Let $A$ be an elementary Abelian uncountable $p$-group. Is it known if there is an action of a Prufer $q$-group (here $q$ is a prime not necessarily distinct from $p$) $C\_{q^{\infty}}$ onto $A$ such that $A$ does not contain proper uncountable $C\_{p^{\infty}}$-invariant subgroups?



Uncountable cardinals and Prufer $p$-groups
The answer is no:



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> For all primes $p,q$, for every uncountable elementary abelian $p$-group, for every action of the Prüfer $q$-group $C\_{q^\infty}\simeq\mathbf{Z}[1/q]/\mathbf{Z}$ on $A$ by group automorphisms, there exists a proper uncountable $C\_{q^\infty}$-invariant subgroup in $A$.
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There are 2 distincts proofs: for $q=p$ and $q\neq p$.


When $q=p$:
Let $(x\_n)\_{n\ge 0}$ be generators of $C\_{p^\infty}$ with $x\_0=1$ and $x\_{n+1}^p=x\_n$ for all $n$. Define $y\_n=x\_n-1$ (in the group algebra $\mathbf{F}\_p[C\_{p^\infty}]$). Then $y\_{n+1}^p=y\_n$ for all $n$ and $y\_0=0$. Define $I\_n=y\_nA$; then $I\_n\subset I\_{n+1}$ for all $n$; define $I\_\infty=\bigcup I\_n$. Then modulo $I\_\infty$, every $x\_n$ acts as the identity and hence every $\mathbf{F}\_p$-subspace of $A$ containing $I\_\infty$ is a submodule. Hence if $I\_\infty\neq A$, then $A$ contains proper submodules of at most countable index (hence uncountable if $A$ is uncountable).


If $I\_\infty=A$ and $A$ is uncountable, then $I\_n$ is uncountable for some some $n$. If $I\_n\neq A$, we are done. Otherwise, $y\_nA=A$ and $y\_n$ nilpotent force $A=0$, contradiction.




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Now assume that $q\neq p$.


If $A$ is uncountable and splits as a direct sum of two submodules, one of the two is uncountable and we are done.


Assume otherwise ($A$ is "indecomposable"). Again denote by $x\_n$ the generators of $C\_{q^\infty}$ with $x\_0=1$ and $x\_{n+1}^q=x\_n$. Then $x\_n$ is an operator on $A$ vanished by some polynomial (namely $X^{q^n}-1$) with simple roots, hence for each $n$, we have a decomposition $A=\bigoplus\_{P\in\mathcal{P}\_n}A\_P$, where $\mathcal{P}\_n$ is the set of monic irreducible divisors of $X^{q^n}-1$ and $A\_P=\mathrm{Ker}(P(x\_n))$. Since $A$ is indecomposable, there exists a unique $P\in\mathcal{P}\_n$ such that $A\_P\neq 0$; denote it by $P\_n$. So $P\_n$ is irreducible and $P\_n(x\_n)=0$. Thus the subring $R\_n$ of $\mathrm{End}\_{\mathbf{F}\_p}(A)$ generated by $x\_n$ is a (finite) field. We have $R\_n\subset R\_{n+1}$ for all $n$ and hence $K=\bigcup R\_n$ is a field as well; it is at most countable. On the other hand, the image of $C\_{q^\infty}$ in $\mathrm{End}\_{\mathbf{F}\_p}(A)$ is contained in $K$. Hence any decomposition of $A$ as a $K$-module is $C\_{q^\infty}$-invariant. In particular, we deduce that if $A$ is uncountable, then it is decomposable (as a direct sum of uncountably many countable submodules), contradiction.


