When is a group the direct sum of its normal subgroups?

[nonequilibrium]

I was sad to find out that if H is a normal subgroup of G, we can't say $$G \cong H \oplus G/H$$. Now I'm wondering: in which cases does this equality hold?

 

[micromass]

This doesn't really answer the question well, but if gcd(|N|,|G/N|)=1, then G is the semidirect product of N and G/N. This is the celebrated theorem of Schur-Zassenhaus.

So I guess, if gcd(|N|,|G/N|)=1 and if G is abelian, then $$G=N\oplus G/N$$.

 

[nonequilibrium]

Hm, I'm not familiar with the term "semidirect", what does it mean? I'm guessing you're demanding G to be abelian because of what semidirect means? Because if the theorem would say "direct sum" instead, then I don't see why we would need G to be abelian.

 

[micromass]

See http://en.wikipedia.org/wiki/Semidirect_product
The semidirect product is a very handy generalization of the direct product. It is defined for every kind of group (not just abelians). If you're taking a course on group theory, then I'm pretty sure that this notion will pop up someday.

The semidirect product is, in general, a nonabelian group. The only situation when a semidirect product yields a abelian group, is when the semidirect product is in fact the direct product.

If you're interested, I suggest picking up a good group theory book and read about it. I recommend fully the book "The theory of finite groups" by Kurzweil and Stellmacher.

 

[nonequilibrium]

Hm, now I'm confused, why are direct products only defined for abelian groups?

If one defines $$(a,\alpha) * (b,\beta) = (a*b,\alpha * \beta)$$ with a and b out of a certain group G and alpha and beta out of a certain group H, then we have a new group, don't we? (it's associative, has an inverse for every element and a neutral element)

 

[micromass]

Nonono, direct products are defined for any group, not just abelian groups. Semidirect products are also defined for all groups. Semidirect products are just a generalization of direct products.

What I said was, that if a semidirect product is abelian, then it had to be a direct product. That does certainly not mean that the direct product is always abelian or that it is only defined for abelian groups.

It's just that: the direct product of abelian groups is always abelian. But the semidirect product is never abelian (unless it was a direct product).

If you're confused, just forget everything I've said

 

[mathwonk]

on my page, http://www.math.uga.edu/~roy/

the free notes 3.a, page 47, and 6.d p. 6, discuss semi direct products.

 

[mathwonk]

it is useful to look at some simple examples. e.g. the integers Z have a subgroup 2Z which does not split this way. What do you think happens for the subgroup {0,3} of Z/6Z?