70

Let $A$ be the additive group of bounded sequences of elements of $\mathbb{Z}[\sqrt{2}]$. Clearly $A\cong A\oplus\mathbb{Z}[\sqrt{2}]\cong A\oplus\mathbb{Z}^2$ as abelian groups, so we just need to show that $A\not\cong A\oplus\mathbb{Z}$.
Let $A_i\cong\mathbb{Z}[\sqrt 2]$ be the subgroup of $A$ consisting of sequences whose terms are all zero, apart from ...

answered Jan 1 '16 at 17:07

Jeremy Rickard

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30

(Essentially the same answer as Neil Strickland's:) Since a cyclic group of order $n$ has $\varphi(n)$ generators, your sum equals
$$ \DeclareMathOperator{\ord}{ord}
\nu(G) = \sum_{g\in G} \frac{ \ord(g) }{ \varphi(\ord(g)) } .
$$
For elements $g$ of $p$-power order, we get $\ord(g)/ \varphi( \ord(g) ) = p/(p-1)$ and thus if $G$ is a $p$-group, $\nu(G) = 1 ...

28

Assuming the axiom of choice, yes. Choose a non-principal ultrafilter on $\mathbb N$. This gives a consistent way to, given a function from $\mathbb N$ to a finite set, choose an element of the finite set, that doesn't depend on any finite subset of $\mathbb N$.
You can take an element of $\prod_{n\geq 1} \mathbb Z/p^n \mathbb Z$ and by modding out by $p^k$ ...

26

This is true more generally for finitely generated modules over a noetherian ring. Your question is equivalent to asking whether the sequence
$$0\rightarrow \operatorname{Hom}(B,A)\rightarrow \operatorname{Hom}(A\oplus B,A)\rightarrow \operatorname{Hom}(A,A)$$
is surjective on the right. To prove this, it suffices to localize and then complete at an ...

answered May 20 '14 at 16:43

Steven Landsburg

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23

Here's an elementary proof that doesn't require ultrafilters, but uses axiom of choice.
The group
\begin{equation*}
\prod_{n=1}^\infty \mathbb{Z} / \bigoplus_{n=1}^\infty \mathbb{Z}
\end{equation*}
clearly surjects onto
\begin{equation*}
\prod_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}) / \bigoplus_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}).
\end{equation*}
The ...

20

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 proper direct summand of $A$ then it would be contained in a cyclic subgroup of $A$ of order $8$, and so $(2,1)$ would be equal to $2a$ for some $a\in A$, ...

answered Jul 19 '20 at 11:13

Jeremy Rickard

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17

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 of modules in direct sums over Dedekind domains" by L. Fuchs and F. Loonstra,
Indagationes Mathematicae, Volume 74, (1971), 163-169
(link)

answered Feb 24 '15 at 12:44

Jeremy Rickard

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17

$\mathbb{Z}$ 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 if $A$ is an abelian group with $A\oplus\mathbb{Z}\cong A\oplus\mathbb{Z}^2$, then $A\cong A\oplus\mathbb{Z}$.

answered Feb 5 '17 at 11:22

Jeremy Rickard

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15

The answer to the question in the title is No. You can prove this from the work of Wanda Szmielew on the elementary properties of abelian groups. This answer works for any kind of nonzero, bi-additive, binary multiplication (associative or not, commutative or not, unital or not).
In particular, an abelian group $A$ is elementarily equivalent to our favorite ...

15

Yes! In fact, more generally, for any rig $R$ in which $0 \neq 1$, the map $X \mapsto R[X]$ is injective (where $R[X]$ denotes the free $R$-module on a set $X$).
Specifically: I claim we can construct $R[X]$ as a quotient of $\newcommand{\List}{\mathrm{List}}\List(R \times X)$, by an equivalence relation defined as follows:
Say lists $a$, $b$ are similar ...

