number of subgroups of $\mathbb{Z}_p \oplus \mathbb{Z}_p \oplus \mathbb{Z}_p$

Question

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.?

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 — jnyan, Dec 15 '17 at 18:24
see https://en.wikipedia.org/wiki/Gaussian_binomial_coefficient — Angina Seng, Dec 15 '17 at 18:28

score 1 · Answer 1 · answered Dec 15 '17 at 19:46

1

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?

answered Dec 15 '17 at 19:46

TomGrubb

12,909

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. – jnyan Dec 16 '17 at 04:28

number of subgroups of $\mathbb{Z}_p \oplus \mathbb{Z}_p \oplus \mathbb{Z}_p$

1 Answers1