# Ways of building groups besides direct, semidirect products?

Let's say we have a group G containing a normal subgroup H. What are the possible relationships we can have between G, H, and G/H? Looking at groups of small order, it seems to always be the case that G = G/H x H or G/H x| H. What, if any, other constructions/relations are possible? And why is it the case that there are or aren't any other possible constructions/relations(if this question admits a relatively elementary answer)?

-
There are other possibilities. Take $G = \mathbb{Z}/4\mathbb{Z}, H = 2 \mathbb{Z}/4\mathbb{Z}$. Then $G/H = \mathbb{Z}/2\mathbb{Z}$ is not a subgroup of $G$, so $G$ can't be either a direct or semidirect product of $H$ and $G/H$. – Qiaochu Yuan Jul 16 '11 at 22:30
Anyway, my understanding is that this is actually fairly nontrivial. See en.wikipedia.org/wiki/Group_extension#Extension_problem . For finite groups, see en.wikipedia.org/wiki/Schur%E2%80%93Zassenhaus_theorem . – Qiaochu Yuan Jul 16 '11 at 22:32
@Jason: you should go ahead and look at groups of order at least 4: they're much more interesting than groups of "small order". – Pete L. Clark Jul 16 '11 at 22:38

I don't believe that this question admits elementary answer. The two ways, direct product and semidirect product, give various groups but not all.

As per my experience with small groups, complexity of constructions of groups lies mainly in $p$-groups. For $p$-groups of order $p^n$, $n>4$ (I think) there are always some groups which can not be semidirect products of smaller groups.

One method, is using generators and relations.

Write generators and relations of normal subgroup $H$, and quotient $G/H$.

Choose some elements of $G$ whose images are generators of $G/H$; make single choice for each generators.

So this pullback of generators and generators of $H$ gives generators of $G$. We only have to determine relations. Relations of $H$ are also relations of $G$. Other possible relations are obtained by considering relations of $G/H$ and their pullbacks.

Not all pullbacks of relations of $G/H$ give groups of order equal $|G|$; but order may become less. Moreover, different relations may give isomorphic subgroups.

For best elementary examples by this method (generators and relations), see constructions of non-abelian groups of order $8$; (Ref. the excellent book "An Introduction to The Theory of Groups : Joseph Rotman")

-