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The question is to draw the complete lattice of subgroups of $(\mathbb{Z}/32\mathbb{Z})^{\times }$ and for each proper subgroup, identify the isomorphism type.

(Accoding to definition, $(\mathbb{Z}/32\mathbb{Z})^{\times }=\left \{ {\overline{1},\overline{3},\overline{5},\overline{7},\overline{9},...,\overline{31}}\right \}$, but what then? Can these elements form any subgroup? how to draw the lattice and for each proper subgroup, identify the isomorphism type?)

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A general result is that $\mathbb{Z}/2^n\mathbb{Z}^\times$ is generated by the residue classes of $5$ and $-1(\equiv31$ in this case). You can find many proofs for that fact on this site, and use that as a starting point. – Jyrki Lahtonen Feb 26 '13 at 22:12
To me, it's more straightforward to first figure out the isomorphism type of $(\mathbb Z/32\mathbb Z)^\times$ (Jyrki's comment indicates how), and then you'll know what possible isomorphism types its subgroups will have. – Greg Martin Feb 26 '13 at 23:06

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