# How is a group made up of simple groups?

I’ve read more than once the analogy between simple groups and prime numbers, stating that any group is built up from simple groups, like any number is built from prime numbers.

I’ve recently started self-studying subgroup series, which is supposed to explain the analogy, but I’m not completely sure I understand how “any group is made of simple groups”.

Given a group $G$ with composition series

then $G$ has associated the simple factor groups $H_{i+1}=G_{i+1}/G_i$. But how is it “built” from them?

Well, if we have those simple groups $H_i$ then we can say the subnormal subgroups in the composition series can be recovered by taking certain extensions of $H_i$:

where $H_i = G_i/G_{i-1}$, $K_i\simeq G_{i-1}$.

Then $G$ is built from some uniquely determined (Jordan-Hölder) simple groups $H_i$ by taking extensions of these groups.

Is this description accurate?

The question now is: this description seems overly theoretical to me. I don’t know how the extensions of $H_i$ look like, and I don’t understand how $G$ puts these groups together. Can we describe more explicitly how a group $G$ is made of simple groups?

EDIT: I forgot a (not-so-tiny) detail. The previous explanation works for finite groups, or more in general for groups with a composition series. But what about groups which don’t admit a composition series? Is it correct to say that they are built from simple groups?

Everything you say is correct: the sense that a finite group is “built” from its simple Jordan-Hölder factors is by repeated extensions. But this “building” process is much more complicated for groups than the analogous process of building integers from prime numbers because given a (multi-)set of building blocks — i.e., a finite list $\mathcal{H} = \{ \{H_1,\ldots,H_n\} \}$ of finite simple groups — there will be in general several (finitely many, obviously, but perhaps a large number) nonisomorphic groups $G$ with
composition factors $\mathcal{H}$. The simplest example of this has already been given by Zhen Lin in a comment: if

$\mathcal{H} = \{ \{ C_2, C_2 \} \}$,

then the two groups with these composition factors are $C_4$ and $C_2 \times C_2$.

It seems to be a working assumption of experts in the field that it is hopeless to expect a nice solution to the extension problem. For instance, consider the special case $\mathcal{H} = \{ \{ C_p,\ldots,C_p \} \}$, in which every composition factor is cyclic of order $p$ — i.e. a finite $p$-group. It is known that the function $f(p,n)$ which counts the number of isomorphism classes of finite groups of order $p^n$ grows very rapidly as a function of $n$ for any fixed $p$. For instance, see here for a reference to the fact that $f(2,9) = 10494213$.

Nevertheless the group extension problem is an important and interesting one — it is one of the historical sources for the field of group cohomology and still plays a major role — and in many special cases one can say something nice. But the general “program” of classifying all finite groups by (i) classifying all simple groups and (ii) determining all finite groups with a given set $\mathcal{H}$ of composition factors does not seem realistic: step (i) was amazingly hard but in the end doable. It looks very easy compared to step (ii)!

Finally, you ask about infinite groups. Here the Jordan-Hölder theory extends precisely to groups $G$ which admit at least one composition series, and a standard (necessary and sufficient) criterion for this is that there are no infinite sequences of subgroups

$H_1 \subsetneq H_2 \subsetneq \ldots$

with each $H_i$ normal in $H_{i+1}$

or

$H_1 \supsetneq H_2 \supsetneq \ldots$

with each $H_{i+1}$ normal in $H_i$.

So for instance an infinite cyclic group $\mathbb{Z}$ does not satisfy the descending chain condition on subgroups and there is no sense (known to me, at least) in which $\mathbb{Z}$ is built up out of simple groups.