Finite Abelian Group: Proving # Subgps Order p = # Subgps Index p

[fightalgebra]

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!

 

[mighty2000]

Thank you for your post. This is a very interesting problem that requires a deep understanding of abelian groups and their subgroups. Let me provide you with a hint that may help you in your proof.
First, consider the case where A is a cyclic group of order p^n, where p is a prime. In this case, it is easy to see that the number of subgroups of order p is p^n - 1, since there are p^n elements in A and only one of them is the identity element.
Next, consider the case where A is a direct product of two cyclic groups of order p^n and p^m, where n > m. In this case, it can be shown that the number of subgroups of order p is (p^n - 1)(p^m - 1), which is equal to the number of subgroups of index p.
Using these two cases, you can try to generalize the result for any finite abelian group A by using the fundamental theorem of finite abelian groups, which states that any finite abelian group is isomorphic to a direct product of cyclic groups.
I hope this helps. Good luck with your proof!