G finite abelian group

1 subgroup of order d for each divisor d of |G|.

Prove G cyclic.

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- Nov 2nd 2010, 07:47 PMstumped765finite abelian groups
G finite abelian group

1 subgroup of order d for each divisor d of |G|.

Prove G cyclic. - Nov 2nd 2010, 08:00 PMDrexel28
What if we proved the contrapositive. Suppose that $\displaystyle G$ is not cyclic. Try constructing subgroups of $\displaystyle G$ which are distinct but have the same order (hint: you're going to want to construct these subgroups to be cyclic groups [math}C_n,C_m[/tex] say with some prime $\displaystyle p\mid m,n$)