# "Universal" abelian p-groups

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?

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.

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