# Tag Info

## New answers tagged abelian-groups

Accepted

### 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 $A$....
• 7,421
Accepted

### 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 ...
• 229k

### 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: ...
• 36.9k

### 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

• 26.6k
Accepted

### 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 ...
• 8,226
Accepted

### 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
### 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 ...