Search Results

There's a construction of a rank two (and therefore not locally cyclic) abelian group with endomorphism ring $\mathbb{Z}$, and therefore automorphism group cyclic of order 2, in "On the cancellation …

Yes, it must. And $G$ doesn't need to be countable. Let $H$ be the $p$-primary component of the torsion subgroup of $G$. Then the natural map $H/pH\to G/pG$ is injective, so $H$ also satisfies (1), an …

In fact, more is true. Every direct product of cotorsion-free abelian groups is cotorsion-free. This is clear because the cotorsion-free abelian groups are precisely those that have no nonzero homomor …

Here’s a quick homological proof. Suppose $F$ is finite and $H$ torsion free. Then $F\cong\text{Hom}(F,\mathbb{Q}/\mathbb{Z})$, so $$\text{Ext}^1(H,F)\cong\text{Ext}^1\left(H,\text{Hom}(F,\mathbb{Q}/ …

As suggested in my comment, define $h^*_p(a)$, as Fuchs does, to be the smallest ordinal $\sigma$ with $a\not\in p^{\sigma+1}A$ if there is such a $\sigma$, but if there is no such $\sigma$ then defin …

"Groups with a small number of automorphisms" by H. de Vries, A. B. de Miranda (Math Zeitschrift (1957/58) Volume 68, Issue 1, pp 450-464) link gives examples with cyclic automorphism groups of order …

Arturo Magidin's answer is absolutely correct, but there's an earlier counterexample than Corner's. This question is Kaplansky's first "test problem" in his 1954 book on Infinite Abelian Groups, and …

I'll start by describing the notation that I'll use. I'll think of elements of $\mathbb{Z}^\mathbb{N}$ as infinite column vectors $${\bf x}=\begin{pmatrix} x_0\\x_1\\x_2\\\vdots \end{pmatrix}$$ of in …

For the first question, $$\bigoplus_{p\text{ prime}}\mathbb{Z}/p\mathbb{Z}$$ is a counterexample.

The answer to Question 1 is no. Let $A=\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ and let $B$ be the subgroup generated by $(2,1)$. Since $B$ is cyclic of order $4$, if it were contained in a …

The Baer-Specker group $B$, the direct product of countably many copies of $\mathbb{Z}$, is semi-free but not free. It is semi-free, because for any nonzero element $x\in B$ there is some projection $ …

First, just a quick comment about terminology. It's possible that there are varying conventions, but I have seen "almost free" used to mean that all subgroups of an abelian group $G$ that have cardina …

$\mathbb{Z}$ is cancellable for abelian groups. This was proved in the 1950s by Walker and Cohn (independently) and is often called "Walker's cancellation theorem". The proof is only a few lines. So …

It is proved in Joel M. Cohen and Herman Gluck, MR 254028 Stacked bases for modules over principal ideal domains, J. Algebra 14 (1970), 493--505, that the answer is yes. They credit Kaplansky for as …

I think it's true that $\varprojlim A_n\to\varinjlim A_n$ is always injective. We may as well assume that $A_1\leftarrowtail A_2\leftarrowtail A_3\leftarrowtail A_4\leftarrowtail \cdots$ is a sequ …