Let $p$ be a prime number. I am interested in the abelian groups $G$ with the following property:

(U) every finite abelian $p$-group $A$ admits a monomorphism $A\hookrightarrow G$.

In other words, every finite abelian $p$-group is a subgroup of $G$.

For instance, denoting by $C_n$ the cyclic group of order $n$, the group $$ G_p:= \bigoplus_{k=1}^\infty C_{p^k} $$ has this property.

I am interested whether (U) is equivalent to the following property:

(U') there is a monomorphism $G_p\hookrightarrow G$.

Clearly, (U')$\Rightarrow$(U), but the opposite implication does not look obvious to me (perhaps it is wrong).

Another related question: does it exist a "nice" characterization of the abelian groups $G$ such that both monomorphisms $G\hookrightarrow G_p$ and $G_p\hookrightarrow G$ exist? Clearly, groups of the shape $$ \bigoplus_{k=1}^\infty C_{p^k}^{m_k}, $$ (where $(m_k)$ is a sequence of non-negative integers with infinitely many strictly positive terms) have this property. Are there others?


1 Answer 1


There are two questions; probably they should be asked separately.

The first has a positive answer. Suppose that every finite abelian $p$-group embeds into $G$ (equivalently, $C_{p^n}^m$ embeds into $G$ for all $n,m$). Let us prove that this is equivalent to the a priori stronger condition: $\bigoplus_n C_{p^n}$ embeds into $G$.

Write $H[p]=\{h\in H: ph=0\}$.

Claim: let $F$ be a finite subgroup of $G$. Then there for every $n$ there exists a subgroup $L$ of $G$ isomorphic to $C_{p^n}$ with $L\cap F=\{0\}$.

Proof: let $m$ be the dimension of $F[p]$ over $\mathbf{Z}/p\mathbf{Z}$. Let $H_0$ be a subgroup of $G$ isomorphic to $C_{p^n}^J$ with $|J|=m+1$. If $H\cap F=0$ we are done. Otherwise $H\cap F$ contains a nontrivial element, which, viewed as element of $C_{p^n}^J$, has some $j\in J$ in its support. Define $H_1=C_{p^n}^{J-\{j\}}$. Then $(H_1\cap F)[p]$ has dimension $\le m-1$. If trivial, we are done, otherwise, reiterate. Eventually, for some $i\le m$ we get $H_i\simeq C_{p^n}^{m+1-i}$ and $H_i\cap F=\{0\}$. This proves the claim.

Using the claim iteratively, we deduce the existence of a copy of $G_p$ as desired.

PS: "universal abelian $p$-groups" would be a bad terminology, because infinite abelian $p$-groups (those in which every element has $p$-power order) are widely considered in the literature.

  • $\begingroup$ Thanks, great! I do not suggest to name these groups "universal", I just wanted to have an attractive title of my post. $\endgroup$
    – Yuri Bilu
    Commented Jun 18 at 18:16
  • 2
    $\begingroup$ @YuriBilu Sure. Maybe "Abelian groups that are universal for finite abelian $p$-groups" would make sense (but it's indeed longer). $\endgroup$
    – YCor
    Commented Jun 18 at 18:33
  • $\begingroup$ My feeling is that the second question has a similar simple answer. $\endgroup$
    – Yuri Bilu
    Commented Jun 18 at 19:22

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