1. The problem statement, all variables and given/known data A be a finite abelian group, prove # of subgps of order p = # of subgps of index p, p is a prime. 3. The attempt at a solution I have thought about this probably very easy problem for 2 hours and could not find a satisfying proof. I have tried bijective proof but failed (sending <x> |-> A/<x> is fruitless), and I tried elementary divisor decomposition (writing A as G x Z_(p^(a_1))x... x Z_(p^(a_n)) x H, focusing on the cyclic p-group part, and I have found that the number of order p subgroups must be p^n - 1 in any such A) which I think is the right direction but still can not work it out..T_T...please someone please give me some hint please...please don't refer to too advanced a theorem apart from the two abelian group decomposition theorems...thank you so much!!