. 80 (1980), 386-388

[Ge83] Geyer, WD. Jede endliche Gruppe ist Automorphismengruppe einer endlichen Erweiterung $K|\mathbb{Q}$. Arch. Math 41, 139–142 (1983)

[Wat74] William C. Waterhouse."Profinite groups are Galois groups".Proc. Amer. Math. Soc. 42 (1974), 639-640

EDIT: Apologies for the delay. I was too tired yesterday for anything more than a few comments. So let me turn this into a proper answer for everyone's sake, with a few more details added, and recap the discussion from yesterday.

1. Case: $G$ finite

It follows from Artin's theorem that every finite group $G$ is the Galois group of some finite Galois extension $L/L^G$, but we have no control over $L^G$ (this would be the inverse Galois problem). As Keith Conrad points out in the comments, however, this is a non-problem in the context of OP's question, because every finite group is the automorphism group of some finite, not necessarily Galois extension $L/\mathbb{Q}$ as shown in [Fr80]. A somewhat simpler proof of this is also given in [Ge83] (in German). This settles OP's question in the finite case.

2. Case: $G$ infinite

An infinite Galois group in the sense of infinite Galois theory is necessarily profinite. Conversely, every profinite group is the Galois group (in the sense of infinite Galois theory) of some extension $L/L^G$ as shown in [Wat74], alas once again we have no control over $L^G$.

But once again this is a non-problem in the context of OP's question. Even more generally it is proved in [DugGöb87] that for every prescribed infinite (not just profinite) group $G$ and every prescribed base field $K$ there exists an extension $L = L(G,K) / K$ (in their notation $R(K,G)$) such that $\operatorname{Aut}_K(L) \cong G$. Taking $K:=\mathbb{Q}$, we get $G \cong \operatorname{Aut}(L)$ (the full automorphism group of $L$), which settles OP's question in the infinite case.

Note that, as pointed out by Keith Conrad in the comments, the class of infinite groups is much wider than that of profinite groups, for example because profinite groups are either finite or uncountably infinite. I implicitly (and incorrectly) assumed from the paper's title that in the profinite case their construction would in fact produce a Galois extension of $K$. But firstly, this was overly optimistic, given that we don't even know the full answer for a general finite $G$ and $K:=\mathbb{Q}$, and secondly, we can simply take $K \cong \bar{K}$ algebraically closed (which in turn of course does not admit proper algebraic extensions). So, beware that their usage of the term "Galois group" is very non-standard as far as infinite Galois theory is concerned, as pointed out by Keith Conrad in the comments.

I hope I haven't forgotten anything.

References:

[DugGöb87] Manfred Dugas and Rüdiger Göbel. “All Infinite Groups Are Galois Groups Over Any Field.” Transactions of the American Mathematical Society, vol. 304, no. 1, 1987, pp. 355–84

[Fr80] M. Fried."A note on automorphism groups of algebraic number fields".Proc. Amer. Math. Soc. 80 (1980), 386-388

[Ge83] Geyer, WD. Jede endliche Gruppe ist Automorphismengruppe einer endlichen Erweiterung $K|\mathbb{Q}$. Arch. Math 41, 139–142 (1983)

[Wat74] William C. Waterhouse."Profinite groups are Galois groups".Proc. Amer. Math. Soc. 42 (1974), 639-640

EDIT: Apologies for the delay. I was too tired yesterday for anything more than a few comments. So let me turn this into a proper answer for everyone's sake, with a few more details added, and recap the discussion from yesterday.

1. Case: $G$ finite

It follows from Artin's theorem that every finite group $G$ is the Galois group of some finite Galois extension $L/L^G$, but we have no control over $L^G$ (this would be the inverse Galois problem). As Keith Conrad points out in the comments, however, this is a non-problem in the context of OP's question, because every finite group is the automorphism group of some finite, not necessarily Galois extension $L/\mathbb{Q}$ as shown in [Fr80]. A somewhat simpler proof of this is also given in [Ge83] (in German). This settles OP's question in the finite case.