answered Jun 11 '18 at 8:23

Peter LeFanu Lumsdaine

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14

This may well be the proof that you have already.
Reduce to the case of $p$-groups as in the question. Put
\begin{align*}
n(x) &= |\langle x\rangle| \\
m(x) &= |\{x' : \langle x'\rangle = \langle x\rangle\}|
\end{align*}
Then
$$ \nu(G) = \sum_{H \text{cyclic}} |H| = \sum_{x\in G} n(x)/m(x). $$
Now, if $x=1$ then $m(x)=1$. On the other hand, ...

answered Jun 3 '16 at 11:02

Neil Strickland

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14

Abelian group $A$ is cotorsion if $\rm{Ext}(F, A) = 0$ for every flat $F$, or, equivalently, $\rm{Ext}(\Bbb Q, A) = 0$
Every $\varprojlim^1$ of an inverse system of abelian group is cotorsion, and, conversely, every cotorsion group is a $\varprojlim^1$. Proof can be found in Warfield, Huber. On the values of the functor $\varprojlim^1$.
If every group in the ...

13

A nice set of generators for the automorphism group of a finite abelian group is described by Garrett Birkhoff in his paper titled "Subgroups of abelian groups", Proc. London Math. Soc., s2-38(1):385-401, 1935 (MR).
Note that each finite abelian group is the product of its $p$-primary subgroups. An automorphism preserves the primary parts. So it is ...

13

No. A classic result of Corner (On a conjecture of Pierce concerning direct decompositions of Abelian groups. 1964 Proc. Colloq. Abelian Groups (Tihany, 1963) pp. 43–48, MR0169905 (30 #148)) shows that for any positive integer $r$, there exist a countable torsion free abelian group $G$ such that the direct sum of $m$ copies of $G$ is isomorphic to the direct ...

13

Here are a few general remarks (concerning finite groups). The relevant results can mostly be found in the 2006 Journal of Algebra paper by Bob Guralnick and myself (a link to a preprint version is given in one of Benjamin Steinberg's comments).
An extra-special $p$-group of order $p^{3}$ has commuting probability $\frac{p^{2}+p-1}{p^{3}},$ which is only ...

answered Sep 8 '18 at 8:34

Geoff Robinson

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12

One can skip the ultraproduct part: fix any nonprincipal ultrafilter $F$, as in the SJR's construction. For each $(a_n)\in R$, split $\mathbb{N}$ into $N_0,N_1,\ldots,N_{p-1}$ so that $n\in N_k$ iff $a_n\cong k \mathop{\rm mod} p$. Then send $(a_n)$ to $k$ such that $N_k\in F$. But if by "explicit construction" you mean without the axiom of choice - I ...

answered Apr 27 '14 at 17:16

Adam Przeździecki

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12

Let me elaborate on my comment. I think freeness is a red herring. The content of the standard topological proof is that a covering space of a $1$-dimensional CW complex is again a $1$-dimensional CW complex. Of course this naturally generalizes to the case where $1$ is replaced by any positive integer $k$, which suggests the following definition.
Say that ...

answered Jan 28 '15 at 9:31

Qiaochu Yuan

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12

Yes; this is given in Mines, Richman, Ruitenburg, A course in constructive algebra, on p.54, stated just before Lemma 4.1, and proved in that lemma by a Church-Rosser argument. (Thanks to Thierry Coquand for pointing me to this proof.)

answered Jun 11 '18 at 8:26

Peter LeFanu Lumsdaine

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12

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
$$\text{Ext}^1(H,F)\cong\text{Ext}^1\left(H,\text{Hom}(F,\mathbb{Q}/\mathbb{Z})\right)
\cong\text{Hom}\left(\text{Tor}_1(H,F),\mathbb{Q}/\mathbb{Z}\right),
$$
which is zero since torsion free abelian groups are flat.

answered Jan 12 '19 at 12:22

Jeremy Rickard

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11

Let $F$ be any non-principal ultrafilter on $\mathbb{N}$. Let $I$ be the subset of the ring $R:=\prod_1^{\infty}\mathbb{Z}$ consisting of all functions that vanish on some element of $F$. Then $I$ is an ideal containing $J:=\coprod_1^{\infty} \mathbb{Z}$, whence we get a surjective homomorphism $R/J\to R/I$. But $R/I$ is isomorphic to the ultrapower of the ...

