How are simple groups the building blocks?

I know a bit about simple groups. A finite Abelian group is a (direct) product of finite cyclic groups. The simple finite Abelian groups are exactly $\mathbb{Z}_p$ for $p$ a prime. And so, I understand how all finite Abelian groups are made up of finite simple groups.

But, from what I understand, all finite groups are in some way made up of (finite) simple groups.

My question is: how does that work? What does it (more precisely) mean that a finite group is made up of simple groups?

Edit: Thanks to Stefan for directing me to questions that have basically already have the answer. I have done a bit more of research on this and I think I can narrow my question a bit. I would like to understand how simple groups are the building blocks of all finite groups. That is, I would like to understand how, given a finite group $G$, one can find (or show there exists) simple groups $G_1, \dots, G_n$ such that $G$ is [insert something] of $G_1,\dots G_n$. From here, I understand now that it somehow has to do with composition series and Jordan-Holder's Theorem. I think I understand the definition of a short exact sequence. From that same question:

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

I still don't get how this group $G$ is determined by the simple groups.

I guess I am looking for more details basically putting together how one starts with a finite group $G$, "finds" simple groups $G_1, \dots G_n$ and then says that $G$ is isomorphic to something in terms of the simple groups.

• Actually a finite abelian group is a product of cyclic groups but not necessarily simple groups. $\Bbb Z/4\Bbb Z$ for example is not $\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$ Jan 30, 2016 at 14:11
• As far as I know, this "making up" refers to the composition series. Every finite group has unique composition series, in which the factors are simple groups. These factors alone do not determine the group uniquely (far from it, actually), but they do allow us to write every finite group as the result of successive extensions by simple groups, starting from the trivial group. There may be some more content to it, so I will refrain from posting this as an answer (not being a specialist). Jan 30, 2016 at 14:13
• @GregoryGrant: ${\bf Z}/4{\bf Z}$ is not the direct product of ${\bf Z}/2{\bf Z}$, but it is an extension of ${\bf Z}/2{\bf Z}$ by ${\bf Z}/2{\bf Z}$. Jan 30, 2016 at 14:16
• @tomasz: Would you be able to write this as an answer (maybe with more details?) Jan 30, 2016 at 14:18
• @JohnDoe: ping me tomorrow. I am myself curious about whether or not there is a better answer, and I don't want to deter others by making this question answered. Jan 30, 2016 at 14:21

Every finite group could be decomposed as a finite number of extensions of simple groups. What is meant by extension could be read here; this is a different form as for example to decompose solely as a direct product (but a direct product could be seen as a special case of a group extension). By repeated extensions by simple groups you get a so called composition series, and these are unique up to a permutation of their composition factors (the quotient groups of successive groups in a series) by Jördan-Hölder. If you are interested there is also a theory for decompositions based on direct products, see Krull-Schmidt theorem.

This question appeared here already in a more or less different flavour, take a look at these posts and their answers:

Let $G=G_0$ be a finite group.

Consider the set of proper nontrivial normal subgroups of $G_0$. If this set is empty, then $G_0$ is simple. Otherwise, the set is ordered by inclusion and we may choose a maximal element $G_1$.

Note that by the correspondence theorem $G_0/G_1$ must be simple and we obtain a short exact sequence $$1\to G_1\to G_0\to G_0/G_1\to 1.$$ This says that $G_0$ is built out of $G_1$ and $G_0/G_1$ (or, $G_0$ is an extension of $G_0/G_1$ by $G_1$).

Now, suppose that $i\geq1$ and we have constructed $$G_{i}< G_{i-1}<\cdots< G_1< G_0$$ with $G_j\lhd G_{j-1}$ and $G_{j-1}/G_j$ simple for $0\leq j\leq i$.

Consider the set of proper nontrivial normal subgroups of $G_{i}$. If this set is empty, then $G_i$ is simple. Otherwise, the set is ordered by inclusion and we may choose a maximal element $G_{i+1}$. Again, by the correspondence theorem we have that $G_i/G_{i+1}$ is simple and we have an exact sequence $$1\to G_{i+1}\to G_i\to G_i/G_{i+1}\to 1.$$ As before, this means the $G_i$ is an extension of $G_i/G_{i+1}$ (a simple group) by $G_{i+1}$.

Now, since $G$ is finite, this process must terminate and we obtain a sequence $$1=G_{n+1}<G_n<\cdots<G_2<G_1<G_0=G$$ with $G_{i+1}\lhd G_i$ and $G_i/G_{i+1}$ simple (in particular, $G_n$ is simple). This is called a composition series for $G$. The subquotients $G_i/G_{i+1}$ are called the composition factors.

