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:
- $(xy)^3=x^3y^3$, so the map $x\mapsto x^3$ is a homomorphism
- $x^3=1\implies x=1$ (since $x$ is real), so the kernel of the homomorphism is trivial
- 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.
Best Answer
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 $\mathbf{Q}^*$.