Let be a finite abelian group and let be a prime. Prove that the number of subgroups of of order equals the number of subgroups of of index .
although this is not "quite" the same problem, i believe it is similar enough to help you: finite abelian group
I can prove that has a total of distinct subgroups of order .
To solve the original question(post #1) by the approach you have suggested i need to find the number of subgroups( of course distinct) of order of abelian groups like etc. What i have found out (which i am not 100% sure is correct) that number of subgroups of order of is same as the number of subgroups of order of etc. is this correct??
Now i couldn't figure out how to find the number of subgroups having index . Can you please help on this one??
I have to say that I thought I knew the answer to this problem when I posted the hint. But I realised later that I didn't. Anyway, I can say that has more subgroups of order p than because the latter has only , and , while the first has at least and for k<p. Plus the analogous on the side of
If I would have to resolve this problem, I would try to find, with specific examples, such as , the group of index 5 for each group of order 5 and thus understand the one to one correspondence.