I know that classification of finite simple group is completed.

From the fact, can we say that classification of finite group is completed?

I know a few relations of finite group and finite simple group, but can't think that the relation is so powerful.

For example, can we enumerate the group of order $n$ for any natural number $n$?

  • $\begingroup$ In general, the complexity of enumeration of groups of order $n$ depends on the factorisation of $n$. For some $n$, there is a theoretical knowledge, for some there are algorithms and implementations, but some others are really hard. The smallest $n$ for which the number of groups of order $n$ is unknown is 2048. I've added group-enumeration tag - see some questions over there, and also math.stackexchange.com/questions/1607517/…. $\endgroup$ – Alexander Konovalov Jun 14 '16 at 20:56

Not directly.

Recall a group is simple if it has no (proper, nontrivial) normal subgroups. Equivalently, it has no homomorphic images (or quotient) besides the obvious two (itself and the trivial group).

We may ask if a group $G$ is simple. If not, we may write down the chain $1\triangleleft N\triangleleft G$ (where $\triangleleft$ means "is a proper normal subgroup of"). We may also take the quotient to obtain $G/N$, then we can ask the same question of $G/N$: is it simple? If not, we can quotient it further, by one of its own normal subgroups, which (by the lattice isomorphism theorem) corresponds to some normal subgroup $H\le G$ in which $N$ is normal. We can encode this information by the chain: $1\triangleleft N\triangleleft \color{Green}{H}\triangleleft G$. We have effectively inserted an $H$ in between the $N$ and $G$ in the original chain $1\triangleleft N\triangleleft G$.

On the other side of $N$, we can also insert any normal subgroup of $N$ itself if $N$ is not itself a simple group. For example, if $K\le G$ is normal within $N$, then we would at this point have:

$$1\triangleleft K\triangleleft N\triangleleft H\triangleleft G.$$

(Keep in mind $A\triangleleft B$ and $B\triangleleft C$ does not imply $A\triangleleft C$; the above subgroups are not all normal in $G$, they are only each normal within the next subgroup in the chain.)

We can proceed with this process until we hit a point where we cannot refine the chain any further, in which case we have a composition series. Equivalently, $G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_\ell$ is a composition series if (besides each subgroup being normal in the next), the head and tail are the trivial and proper groups ($G_0=1$ and $G_\ell=G$), and each "composition factor" $G_{i+1}/G_i$ is a simple group (which by the lattice isomorphism theorem is equivalent to there being no group one can insert between $G_i$ and $G_{i+1}$ in the chain).

The Jordan-Holder theorem states that the multiset of (isomorphism classes of) composition factors, $\{G_1/G_0,\, G_2/G_1,\, \cdots,G_\ell/G_{\ell-1}\}$ is uniquely determined by $G$ itself. There are in general going to be many different composition series, and in particular the collection of composition factors can come out in different orders, but the collection itself is the same. It is for this reason that all finite groups are made up of simple groups, similar to how natural numbers are made of prime numbers and molecules are made of atoms.

The prime numbers bit is not just an analogy, in fact the fundamental theorem of arithmetic (stating the existence and uniqueness of factorization of natural numbers into prime numbers) can be interpreted as a corollary of the Jordan-Holder theorem. The only simple abelian groups are $\mathbb{Z}_p$ for $p$ prime, therefore the composition factors of $\mathbb{Z}_n$ for arbitrary integers $n$ will be a bunch of $\mathbb{Z}_p$s for primes $p$, and this exactly matches $n$s factorization into primes.

For instance $\mathbb{Z}_1\triangleleft\mathbb{Z}_2\triangleleft\mathbb{Z}_6\triangleleft\mathbb{Z}_{12}\triangleleft\mathbb{Z}_{24}$ is a composition series for $\mathbb{Z}_{24}$ and the corresponding composition factors in order are $\{\mathbb{Z}_2,\mathbb{Z}_3,\mathbb{Z}_2,\mathbb{Z}_2\}$, corresponding to $24=2\cdot3\cdot2\cdot2$.

The simplicity of integers can be misleading though. There is only ever one way to multiply to numbers. But in general, there is more than one way to "put together" two simple groups. That is, if $N$ and $Q$ are any pair of simple groups, there can be more than one group $G$ with $1\triangleleft N\triangleleft G$ and quotient $G/N\cong Q$. Solving for such a $G$ is called the group extension problem.

This is true even at the level of abelian groups, for which there exists a complete classification. For instance let's consider putting together $\mathbb{Z}_p$ and $\mathbb{Z}_p$. One could speak of either $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_p\oplus\mathbb{Z}_p$; both of them have a normal subgroup $\cong\mathbb{Z}_p$ with corresponding quotient $\cong\mathbb{Z}_p$. One could consider $\mathbb{Z}_{p^2}$ to be "the same" as $\mathbb{Z}_p\oplus\mathbb{Z}_p$, except with "carrying" in the last "digit" (second coordinate). This interpretation corresponds to the notion of a cocycle and is one way to motivate group cohomology (or at least I heard this from someone I trust; blame him if this is wrong).


The classification of all finite groups depends on solving the group extension problem, which has been heavily studied but not yet fully solved.


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