Abelian finite groups and their subgroups

This has been, in some sense, answered in this recent MO question: https://mathoverflow.net/q/366008

The short answer is that if you consider exponent $p^n$ for $n\geq 7$, there are wildly many indecomposable pairs. There are infinitely many for $n=6$, according to Jeremy Rickard in that answer.

I wondered why I was having so much trouble even with the case $H$ cyclic! I do have a partial result in that case if it's of interest, but it gets very messy as the number of summands of $G$ increases. Everything is nice for two summands, becomes complicated for three summands, and then I can only write down heuristics after that. For three summands you need $n\geq 6$, which chimes with Jeremy's statement.

Edit: I will write down what I found for the cyclic case, although it's not so obvious now it's of any value.

Notice that being decomposable in the sense of the question is, for cyclic $H$, the same as being contained in an overgroup that has a complement. Call such an element a co-element (complemented overgroup), and otherwise an nco-element. If $x$ is a generator for $H$, it means that $x$ can be written as $(x_1,\dots,x_n)$, where $x_i$ are coefficients in a basis for $G$, and there exists a basis such that some $x_i$ is $0$.

So we can take the projection of $x$ onto some summand of $G$ and work with that. This immediately shows the following:

If $G_1$ has no nco-elements, and $G_1$ is a summand of $G$, then $G$ has no nco-elements.

From now on we will hunt for nco-elements.

We next see that $G=C_{p^n}\times C_{p^n}$ has no nco-elements. This follows easily: every element of $G$ has a root of order $p^n$, and elements of maximal order in abelian groups are complemented. (This is the start of a standard proof of the cyclic decomposition theorem for abelian groups.)

Thus $G=C_{p^{a_1}}\times \cdots \times C_{p^{a_r}}$, and all $a_r$ are distinct. Arrange so that $a_i>a_{i+1}$.

If $G=C_{p^a}\times C_{p^b}$ ($a>b$) and $x$ has no root of order greater than $p^b$ then $x$ is a co-element. Certainly we can take a root of $x$ of order $p^b$, so assume that $o(x)=p^b$. Note that $x=(x_1^{p^{a-b}\alpha},x_2^\beta)$. Such an element generates a complement to a subgroup of order $p^a$ unless $\beta$ is a multiple of $p$. But then $x$ has a $p$th root, a contradiction.

So $x$ can be chosen to have order greater than $p^b$, and cannot have order $p^a$ (as maximal-order elements are co-elements). Thus $a\geq b+2$.

Now there are lots of such elements. We just need to choose an element of order greater than $p^b$ that does not have a root of order $p^a$. For example, $(x_1,px_2)$ will do.

This solves the case of two summands for $G$. In general, we can choose any pair inside a decomposition of $G$ into summands and such an element must be an nco-element, otherwise we can complement it in that summand, and yield a complement in the whole group.

(This next bit is corrected, as noted by Jeremy in the comments.)

So the smallest possible case allowed by this is $C_{p^5}\times C_{p^3}\times C_{p}$, generated by $a,b,c$, and where $x$ should be something like $p^2a+pb+c$. That works for each of $\langle a,c\rangle$, $\langle b,c\rangle$ and $\langle a,b\rangle$. This gives all such elements for exponent $p^5$.

This yields:

Suppose that $G$ has an nco-element, and that $G$ is $d$-generator for $d\geq 3$. Then $G$ has exponent at least $p^5$. If $G$ has exponent $p^5$ then $d=3$ and the examples are exactly as above.

To generalize this you need a step of at least two in between the orders of the summands, and an increasing power of $p$ in the representation of the element $x$.