Let $k$ be a field of characteristic $0$. Fix an integer $m$, and assume $k$ contains $m$ distinct roots of unity. Consider all finite extension of $k$ inside some fixed algebraic closure of $k$.
(1) Let $P$ be a multiplicative subgroup of $k^*$ containing $(k^*)^m=\{a^m\,\,|\,\,a\in k^*\}$.
(2) Let $E$ be the smallest subfield of $\overline{k}$ containing all roots of $x^m-a$ for all $a\in P$.
(3) Then $E$ is abelian (Kummer) extension of $k$, of exponent $m$. Let $G:={\rm Gal}(E/k)$.
(4) Define $G\times P\rightarrow k^*$ by $(\sigma,a)\mapsto \frac{\sigma(\alpha)}{\alpha}$ for any $\alpha\in E$ satisfying $\alpha^m=a$. [This is well-defined.]
(5) The map in (4) is bimultiplicative. If $(\sigma,a)\mapsto 1$ for all $a\in P$ then $\sigma=1$.
(6) If $(\sigma,a)\mapsto 1$ for all $\sigma\in G$ then for any $\alpha$ with $\alpha^m=a$, $\sigma(\alpha)=\alpha$, so $a\in k^*$, i.e. $a\in (k^*)^m$.
(7) It follows that $P/(k^*)^m$ and $G$ are dual to each other; in particular, $P/(k^*)^m\cong {\rm Hom}(G,k^*)$ whenever either side is finite.
Q. This is from Basic Algebra by Cohn p.442-443. I didn't understand (7) (almost nothing in it!)
[When do we say two groups are dual to each other? Also in (7), the particular case of isomorphism is only when either side is finite; where this fails for infinite size? I didn't understand it. In the infinite size of either of $P/(k^*)^m$ and ${\rm Hom}(G,k^*)$, I think, $P/(k^*)^m$ to ${\rm Hom}(G,k^*)$ we get injective hmomorphism, but not necessarily surjective. Why this can happen? ]