11

"Subobjects of free algebras are free" is satisfied comparatively rarely for algebraic theories. I'm going to start a list and people should feel free to add. I'm making it CW.
Before starting, let me say that IMHO a more interesting general question to consider is: when are retracts of free objects free? That can be a very tough question. We had some ...

11

Let me give a bit of an algebro-geometric perspective on Will's excellent answer, which I note gives not just a splitting of abelian groups but actually a ring homomorphism as well. Let's write $R=\prod \mathbb{Z}/p^n$ and try to understand $\operatorname{Spec} R$. Suppose we have a field $k$ and a homomorphism $f:R\to k$. For any subset $A\subseteq \...

11

I suggest you to have a look at the following paper and references therein:
V.K. Jain, P.K. Rai, M.K. Yadav: On finite $p$-groups with abelian automorphism group. Internat. J. Algebra Comput. 23 (2013), no. 5, 1063--1077.

10

Let $a$ be an algebraic number, $K = \mathbb Q(a)$ the associated number field, $\mathcal O_K$ its ring of integers. Then the ring $\mathcal O_K[a]$ depends only on the set of places of $K$ at which $a$ is not integral.
$\mathcal O_K[a]$ is a good place to start for studying $\mathbb Z[a]$ because $\mathbb Z[a]$ is a finite index submodule of $\mathcal O_K[...

10

A great survey on this and some related topics is Osofsky's "The subscript of $\aleph_n$, projective dimension, and the vanishing of $\varprojlim^{(n)}$." As far as I am aware, this 1974 paper still describes the state of the art on the matter.
Osofsky (exposing material from Mitchell's book on rings with many objects) gives the following example ...

10

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), and clearly satisfies (2). So, replacing $G$ by the subgroup $H$, we shall assume that $G$ is a $p$-group.
Let $X$ be a finite subset of $G$ that generates $G/pG$,...

answered Dec 29 '20 at 9:27

Jeremy Rickard

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10

I'll abbreviate $\mathbb{Z}/p^k \mathbb{Z}$ to $L_k$. Let $G = \bigoplus_{i=1}^M L_i^{a_i}$. I claim that $\bigoplus_{i=1}^N L_i^{b_i}$ is in the quotient series of $G$ if and only if the Littlewood-Richardson coefficient
$$c_{(1^a_1),\ \ (2^{a_2}),\ \ \dotsc,\ \ (M^{a_M})}^{(1^{b_1} 2^{b_2} \cdots N^{b_N})}$$
is nonzero. Note the parentheses and commas: We ...

answered May 10 at 12:55

David E Speyer

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9

The category of abelian groups is small-complete, well-powered, and has a cogenerator (e.g., $\mathbb{Q}/\mathbb{Z}$). It follows from the Special Adjoint Functor Theorem that any limit-preserving functor $G: Ab \to Ab$ has a left adjoint $F$. (A proof of the SAFT may be found on this nLab page.) And as Dylan Wilson pointed out in a comment, we then have $G \...

answered Mar 23 '16 at 19:10

9

Note that this equality is as topological groups, where Hom is endowed with uniform convergence on compact subsets.
Every such homomorphism is zero on $p^n$ for some $n$. Hence the given homomorphism group $G=\mathrm{Hom}(\mathbf{Q},\mathbf{Z}(p^{\infty}))$ can be written as $\bigcup G_n$, where $G_n$ is the set of homomorphisms that vanish on $p^n$. Note ...

9

Suppose $r \geq 3$. Then one can show that $\Gamma=\mathrm{SL}(r,A)$ is a lattice in the group $G=\mathrm{SL}\big(r, {\mathbb F}_q(\!(1/t)\!)\big)$. The group $G$ has Kazhdan's property T and hence $\Gamma$ itself has property T. In particular, it is finitely generated and its abelianization is finite. The same conclusion holds for finite index subgroups of $...

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