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 $$ \{e\}=G_0 \triangleleft G_1\triangleleft \dots \triangleleft G_{r-1} \triangleleft G_r=G$$

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 $H_i$: $$ 1 \to K_i \to G_i \to H_i \to 1$$

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?

  • 5
    $\begingroup$ You should read about the extension problem: en.wikipedia.org/wiki/Group_extension Citing: "but no theory exists which treats all the possible extensions at one time." $\endgroup$
    – Myself
    Commented Mar 6, 2011 at 17:24
  • $\begingroup$ Yikes. Surely some advances (albeit partial) have been made in solving the problem? $\endgroup$ Commented Mar 6, 2011 at 17:31
  • $\begingroup$ of course there are partial results. But the problem itself is known to be very difficult. 2-dimensional cohomology classifies certain central extensions of groups and is known to be impossible to compute for certain finitely-presented groups. So looking for certain types of general solutions is a futile task. $\endgroup$ Commented Mar 6, 2011 at 23:29

4 Answers 4


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 $H_{i+1}$


$H_1 \supsetneq H_2 \supsetneq \ldots$

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.

  • 3
    $\begingroup$ The chain conditions should be on subnormal subgroups, rather than on general subgroups. PSL(2,C) has a composition series (being simple), but it also contains an infinite cyclic group (image of [1,1;0,1]), which has an infinite descending sequence of subgroups. $\endgroup$ Commented Mar 7, 2011 at 12:36
  • 1
    $\begingroup$ @Jack: thanks. (I am currently teaching a course on commutative algebra and very much have modules on my mind, in which case these results hold but the subnormal condition is vacuous...) $\endgroup$ Commented Mar 7, 2011 at 14:31

Of course, Jordan–Hölder does show how to build groups from simple composition factors. However, there are other ways to dissect groups (finite groups, at least), which locate the "essential part" of a finite group in a different way. For example, let $G$ be a finite solvable group. Then $G$ has a unique largest nilpotent normal subgroup, its Fitting subgroup $F(G)$. The Fitting subgroup satisfies $C_{G}(F(G)) = Z(F(G)),$ and it follows that the factor group $G/F(G)$ is isomorphic to to a subgroup of the outer automorphism group of $F(G)$. Since $F(G)$ is a direct product of $p$-groups, its outer automorphism group is a direct product of outer automorhism groups of $p$-groups. In fact, $G/F(G)$ is isomorphic to a subgroup of a group of the form $X_1 \times \ldots \times X_t,$ where each $X_i$ is isomorphic to a subgroup of a completely reducible solvable subgroup of ${\rm GL}(n_i,p_i)$ for some integer $n_i$ and prime $p_i$.
For non-solvable groups, the situation is more complicated, and its understanding came later. H Bender introduced the generalized Fitting subgroup $F^{*}(G)$ of a general finite group $G$. It satisfies $C_{G}(F^{*}(G)) = Z(F(G))$, and it is still the case that $G/F^{*}(G)$ is isomorphic to a subgroup of the outer automorphism group of $F^{*}(G)$. The group $F^{*}(G)$ is the product of a pair of normal subgroups $F(G)$ (the usual Fitting subgroup) and $E(G)$. The groups $E(G)$ and $F(G)$ centralize each other, and are characteristic in $G$. A component of $G$ is a subnormal subgroup $L$ such that $L = [L,L]$ and $L/Z(L)$ is simple. It turns out that distinct components centralize each other. The group $E(G)$ is the central product of all the components of $G$ (and $G$ permutes its components by conjugation). The automorphism group of $E(G)$ has a normal subgroup $K$ consisting of the automorphisms which fix every component, and ${\rm Aut}(E(G))/K$ is a permutation group of degree $n,$ where $G$ has $n$ components. Also, $K/E(G)$ is isomorphic to a subgroup of a direct product of outer automorphism groups of finite simple groups. Thus the structure of $F^{*}(G)$ controls the structure of $G$ to a large extent. This viewpoint was very powerful in the later stages in the classification of the finite simple groups, and figures prominently in attempts to revise and simplify its proof. Note that $F^{*}(G)$ is built in a very transparent way from its own composition factors, which are among the composition factors of $G$.

