Isomorphic subgroups of the multiplicative group of a field

Question

Let $F$ be a field. Let $H,K\subseteq F^\times$ be subgroups of the multiplicative group $F^\times$. Suppose $H$ and $K$ are isomorphic as groups. Must there necessarily be a group automorphism $\sigma$ of $F^\times$ (or perhaps $\sigma$ is even a field automorphism of $F$) such that if we let $\phi=\sigma\big\vert_H$ (the restriction of $\sigma$ to $H$), then $\phi:H\rightarrow K$ is an isomorphism between $H$ and $K$?

If $H$ and $K$ are finite, then this is trivial: for every $n$, there is at most one subgroup of $F^\times$ that is of order $n$: the set of $n$-th roots of unity. So if $H$ and $K$ are finite and isomorphic, then they are identical. Furthermore, since in this case $H=K$ is the unique finite subgroup of $F^\times$ of its order, it is characteristic in $F^\times$, and thus it is mapped into itself by every automorphism of $F^\times$.

Also, just a note to myself: I only just realized that there can be group automorphisms of the multiplicative group of a field which do not arise from field automorphisms of the field.

For example, the map $x\mapsto x^3$ is certainly not a field automorphism of $\mathbb{R}$, but it is a group automorphism of $\mathbb{R}^\times$. That's because:

1. $(xy)^3=x^3y^3$, so the map $x\mapsto x^3$ is a homomorphism
2. $x^3=1\implies x=1$ (since $x$ is real), so the kernel of the homomorphism is trivial
3. For every $x\in\mathbb{R}$, there is $a\in\mathbb{R}$ such that $x=a^3$; that is to say, the map $x\mapsto x^3$ is surjective.

Answer 1

Let us take the $F$ to be $\mathbf{Q}$ (or a number field like $\mathbf {Q}{(\sqrt2)}$ which is the field of fractions of a principal ideal domain $R$). By unique factorization theorem, the group $\mathbf{Q}^*$ is isomorphic to $\{\pm1\}\times A$, where $A$ is the free abelian group on the set of all the prime numbers.

Let $H,K$ be multiplicative subgroups generated by $2$ and $3$ respectively. Both these groups are isomorphic (because they are isomorphic to the additive group $\mathbf Z$). As any field automorphism has to be identity on $\mathbf{Q}$ no isomorphism of $H$ with $K$ can be extended to a field automorphism of $\mathbf Q$.

Now set up a bijection from the set of all odd primes to all the primes . This gives a surjective homomorhism, from a proper subgroup of $\mathbf{Q}^*$ (generated by odd primes) to the whole, and any extension of this will fail to be injective, so does not give any group automorphism of $\mathbf{Q}^*$.

Answer 2

I haven't checked all the details (edit: I did in the edit), but I imagine the answer is no.

Take $\Bbb{Q}_p$, the field of $p$-adic numbers, and consider the multiplicative subgroups $1+p^n\Bbb{Z}_p$ and $1+p^m\Bbb{Z}_p$ for $n,m\geq 1$ and $m\neq n$. These are "balls" around $1$, and due to the non-archimedean nature of the $p$-adics, they are closed with respect to multiplication. They are isomorphic as groups, by sending $1+\alpha p^n\mapsto 1+\alpha p^m$, but this map cannot be extended to a field automorphism of $\Bbb{Q}_p$. For example, we could have $\alpha = p^n$.

I have not entirely checked that there could not be another group automorphism that would provide an isomorphism, but I imagine it isn't too hard.

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Edit: As you are not familiar with $p$-adic numbers, let me give all the details.

Let $B_k:=1+p^k\Bbb{Z}_p$, and let $m>n$. Assume for contradiction there exists a group automorphism $\phi:\Bbb{Q}_p\to\Bbb{Q}_p$ such that $\phi|_{B_n}$ is a group isomorphism of $B_n$ and $B_m$. We note that $1+p^n$ and $1+p^{n+1}$ lie in $B_n$. By being a group homomorphism, $\phi$ is completely determined on $\Bbb{Z}$ by $\phi(1+p^n)=1+\alpha p^m$ for some $\alpha\in\Bbb{Z}_p$, so we get $\phi(z)=z\frac{1+\alpha p^m}{1+p^n}$ for $z\in\Bbb{Z}$.

Now we will show that $\alpha\notin \Bbb{Z}_p$, a contradiction. Let $$D:=(1+p^n)(1+p^{n+1})=1+p^n+p^{n+1}+p^{2n+1}.$$ We see that, by the above, and using that $\phi|_{B_n}$ is a group isomorphism, we can compute the image of $D$ in two ways:

$$\phi(D)= D\frac{1+\alpha p^m}{1+p^n},$$ $$\phi(D) = \phi(1+p^n)\phi(1+p^{n+1}) = (1+\alpha p^m)^2\frac{1+p^{n+1}}{1+p^n}.$$ Putting these identities together and reducing, we get $$1+p^n+p^{n+1}+p^{2n+1} = D = (1+\alpha p^m)(1+p^{n+1}) = 1+\alpha p^m+p^{n+1}+\alpha p^{m+n+1}.$$ After cancellation and moving around, we get $$\alpha = p^{n-m}\notin \Bbb{Z}_p.$$