0

Consider the short exact sequence $0 \rightarrow \mathbb{Z}_n \rightarrow G \rightarrow \mathbb{Z}_m \rightarrow 0$, where

  1. $\mathbb{Z}_n \simeq$ normal subgroup $N$ of $G$
  2. $\mathbb{Z}_m \simeq G/N$
  3. $n$ and $m$ are primes

What can we say about classification of all non-abelian extension groups of $\mathbb{Z}_n$ by $\mathbb{Z}_m$? Is there a method to find all non-abelian extension groups without cohomology theory?

Ben Sheller
  • 4,085
  • Split into two cases, $n=m$ and $n \ne m$. Both cases have been asked lots of times already. See for example http://math.stackexchange.com/questions/365261 for the case $n \ne m$. – Derek Holt Oct 20 '15 at 19:48

1 Answers1

0

Any group of order $nm$ is of such type. Let $n\le m$, and take an element of order $m$, by Cauchy's theorem. Then we have that the cyclic subgroup generated by this element is normal, since subgroups of index the smallest prime are normal. The quotient is a group of order $n$, and thus since all groups of prime index are cyclice, we have produced an exact sequence of your type. Thus what you are looking for is a classification of non-abelian groups of order $pq$. This has been done as in the question linked in the comment.

Pax
  • 5,762