2. Case: $G$ infinite

An infinite Galois group in the sense of infinite Galois theory is necessarily profinite. Conversely, every profinite group is the Galois group (in the sense of infinite Galois theory) of some extension $L/L^G$ as shown in [Wat74], alas once again we have no control over $L^G$.

But once again this is a non-problem in the context of OP's question. Even more generally, it is proved in [DugGöb87] that for every prescribed infinite group $G$ and every prescribed base field $K$ there exists an extension $L = L(G,K) / K$ (in their notation $R(K,G)$) such that $\operatorname{Aut}_K(L) \cong G$. Taking $K:=\mathbb{Q}$, we get $G \cong \operatorname{Aut}(L)$ (the full automorphism group of $L$), which settles OP's question in the infinite case.

Note that, as pointed out by Keith Conrad in the comments, the class of infinite groups is much wider than that of profinite groups, for example because profinite groups are either finite or uncountably infinite. I implicitly (and incorrectly) assumed from the paper's title that in the profinite case their construction would in fact produce a Galois extension of $K$. But firstly, this was overly optimistic, given that we don't even know the full answer for a general finite $G$ and $K:=\mathbb{Q}$, and secondly, we can simply take $K \cong \bar{K}$ algebraically closed (which in turn of course does not admit proper algebraic extensions). So, beware that their usage of the term "Galois group" is very non-standard as far as infinite Galois theory is concerned, as pointed out by Keith Conrad in the comments.

I hope I haven't forgotten anything.

References:

[DugGöb87] Manfred Dugas and Rüdiger Göbel. “All Infinite Groups Are Galois Groups Over Any Field.” Transactions of the American Mathematical Society, vol. 304, no. 1, 1987, pp. 355–84

[Fr80] M. Fried."A note on automorphism groups of algebraic number fields".Proc. Amer. Math. Soc. 80 (1980), 386-388

[Ge83] Geyer, WD. Jede endliche Gruppe ist Automorphismengruppe einer endlichen Erweiterung $K|\mathbb{Q}$. Arch. Math 41, 139–142 (1983)

[Wat74] William C. Waterhouse."Profinite groups are Galois groups".Proc. Amer. Math. Soc. 42 (1974), 639-640

EDIT: Apologies for the delay. I was too tired yesterday for anything more than a few comments. So let me turn this into a proper answer for everyone's sake, with a few more details added, and recap the discussion from yesterday.

1. Case: $G$ finite

It follows from Artin's theorem that every finite group $G$ is the Galois group of some finite Galois extension $L/L^G$, but we have no control over $L^G$ (this would be the inverse Galois problem). As Keith Conrad points out in the comments, however, this is a non-problem in the context of OP's question, because every finite group is the automorphism group of some finite, not necessarily Galois extension $L/\mathbb{Q}$ as shown in [Fr80]. A somewhat simpler proof of this is also given in [Ge83] (in German). This settles OP's question in the finite case.

2. Case: $G$ infinite

An infinite Galois group in the sense of infinite Galois theory is necessarily profinite. Conversely, every profinite group is the Galois group (in the sense of infinite Galois theory) of some extension $L/L^G$ as shown in [Wat74], alas once again we have no control over $L^G$.

But once again this is a non-problem in the context of OP's question. Even more generally, it is proved in [DugGöb87] that for every prescribed infinite group $G$ and every prescribed base field $K$ there exists an extension $L = L(G,K) / K$ (in their notation $R(K,G)$) such that $\operatorname{Aut}_K(L) \cong G$. Taking $K:=\mathbb{Q}$, we get $G \cong \operatorname{Aut}(L)$ (the full automorphism group of $L$), which settles OP's question in the infinite case.

Note that, as pointed out by Keith Conrad in the comments, the class of infinite groups is much wider than that of profinite groups, for example because profinite groups are either finite or uncountably infinite. I implicitly (and incorrectly) assumed from the paper's title that in the profinite case their construction would in fact produce a Galois extension of $K$. But firstly, this was overly optimistic, given that we don't even know the full answer for a general finite $G$ and $K:=\mathbb{Q}$, and secondly, we can simply take $K \cong \bar{K}$ algebraically closed (which in turn of course does not admit proper algebraic extensions). So, beware that their usage of the term "Galois group" is very non-standard as far as infinite Galois theory is concerned, as pointed out by Keith Conrad in the comments.

