I've difficulty in understanding the last steps of the AHSP algorithm. Let $G$ be an abelian group and $f$ be the function which hides the subgroup $H$. Let $G^*$ represent the dual group of $G$.Here are the steps of the algorithm First prepare the state, $\qquad \displaystyle I=\frac{1}{|G|} \sum_{g \in G} |g\rangle|0\rangle$.Then apply the quantum oracle which evaluates $f$ on $I$, we get $\qquad \displaystyle I'=\sum_{g \in G} |g\rangle|f(g)\rangle$.Now measure the the second qubit of $I'$, we get $\qquad\displaystyle I'= \left(\frac{1}{|H|}\Sigma_{g \in H} |rh\rangle\right) \otimes |f(rh)\rangle$ for some $r \in G$.Now we  apply the quantum fourier transform on the first qubit, we get $\qquad \displaystyle I_m = \frac{1}{|H^*|}\sum_{\chi \in H^*} |\chi\rangle$, where $H^*= \{\chi \in G^*: \chi(h)=1 ,\forall h \in H\}$.Now from the state $I_m$ how can we get generators of the group $H$?
Difficulty in understanding the quantum algorithm for the abelian hidden subgroup problem
ds.algorithms;quantum computing;gr.group theory