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Let $G$ be an abelian group of order $p^{r}$. Is there a way to count how many subgroups of $G$ with order $p^{r - 1}$ are there?

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Every such subgroup contains $pG$, so you can mod that out and count in $G/pG$ the maximal subgroups. That quotient is a vector space of $\mathbb F_p$, so that is easy. You need only find the dimension of $G/pG$. Can you do that from the elementary factors, for example? – Mariano Suárez-Alvarez Nov 4 '11 at 0:16
up vote 4 down vote accepted

If your p-group is generated by $d$ elements, then it has a well-defined quotient which looks like the $d$-dimensional vector space over $\mathbb{Z}/p\mathbb{Z}$. This quotient is $G/\Phi(G)$, so all maximal subgroups are preserved. Specifically, your $G$ has the same number of maximal subgroups as this vector space does, and those are easy to count: there are $\dfrac{p^d-1}{p-1}$.

None of this requires $G$ to be abelian. However, if $G$ is abelian, your $d$ is the number of factors in the usual cyclic decomposition.

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:Why the number of maximal subgroups of that vector space is $\frac{p^{d}-1}{p-1} \ $? – WLOG Dec 26 '11 at 19:40
@stef: Vector spaces have this nice reducibility property: if I give you one subspace, you can find a complement. Counting maximal subgroups is then the same as counting minimal subgroups, or 1-dimensional subspaces... – user641 Dec 28 '11 at 8:37

Not without more information. The cyclic group of order $p^r$ has only one subgroup of order $p^{r-1}$; the external direct sum of $r$ copies of $\mathbb{Z}/p\mathbb{Z}$ has $r$ of them obtained by fixing the projection on any one factor to be the identity element.

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This is not right. Think of $p=2$, $r=2$; how many subgroups of order 2 are there? – user641 Nov 4 '11 at 0:47
@Steve: The cyclic group of order $4$ has one. The external direct sum of $2$ copies of $\mathbb{Z}/2\mathbb{Z}$ has $2$ that are obtained by fixing the projection on one factor to be $0$. I didn’t say that those were the only ones: I was merely pointing out that the answer wasn’t determined by the information given. There is no comma between them and obtained, so the obtained specification is restrictive, not non-restrictive. – Brian M. Scott Nov 4 '11 at 1:00

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