I hope I haven't forgotten anything.

References:

[DugGöb87] Manfred Dugas and Rüdiger Göbel. “All Infinite Groups Are Galois Groups Over Any Field.” Transactions of the American Mathematical Society, vol. 304, no. 1, 1987, pp. 355–84

[Fr80] M. Fried."A note on automorphism groups of algebraic number fields".Proc. Amer. Math. Soc. 80 (1980), 386-388

[Ge83] Geyer, WD. Jede endliche Gruppe ist Automorphismengruppe einer endlichen Erweiterung $K|\mathbb{Q}$. Arch. Math 41, 139–142 (1983)

[Wat74] William C. Waterhouse."Profinite groups are Galois groups".Proc. Amer. Math. Soc. 42 (1974), 639-640

Every infinite group is the Galois group over any field.

Reference: https://www.jstor.org/stable/2000718

PS: I'll add some details when I get home, it's hard to type atm.

EDIT: Apologies for the delay. I was too tired yesterday for anything more than a few comments. So let me turn this into a proper answer for everyone's sake, with a few more details added, and recap the discussion from yesterday.

1. Case: $G$ finite

It follows from Artin's theorem that every finite group $G$ is the Galois group of some finite Galois extension $L/L^G$, but we have no control over $L^G$ (this would be the inverse Galois problem). As Keith Conrad points out in the comments, however, this is a non-problem in the context of OP's question, because every finite group is the automorphism group of some finite, not necessarily Galois extension $L/\mathbb{Q}$ as shown in [Fr80]. A somewhat simpler proof of this is also given in [Ge83] (in German). This settles OP's question in the finite case.

2. Case: $G$ infinite

An infinite Galois group in the sense of infinite Galois theory is necessarily profinite. Conversely, every profinite group is the Galois group (in the sense of infinite Galois theory) of some extension $L/L^G$ as shown in [Wat74], alas once again we have no control over $L^G$.

But once again this is a non-problem in the context of OP's question. Even more generally, it is proved in [DugGöb87] that for every prescribed infinite group $G$ and every prescribed base field $K$ there exists an extension $L = L(G,K) / K$ (in their notation $R(K,G)$) such that $\operatorname{Aut}_K(L) \cong G$. Taking $K:=\mathbb{Q}$, we get $G \cong \operatorname{Aut}(L)$ (the full automorphism group of $L$), which settles OP's question in the infinite case.

Note that, as pointed out by Keith Conrad in the comments, the class of infinite groups is much wider than that of profinite groups, for example because profinite groups are either finite or uncountably infinite. I implicitly (and incorrectly) assumed from the paper's title that in the profinite case their construction would in fact produce a Galois extension of $K$. But firstly, this was overly optimistic, given that we don't even know the full answer for a general finite $G$ and $K:=\mathbb{Q}$, and secondly, we can simply take $K \cong \bar{K}$ algebraically closed (which in turn of course does not admit proper algebraic extensions). So, beware that their usage of the term "Galois group" is very non-standard as far as infinite Galois theory is concerned, as pointed out by Keith Conrad in the comments.

I hope I haven't forgotten anything.

References:

[DugGöb87] Manfred Dugas and Rüdiger Göbel. “All Infinite Groups Are Galois Groups Over Any Field.” Transactions of the American Mathematical Society, vol. 304, no. 1, 1987, pp. 355–84

[Fr80] M. Fried."A note on automorphism groups of algebraic number fields".Proc. Amer. Math. Soc. 80 (1980), 386-388

[Ge83] Geyer, WD. Jede endliche Gruppe ist Automorphismengruppe einer endlichen Erweiterung $K|\mathbb{Q}$. Arch. Math 41, 139–142 (1983)

[Wat74] William C. Waterhouse."Profinite groups are Galois groups".Proc. Amer. Math. Soc. 42 (1974), 639-640