75
votes

Accepted

### $A$ is isomorphic to $A \oplus \mathbb{Z}^2$, but not to $A \oplus \mathbb{Z}$

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 ...

74
votes

### Results from abstract algebra which look wrong (but are true)

The free group with infinitely many generators is a subgroup of the free group with two generators.

Community wiki

42
votes

### Results from abstract algebra which look wrong (but are true)

Let $G$ be a finite group and $n \mid |G|$.
If $S = \{x \in G : x^n = 1\}$ contains exactly $n$ elements, then $S$ is a subgroup of $G$.
There seems no a priori reason to expect $S$ to be a subgroup ...

Community wiki

38
votes

### Results from abstract algebra which look wrong (but are true)

Every finite simple group can be generated by at most $2$ elements. This is another famous consequence of the classification.

Community wiki

30
votes

Accepted

### Is there a nice explanation for this curious fact about cyclic subgroups?

(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{ \...

29
votes

### If $A$, $B$ are abelian groups such that $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ for all abelian groups $G$, must $A$ and $B$ be isomorphic?

I just stumbled across the answer to this in Fuchs' 2015 book on Abelian Groups.
The papers
Hill, Paul, Two problems of Fuchs concerning tor and hom, J. Algebra 19, 379-383 (1971). ZBL0228.20027.
and ...

26
votes

### Results from abstract algebra which look wrong (but are true)

Every element of a finite simple non-abelian group is a commutator. This is the positive solution to the Ore conjecture (see Liebeck, O’Brien, Shalev, and Tiep - The Ore conjecture) and uses the ...

Community wiki

25
votes

### Results from abstract algebra which look wrong (but are true)

Every finite index subgroup of a finitely generated profinite group is open. The converse is obvious, but this direction was quite surprising to me. This is a result of Nikolov and Segal and uses the ...

Community wiki

24
votes

### Results from abstract algebra which look wrong (but are true)

The Nielsen-Schreier theorem that subgroups of free groups are free might have seemed surprising from am algebraic view given the analogue for many other algebraic structures is false. While this is ...

Community wiki

24
votes

### Results from abstract algebra which look wrong (but are true)

As fields, the algebraic closures of the fields ${\bf Q}_p$ are isomorphic, and are isomorphic to the complex numbers.

Community wiki

23
votes

### Results from abstract algebra which look wrong (but are true)

For a group with finitely many elements of finite order, the set of elements of finite order is a subgroup.

Community wiki

20
votes

Accepted

### Classification of subgroups of finitely generated abelian groups

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 ...

19
votes

Accepted

### Groupoid cardinality of the class of abelian p-groups

This is Corollary 3.8 in Cohen, H.; Lenstra, H. W., Jr. Heuristics on class groups. Number theory (New York, 1982), 26–36, Lecture Notes in Math., 1052, Springer, Berlin, 1984 (MR0750661).
In this ...

19
votes

Accepted

### Finite abelian groups with fewer automorphisms than a subgroup

From the Hiller-Rhea formula
$$|\operatorname{Aut} H_p| = \prod_k (p^{d_k} - p^{k-1}) \prod_j (p^{e_j})^{n-d_j} \prod_i (p^{e_i-1})^{n-c_i+1},$$
given an abelian $p$-group of type $p^{e_1}\cdots p^{...

17
votes

Accepted

### Torsion-free abelian group $A$ such that $A \not \simeq A \oplus \Bbb Z \simeq A \oplus \Bbb Z^2$

$\mathbb{Z}$ 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 ...

16
votes

Accepted

### Is the class of additive groups of rings axiomatizable?

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,...

16
votes

### Constructively, is the unit of the “free abelian group” monad on sets injective?

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 ...

16
votes

### Results from abstract algebra which look wrong (but are true)

In combinatorial group theory, loosely speaking almost any problem one can imagine, in full generality, turns out to be undecidable. This includes the word problem, the isomorphism problem, the ...

Community wiki

15
votes

### Results from abstract algebra which look wrong (but are true)

There exists a finitely-generated infinite group with only two conjugacy classes, a difficult result of Osin.

Community wiki

14
votes

### Is there a nice explanation for this curious fact about cyclic subgroups?

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 =...

14
votes

Accepted

### Which abelian groups are $\varprojlim^1$ groups?

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, ...

13
votes

### Two abelian groups, each being direct factor of the other

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 ...

13
votes

Accepted

### How nearly abelian are nilpotent groups?

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 ...

13
votes

Accepted

### Finite-by-torsion-free abelian groups (or compact abelian groups with finitely many components)

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}/\...

12
votes

### Constructively, is the unit of the “free abelian group” monad on sets injective?

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 ...

11
votes

Accepted

### Example of an uncountable sequence of abelian groups with nonvanishing $\varprojlim^2$?

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 ...

11
votes

Accepted

### Do these properties of a countable abelian group guarantee a Prüfer subgroup?

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), ...

11
votes

### On $p$-groups with abelian automorphism group

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)...

11
votes

Accepted

### Trivial group cohomology induces trivial cohomology of subgroups

For any abelian group $A$ we have a canonical isomorphism $\bigwedge^2A\to H_2(A,\mathbb{Z})$, given by the (anti-symmetric) Pontrjagin product $H_1(A,\mathbb{Z})\times H_1(A,\mathbb{Z}) \to H_2(A,\...

10
votes

### $\mathbb{Z}$-module structure of the subring generated by an algebraic number

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 $...

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