0
$\begingroup$

I want to know the number of non-trivial subgroups of $\mathbb{Z}_{p^2}\times \mathbb{Z}_p$. For example, for $p=2$, we have $6$ non-trivial subgroups. I think there exists only $3$ subgroups of order $p$.

Is there any formula for counting the number of subgroups $\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}\times ... \times \mathbb{Z}_{n_k}$?

1 Answers 1

1

For $\mathbb{Z}_{p^2}\times \mathbb{Z}_p$, we have

  • one subgroup of order $1$
  • $p+1$ subgroups of order $p$, as the elements of order $p$ generate a subgroup of order $p^{2}$
  • the just mentioned subgroup of order $p^{2}$
  • $(p^{3} - p^{2})/(p^{2} - p) = p$ cyclic subgroups of order $p^{2}$
  • the whole group.

In total, including the trivial subgroup and the whole group, we have $$ 2 p + 4 $$ subgroups.

For the general case, I would refer to Goursat's Lemma. It is spelled out in

Counting Subgroups in a Direct Product of Finite Cyclic Group by Joseph Petrillo, The College Mathematics Journal Vol. 42, No. 3 (May 2011), pp. 215-222.

  • 0
    Can we say that every subgroup of order $p^2$ has the element $(p,0)$? – 2017-02-10
  • 1
    Yes, because if $H$ is a subgroup of order $p^{2}$ of $G = \mathbb{Z}_{p^2}\times \mathbb{Z}_p$, then $G/H$ has order $p$, and thus the $p$-th multiple of each element of $G$ is contained in $H$. In particular $p \cdot (1, 0) = (p, 0) \in H$. – 2017-02-10