  • $\begingroup$ Excuse me. It’s an old question and I’m sorry to disturb you. I wonder if you could explain why ${\rm Aut}(E(G))/K$ is permutation group of degree $n$ and why $K/{\rm Inn}(G)$ is isomorphic to a subgroup of a direct product of outer automorphism groups of finite simple groups. I asked a new question including the questions above. Thanks for your nice answer! $\endgroup$
    – user517681
    Commented May 1, 2020 at 12:25
  • $\begingroup$ You mean $K/Inn(E(G))$ I think. Let $E(G)/Z(E(G)) = S_{1} \times S_{2} \times \ldots S_{t}$, where each $S_{i}$ is a non_Abelian simple group. Then $K$ normalizes each $S_{i}$.. Let $X$ be the subgroup of $K$ which acts trivially on each $S_{i}$. Then $[E,X] \leq Z(E)$ where $E = E(G)$. Then $[E,X,E] = 1$ and $[X,E,E] = 1$, so $[E,E,X] = 1$ by the 3-subgroups Lemma. Hence $X$ acts trivially on $E$, so $X = 1.$ Hence $K$ acts on $S_{1} \times S_{2} \times \ldots S_{t} \cong Inn(E)$. Thus $K/Inn(E)$ is isomorphic to a subgroup of Out$(S_{1}) \times $ Out$(S_{2}) \times \ldots $. $\endgroup$ Commented May 1, 2020 at 13:17
  • $\begingroup$ $K$ normalizes $E(G)$ because $E = E(G)$ is characteristic in $G$.Now $E = L_{1}L_{2} \ldots L_{t}$ where each $L_{i}$ is a component of $G$. By the definition of $K$, $K$ normalizes each $L_{i}$, and then $K$ induces automorphisms of $L_{i}/Z(L_{i}) \cong S_{i}.$ But you are right, I should have said $K$ normalizes each $L_{i}.$ But the argument I give above shows that $K/Inn(E)$ induces a faithful group of automorphisms of $S_{1} \times \ldots \times S _{t}.$ $\endgroup$ Commented May 1, 2020 at 14:48
  • $\begingroup$ You think of $X$ as a group of automorphism of $E$, and then $[E,X]$ is defined to be $\langle e^{-1}e^{x}: e, \in E, x \in X \rangle$, where $e^{x}$ denotes the image of $e$ under the automorphism $x$. Then $[E,X]$ is a well-defined subgroup of $E$. I am only sketching arguments in the answer- you may have to do some reading to fill in some details. $\endgroup$ Commented May 1, 2020 at 15:12
  • $\begingroup$ The idea is that $G/F*(G)$ is a subgroup of $Out(F*(G))$ , and this breaks up in terms of $Out(F(G))$ and $Out(E(G))$. I explained somewhat about $Out(E(G))$. It's also the case that $Out(F(G))$ can be described to an extent. So we have broken up the structure of $G$ according to the structure of $F*(G)$ and the outer automorphism group of $F*(G)$. There are issues such as extension problems which are not addressed by this description. $\endgroup$ Commented May 1, 2020 at 15:40

In Dummit & Foote's "Abstract Algebra" they briefly discuss the Hölder Program:

  1. Classify all finite simple groups.
  2. Find all ways of "putting simple groups together" to form other groups.

They write the following on part 2 of the program (the so-called extension problem for finite groups):

Part (2) of the Hölder Program, sometimes called the extension problem, was rather vaguely formulated. A more precise description of "putting two groups together" is: given groups $A$ and $B$, describe how to obtain all groups $G$ containing a normal subgroup $N$ such that $N \cong B$ and $G/N \cong A$. For instance, if $A=B=Z_2$, there are precisely two possibilities for $G$, namely $Z_4$ and $V_4$ [the Klein four grup] and the Hölder Program seeks to describe how the two groups of order 4 could have been built from two $Z_2$'s without a priori knowledge of the existence of the groups of order 4. This part of the Hölder Program is extremely difficult, even when the subgroups involved are of small order. For example, all composition factors of a group $G$ have order 2 if and only if $|G| = 2^n$, for some $n$ (...). It is known, however, that the number of nonisomorphic groups of order $2^n$ grows (exponentially) as a function of $2^n$, so the number of ways of putting groups of 2-power together is not bounded. Nonetheless, there are a wealth of interesting and powerful techniques in this subtle area which serve to unravel the structure of large classes of groups.


Have you already met the notion of a semidirect product?

  • 6
    $\begingroup$ Yes, I have. Are you implying that every group is made of simple groups by taking semidirect products? $\endgroup$ Commented Mar 6, 2011 at 17:22
  • 1
    $\begingroup$ @Bruno: Well, there are some very small counterexamples. We have the composition series $\{e\} \triangleleft C_2 \triangleleft C_4$ but $C_4$ is most definitely not any (semi)direct product of $C_2$ and $C_2$. $\endgroup$
    – Zhen Lin
    Commented Mar 6, 2011 at 18:17
  • $\begingroup$ @Bruno: Sorry, I was in a hurry. No, I did not mean to imply that, I only meant to point you in a direction useful to get a more concrete grasp of some instances of group extensions. $\endgroup$
    – DaG
    Commented Mar 6, 2011 at 20:00
  • 4
    $\begingroup$ @Bruno: Note that the Schur--Zassenhaus theorem states that an extension of two finite groups of coprime order is necessarily a direct product; this is one explanation as to why, when people discuss the complexities of the extension problem (e.g. as in Pete Clark's answer), they focus on extensions of $p$-groups for a fixed prime $p$ --- this is the "opposite" situation to the one considered by Schur--Zassenhaus, and in this case extensions certainly don't have to be semi-direct products. $\endgroup$
    – Matt E
    Commented Mar 7, 2011 at 4:44
  • $\begingroup$ @Matt E: Is necessarily a semidirect product,not necessarily a direct product. $\endgroup$ Commented Mar 26, 2013 at 20:36

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