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2 votes

Cardinal of the injective hull

Yes. It is enough to embed a countable abelian group $A$ into a countable divisible (i.e., injective) abelian group $D$, because such a $D$ will contain an isomorphic copy of the injective hull of $A$....
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5 votes

Proof that all Quotient Groups are Abelian- where is my error?

Look at the first line of your proof: $xy(yx)^{-1}$ is not $xyy^{-1}x^{-1}$ (unless $G$ is itself abelian). Instead $xy(yx)^{-1}=xyx^{-1}y^{-1}$, which is not guaranteed to simplify to $e$. Bringing ...
0 votes

Show that $H = \{a \in G \ | \ ax=xa \ \forall x \in G\}$ is an abelian subgroup of G.

Let $m,n \in H$. Then $n \in G$ ($H \le G \implies H \subseteq G$). Since $$mg=gm$$ for any $g \in G$, it holds in particular when $g := n$, i.e. $$mn=nm$$ Addendum: This set $H$ has a special name: ...
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0 votes

Abelian, divisible and finite subgroup implies trivial subgroup?

The fact that the group is ordered is essential. Take any element $x \in H$. If $x > 0$, then $2x = x + x > x$, right? How about $3x$? Can $H$ be finite and non-trivial?
  • 123
1 vote

Showing group with $p^2$ elements is Abelian

If $G$ is cyclic, then it is abelian. If it isn't cyclic, then every nontrivial element has order $p$. Therefore, $p^2=1+k(p-1)$ for some positive integer $k$($=$number of subgroups of order $p$), ...
  • 1,521
0 votes

Skew-symmetric bicharacters on $A$ and $H^2(A,\mathbb{C}^{\times})$

The answer can be found in proposition 2.6 of this article by D. Tambara. Namely, let $X^2_a(A)$ denote the group of skew-symmetric bicharacters. There is a canonical map $Alt:Z^2(A;\mathbb{C}^{\times}...
  • 431
2 votes

What happens if I add an $\omega$th digit to the $p$-adic numbers?

I've been trying to understand what happens when we add an extra digit so that the digit gets incremented by 1 if there is a cascade of carrying which goes on forever like that. A $p$-adic number ...
  • 37.4k
4 votes

What happens if I add an $\omega$th digit to the $p$-adic numbers?

There is a simple argument for why your idea shouldn't work at least for $p=2$. Let $R$ be a topological ring where $\lim_{n\to \infty} 2^n$ converges to some element $\omega \in R$. $$2 \omega=2\lim_{...
  • 75.9k
1 vote

An exercise from Shapiro's Abstract Algebra

$(a^{n})^{-1} *a^{n}=e(identity)$ $(a^{-1})^{n}*a^{n}=(a^{-1}*a^{-1}*...*a^{-1})*(a*a*...*a)=e(identity) $ using associativity. So $(a^{n})^{-1}=(a^{-1})^{n}$ by uniqueness of inverse.
-2 votes

Abelian group $G$ such that the infinite-order elements form a subgroup with the identity.

R* is multiplication group. And H={x is positive real number} is subgroup of R* . But R* has a elements in order 2 .
9 votes

For abelian groups: does knowing $\text{Hom}(X,Z)$ for all $Z$ suffice to determine $X$?

I've just posted a negative answer (that I came across in the literature) on MathOverflow. If $S=\mathbb{Z}^{(\omega)}$ is the direct sum of countably many copies of $\mathbb{Z}$ and $A=S\oplus\mathbb{...
3 votes

Which cyclic groups are automorphism groups?

The situation about cyclic groups of automorphisms is as follows. If $G$ is an infinite periodic group, then its automorphism group is also infinite (R. Baer). If a cyclic group $A$ is the ...
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2 votes

A group whose automorphism group is cyclic

This MathOverflow question cites this paper, which says there are torsion-free groups of any finite rank of without automorphism group $C_2$, and which in turn cites this paper in which Theorem III ...
  • 2,691
5 votes

For a group $G$, let $a,b$ be arbitrary elements in $G$, if $a^{n}b^{n-1} = b^{n-1}a^{n}$ for all integers $n$, does this imply $G$ is abelian?

First note that the property is true "for all integers $n$" iff it is true for $n=2,$ i.e. iff every square is in the center. This property does not imply abelianity. A counterexample is the ...
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