Of course, at each step in the construction of the composition series, we chose a maximal normal subgroup. Evidently, this choice need not be unique and it begs the question as to what extent these simple subquotients characterize $G$. This is the content of the Jordan-Holder Theorem.

Theorem (Jordan-Holder): Let $$1=G_{n+1}<G_n<\cdots<G_2<G_1<G_0=G$$ and $$1=H_{m+1}<H_m<\cdots<H_2<H_1<H_0=G$$ be two composition series for $G$. Then, $m=n$ and the set of composition factors for the two series are the same (except possibly for the order in which they appear).

The proof of this fact is not so hard (though it benefits from diagrams which are not so easy to produce on this site). One considers the set of all finite groups for which the above theorem fails. If this set is non-empty, it contains a group $G$ of minimal order.

Now, this $G$ has composition series $$1=G_{n+1}<G_n<\cdots<G_2<G_1<G_0=G$$ and $$1=H_{m+1}<H_m<\cdots<H_2<H_1<H_0=G$$ Note that both $G_1$ and $H_1$ are smaller groups, so they satisfy the Jordan-Holder theorem by assumption. The idea is to find some relationship between their composition series.

Consider a composition series for $G_1\cap H_1$: $$1=K_{r+1}<K_r<\cdots<K_3<G_1\cap H_1$$ (bear with me on the indexing).

Since both $G_1,H_1\lhd G$ are maximal, $G_1H_1=G$. It follows that $H_1/G_1\cap H_1\cong G/G_1$ is simple so $G_1\cap H_1$ is maximal in $H_1$. Therefore, $$1=K_{r+1}<K_r<\cdots<K_3<G_1\cap H_1<H_1$$ is a composition series for $H_1$. In particular, $r=m$. Similarly, $G_1/H_1\cap G_1\cong G/H_1$ is simple, so $$1=K_{r+1}<K_r<\cdots<K_3<G_1\cap H_1<G_1$$ is a composition series for $G_1$ and $r=n$. Hence, the length is unique.

As for the composition factors, we know that the $G/G_1\cong G_1/(G_1\cap H_1)$ and $G/H_1\cong H_1/(G_1\cap H_1)$. Therefore, \begin{align*} \{G/G_1,G_1/G_2,G_2/G_3,\ldots,G_n/G_{n+1}\}&=\{G/G_1,G_1/(G_1\cap H_1),(G_1\cap H_1)/K_3,\ldots, K_n/K_{n+1}\}\\ &=\{G_1/(G_1\cap H_1),G/H_1,(G_1\cap H_1)/K_3,\ldots,K_n/K_{n+1}\}\\ &=\{G/H_1,G_1/(G_1\cap H_1),(G_1\cap H_1)/K_3,\ldots,K_n/K_{n+1}\}\\ &=\{G/H_1,H_1/H_2,H_2/H_3,\ldots,H_n/H_{n+1}\} \end{align*} This completes the proof.

A Final Word: The composition factors themselves are not enough to determine a group. This is easy to see since $\mathbb{Z}/2\times\mathbb{Z}/2$ and $\mathbb{Z}/4$ have the same composition factors (or $\mathbb{Z}_6$ and $S_3$, or $D_8$ and $Q_8$, etc.). To really understand the group, you need to know not only the composition factors, but also the various extensions $$1\to G_{i+1}\to G_i\to G_i/G_{i+1}\to 1.$$

While there is most likely no hope of giving a complete solution to this problem, it has lead to some pretty interesting mathematics.

• Thanks for the answer. Would you be able to elaborate a bit on the statement that "This says that $G_0$ is built out of $G_1$ and $G_0/G_1$"? Feb 10, 2016 at 14:06
• There are a number of things that can be said here (more than enough for a semester class on group cohomology). When $G_0\cong G_1\times (G_0/G_1)$, it is very clear what this means. More generally, it is often the case that information about $G_1$ and $G_0/G_1$ can be lifted to information about $G_0$. For example, the correspondence theorem says that a large portion of the subgroup lattice for $G_0$ can be recovered from $G_1$ and $G_0/G_1$. Feb 10, 2016 at 16:08
• That sounds interesting. Would you put some of this in our answer? Feb 10, 2016 at 16:10
• Yes, I will try to do this this afternoon. Feb 10, 2016 at 16:11