-1

Some terms before I define my problem:

Definition: A group is abelian if every pair of elements commutes.

Proposition: All subgroups of an abelian group are normal.

Definition: A group $G$ is called metabelian if it has an abelian normal subgroup $N$ such that $G/N$ is also an abelian group.

I am tasked with proving that every abelian group is metabelian, but I have no idea where to start. I am inclined to believe that I should prove that $G/N$ is abelian if $G$ is abelian and $N$ is a normal subgroup. This satisfies the definition of metabelian, and since $G$ and $N$ are arbitrary, shouldn’t that conclude the argument?

Shaun
  • 47,747
V. Isidore
  • 129
  • 2
    Not following. Why can't you just take $N={e}$? All you need one "good" $N$, right? – lulu Dec 21 '22 at 19:56
  • I am not sure what you mean by that, to be honest. How would I use that to prove all abelian groups are metabelian? – V. Isidore Dec 21 '22 at 19:58
  • 2
    Well...can you see that taking $N={e}$ works for any abelian group $G$? Alternatively, take $N=G$. Same principle. – lulu Dec 21 '22 at 20:00
  • By the way, another standard definition is to say that $G$ is metabelian, if it is $2$-step solvable, i.e., if $[G,G]$ is abelian. For an abelian group $G$ this is even more obvious. – Dietrich Burde Dec 21 '22 at 20:10

1 Answers1

4

Suppose $G$ is abelian.

Each subgroup of an abelian group is both abelian and normal in the group; in particular, $G\unlhd G$. Observe that $G/G$ is isomorphic to the trivial group, which is abelian.

Hence $G$ is metabelian.

Shaun
  • 47,747