# When is a group the product of its subgroup and their factor group?

If G is a group and H a normal subgroup of G, under what conditions is it true that $$H \times (G/H) =G ?$$

To be more specific:

1. Characterise groups G which allow proper non-trivial normal subgroups H such that $H \times G/H =G.$
2. Given a group G which allow such a decomposition, characterise (normal) subgroups H of G such that $H \times G/H =G.$

Clearly, simple groups allow no such factorisation except the trivial ones.

And the smallest non-Abelian group $S_3$ cannot be the product of its (normal) subgroup $A_3$ and the quotient $\mathbb{Z}_2$.

Also, it follows from the structure theorem for finitely generated Abelian groups that cyclic groups with order a power of a prime do not allow any non-trivial decomposition. It is also not difficult to see that $\mathbb Z$ is not decomposable, i.e., expressible as the product of two proper non-trivial subgroups. These two facts together with the full force of the structure theorem tells us precisely which finitely generated Abelian groups are decomposable, and also enables us to list all the decompositions.

To summarise, in suggestive language, to what extent is the group-quotient operation an inverse for the group-product operation?

• For general finite groups I don't think you're going to find a better answer to 1 than "groups that nontrivially decompose as a direct product." There is a characterization in terms of exact sequences, but in my opinion that is no more illuminating. Feb 11, 2015 at 3:34
• As Matt Samuel said, this question is very difficult. Here are two articles on the topic. en.wikipedia.org/wiki/… en.wikipedia.org/wiki/Ext_functor Feb 11, 2015 at 3:52
• The problem with questions like this is that they are too general to allow any meaningful answer. Feb 11, 2015 at 11:19
• @MattSamuel, SpamIAm, DerekHolt I realise the point now, that this question is equivalent to decomposability, as I found this question math.stackexchange.com/questions/723707/… which however didn't appear in search before. I think I will post an answer myself making the point. Feb 11, 2015 at 11:25