I am trying to find number of subgroups of $\mathbb{Z}_p \oplus \mathbb{Z}_p \oplus \mathbb{Z}_p$ for prime $p$. I got $p^2+p+1$ subgroups of order $p$, but for subgroups of order $p^2$, I cant see how would it be $p^2 + p+1$ subgroups?

Is there any generalised method for $\mathbb{Z}_p \oplus \mathbb{Z}_p \oplus \mathbb{Z}_p \oplus \mathbb{Z}_p$..... and so on.?

  • $\begingroup$ try $p=2$.${}{}{}$ $\endgroup$ – Andres Mejia Dec 15 '17 at 18:17
  • $\begingroup$ i was able to verify the formula, but I could just not see how was that obtained. I mean for order $p$, i took the number of $p$ order elements. but$p^2$ group is direct product of groups and I got lost $\endgroup$ – jnyan Dec 15 '17 at 18:24
  • $\begingroup$ see en.wikipedia.org/wiki/Gaussian_binomial_coefficient $\endgroup$ – Lord Shark the Unknown Dec 15 '17 at 18:28

You can consider this group a vector space over the field $\mathbb{Z}_p$. Your question thus reduces to this question How to count number of bases and subspaces of a given dimension in a vector space over a finite field?

  • $\begingroup$ ok. thanks. I solved it by figuring out $p^2$ order subgroup has p+1 $p$ order cyclic groups. and then calculating number of ways p+1 cyclic group combination, and taking care of double counting. $\endgroup$ – jnyan Dec 16 '17 at 4:28

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