# Subgroups and quotient groups of finite abelian $p$-groups

Motivation

The fundamental theorem of finite abelian groups gives us a concise description of the isomorphism types of finite abelian $p$-groups $G$ (in the following, $p$ is a fixed prime). The isomorphism type can be encoded in a sequence of numbers (a partition), the type of $G$.

Since subgroups $H$ and quotient groups $G/H$ of $G$ are again finite abelian $p$-groups, this suggests the question which types can appear for $H$ and $G/H$.

Definition

Let $G$ be a finite abelian $p$-group. By the fundamental theorem of finite abelian groups, there is a unique non-increasing sequence $a = (a_0, a_1, a_2,\ldots)$ of non-negative integers such that $$G \cong \mathbb Z/p^{a_0}\mathbb Z \times \mathbb Z/p^{a_1}\mathbb Z \times \mathbb Z/p^{a_2}\mathbb Z \times \ldots$$ This sequence is called the type of $G$ and will be denoted by $\operatorname{type}(G)$. We assume that the set of types is partially ordered by component-wise comparison.

Since $G$ is finite, starting with some position, all entries in the infinite sequence $\operatorname{type}(G)$ are zero. As usual in the theory of partitions, we may drop all those zero entries to represent the type with a finite non-decreasing sequence of positive integers.

Example

Consider $p = 2$ and $G = \mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z \times \mathbb Z/8\mathbb Z$. Then $\operatorname{type}(G) = (3,2,2,0,0,0,\ldots) = (3,2,2)$. This type is smaller than the type $(4,3,2)$, larger than $(3,2)$ and not comparable to $(4,2)$.

Problem

Let $G$ be a finite abelian $p$-group and $H$ be a subgroup of $G$.

Prove or disprove:

1. $\operatorname{type}(H) \leq \operatorname{type}(G)$.
2. $\operatorname{type}(G/H) \leq \operatorname{type}(G)$.

My first feeling was that both statements should be true and not too hard to prove. I've tried to attack the problem by various inductive arguments. Maybe I'm blind, but it didn't work out so far. This leaves me wondering if (one of) the statements might actually be wrong. But I couldn't find a counterexample, either.

To document a failed attempt which came up in the discussion below: By the type of $G$, it is not hard to count the number of elements of a certain order in $G$.

• For example, a group of type $(2,2)$ has $1$ element of order $1$, $p^2 - 1$ elements of order $p$ and $p^4 - p^2$ elements of order $p^2$. A group of type $(1,1,1)$ has $1$ element of order $1$ and $p^3 - 1$ elements of order $p$. Since $p^3 - 1 > p^2 - 1$, this shows that a group of type $(2,2)$ doesn't have a subgroup of type $(1,1,1)$.
• However, the comparison of the number of elements of a certain order is not enough to derive the claimed statement 1. For example, type $(2,1,1,1)$ has $p^4 - 1$ elements of order $p$ and $p^5 - p^4$ elements of order $p^2$. These two numbers are larger than those for type $(2,2)$ above. So for a group of type $(2,1,1,1)$, the subgroup type $(2,2)$ cannot be eliminated in this way.
• Which ordering are you putting on the partitions? – Tobias Kildetoft Oct 16 '13 at 12:12
• perhaps I'm missing something, but isn't a little weird to define that apparently infinite type sequence of a finite abelian p- group in a decreasing order? You must then know what the highest number (exponent of p) appears to begin with the sequence... – DonAntonio Oct 16 '13 at 12:13
• Ahh, I missed that somehow (I usually denote that ordering by $\subseteq$). Then it should be clear that both statements are true (note that any quotient of an abelian group is isomorphic to a subgroup as well, so we only need to do one of them). – Tobias Kildetoft Oct 16 '13 at 12:15
• @DonAntonio If it was to be non-decreasing, you could not let it end in $0$'s. – Tobias Kildetoft Oct 16 '13 at 12:17
• The proof I know goes by showing first that the set of homomorphisms from an abelian group to $\mathbb{C}^{\times}$ is isomorphic to the group itself with usual multiplication (the proof can be done by induction on the number of cyclic groups in the decomposition and by checking that the "obvious" map then becomes injective, and the surjective part then follows by a counting argument). The proof then goes by noticing that given a subgroup, one gets the quotient by that subgroup as a subgroup of the group of homomorphisms (as the set of those homomorphisms vanishing on the given subgroup). – Tobias Kildetoft Oct 19 '13 at 23:59

For the simplicity of symbol, write $A^l=\{ a^{p^l} | a \in A \}$ for the abelian $p$-group. Let $type(G) =(n_1, n_2, \cdots, n_r)$ and $type(G/H)=(m_1, m_2, \cdots, m_r)$. Suppose that for some $t$, $n_i \ge m_i$ ($i =1, \cdots, t$), but $m_{t+1}>n_{t+1}$. Let $l=n_{t+1}$. Now consider $G^l$ and $(G/H)^l$. Clearly $G^l$ is the direct product of $t$ cyclic sugroups, and $(G/H)^l$ is the direct product with at least $t+1$ factors. This means $G^l$ can be generated by $t$ elements, but $(G/H)^l$ can not be generated by $t$ elements. Note that $(G/H)^l=G^lH/H \cong G^l/(G^l \cap H)$, wich can be generated by $t$ elements. This contradiction completes the proof of your problem 2.