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fightalgebra
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Homework Statement
A be a finite abelian group, prove # of subgps of order p = # of subgps of index p, p is a prime.
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!
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