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We know the following isomorphism theorem for fields: Let $F, F'$ be fields. Suppose $E$ is an extension field of $F$ and $\overline{F}$ is the algebraic closure of $F$. Also $\overline{F'}$ is the algebraic closure of $F'$. Suppose $\sigma$ is an isomorphism from $F$ to $F'$. Then we can extend it to an isomorphism $\tau: E \to \tau[E]$.

Is there any analogous "extension" theorem for groups?

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closed as not a real question by Qiaochu Yuan, Charles Siegel, Pete L. Clark, S. Carnahan, Anton Geraschenko Jan 20 '10 at 8:17

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An isomorphism from F to F'? An isomorphism of rings or an isomorphism of k-algebras for some underlying field K? You need to be a little clearer. –  Harry Gindi Jan 20 '10 at 0:29
Your statement for fields didn't ever use the algebraic closures. –  Richard Kent Jan 20 '10 at 0:34
Yeah, this needs to go back to a drawing board. –  Harry Gindi Jan 20 '10 at 0:43
Thomas, I too don't understand what your statement about fields means. In particular: (1) what do you mean by tau[E], and (2) why do you carefully introduce notation for the algebraic closure of a field, but never make use of it? –  Tom Leinster Jan 20 '10 at 0:48
Okay, I see now that you did define E. But you seem a little confused; the notation tau[E] does not really mean what you want it to mean because you have not defined what tau does to elements of E not in F. –  Qiaochu Yuan Jan 20 '10 at 0:50

2 Answers 2

I'm not quite sure if you want to pick the extension first and then ask for an extension, or if you'd like to just extend to anything at all that's larger, but the following may be useful:

$\mathbb{Z}$ embeds in $\mathbb{Q}$. It also embeds in some infinite simple group $E$. So the inclusion of $\mathbb{Z}$ into $\mathbb{Q}$ does not extend to a homomorphism $E \to \mathbb{Q}$.

(More generally, and perhaps of interest: it's an old theorem of Philip Hall that any countable group embeds in a finitely generated simple group.)

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Let me see if I can clear up your confusion. It is true that any algebraic extension of a field $F$ embeds in a given algebraic closure, but this embedding is not unique; you can compose any given embedding with an element of the Galois group of the normal closure, and if the extension isn't itself Galois then you'll even end up with embeddings that have different images. Also, it's usually considered bad form to say "the" algebraic closure of a field because algebraic closures are only unique up to isomorphism and the choice of an algebraic closure cannot be made canonically (to my knowledge). Anyway, here's what I think you wanted to say:

"If I have an extension $E/F$ and an isomorphism $\sigma : F \to F'$, I should be able to find a corresponding extension $E'/F'$ and an isomorphism $\tau : E \to E'$ which restricts to $\sigma$."

This statement is true, but again, $\tau$ is not unique. (And I wouldn't call this an "isomorphism theorem." Those refer to a specific set of theorems.) I also don't know why you give a statement about isomorphisms and then ask for a statement about homomorphisms (in the group setting). What kind of statement, exactly, are you looking for?

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Here's one possible group statement (not exactly analogous to the field one): If $\sigma:H\rightarrow H'$ is an isomorphism of groups, $G\supset H$ and $G'\supset H'$ are "overgroups" (as they are apparently called), and $\sigma':G\rightarrow G'$ is an isomorphism (but $\sigma'$ does not necessarily restrict to $\sigma$), then there is also an isomorphism $\tau:G\rightarrow G'$ which does restrict to $\sigma$. I'm not sure if this is trivially true (or false), just throwing it out there. –  Zev Chonoles Jan 20 '10 at 1:11
That statement's false, Zev. Take the group $E$ from my example. Then $E \times \mathbb{Q}$ is isomorphic to itself, but there is no isomorphism carrying an infinite cyclic subgroup of $E$ into $\mathbb{Q}$. –  Richard Kent Jan 20 '10 at 1:18
Note that $G\supset H$ and $G'\supset H'$ though (and I was assuming that $H\simeq H'$ to begin with). –  Zev Chonoles Jan 20 '10 at 1:31
Yeah, $H$ and $H'$ are both isomorphic to $\mathbb{Z}$, $H$ is in $E$, $H'$ is in the $\mathbb{Q}$ factor, and $G$ and $G'$ are $E \times \mathbb{Q}$. Then there is no homomorphism from $G$ to itself carring $H$ to $H'$. –  Richard Kent Jan 20 '10 at 1:43
Ah, ok - I was confused about your arrangement. –  Zev Chonoles Jan 20 '10 at 2:16

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