2
$\begingroup$

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.?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.