Happy new year!
Group theory warm-up exercise for the year: what are all the groups of order 2024?
Last year we did 2023, which was rather boring because all groups of order are abelian. This year the possibilities are much more numerous and the calculation is somewhat involved.
There are some easy initial reductions using Sylow’s theorem. Let be a group of order . Sylow’s theorem implies that there must be a unique subgroup of order . Similarly, in the quotient of order there must be a unique subgroup of order . Therefore has a unique subgroup of order . Let be a Sylow -subgroup of . Then for some homomorphism .
We can easily list the possibilities for and . There are just two possibilities for : the cyclic group and the unique nonabelian semidirect product . There are five possibilities for : , , , the dihedral group , and the quaternion group .
Next, for each choice of and , we must consider all possibilities for . However, in order to avoid duplication, we must determine when two homomorphisms induce isomorphic semidirect products.
A quick word about notation. Following standard practice in group theory, we denote group actions on the right. If and are elements of a common group , we denote by the conjugate . We also denote by the image of under a given homomorphism . We can then say that the semidirect product is generated by copies of and subject to the natural-looking relation
This relation asserts that the conjugation action of on is given by .
Suppose is an isomorphism from to such that . Then induces an automorphism of as well as an automorphism of . These automorphisms are not arbitrary: they satisfy a certain compatibility relation. Namely, for all and we have
In other words, for all we have, in ,
Written a third way, we have, as elements of ,
where denotes the inner automorphism of induced by . (Alternatively, if somewhat more traditionally we denote function composition in a right-to-left manner, the compatibility relation is .)
Let us denote by the set of all isomorphisms such that , and by the set of all compatible isomorphism pairs , i.e.,
The import of the previous paragraph is that there is a natural map . Moreover this map is surjective (though typically not injective), because if is a compatible isomorphism pair then the map defined by is an isomorphism (easy exercise). In particular, there is an isomorphism from to such that if and only if , i.e., if and only if there is a pair of compatible automorphisms .
We can phrase this conclusion another way. Observe that acts naturally on . The factor acts by precomposition, while the factor acts by conjugation. Isomorphism classes of split extensions are in bijection with orbits of in . (Here we consider and to be isomorphic as extensions if and only if there is an isomorphism such that . In general this is more restrictive than mere isomorphism as groups.)
Now consider the special case in which . In this case is a subgroup of and is a subgroup of , and the natural map is a homomorphism. Let be the kernel. Then consists of all isomorphism that restrict to the identity on and induce the identity on . This implies that there is a map such that for all . Since restricts to the identity on , we have
which implies that takes values in the centre of . Moreover, is a homomorphism , so we must have the relation
This relation means that is a crossed homomorphism from to . The group of crossed homomorphisms is usually denoted . Thus we have a short exact sequence
In particular, if is a characteristic subgroup of , is the full automorphism group, so we have a short exact sequence
In fact can be identified with the subgroup of automorphisms of preserving , so this sequence splits and we find that
Let us put all this into practice. Recall that we have two possibilities for : or . Let us consider the case first. We have
For each choice of among we must tabulate the homomorphisms up to the action of by precomposition and the conjugation action of . Since is abelian, the latter action is trivial. Any such homomorphism must take values in the Sylow -subgroup , so we are reduced to tabulating homomorphisms up to automorphisms of .
Case : There are four. Among these are groups with the structures , , . The last has the structure , and the action of on is fixed-point-free.
Case : Write . Any homomorphism must kill , so factors through . Naively are homomorphisms , but acts by swapping and (while is fixed) and the number of orbits of homomorphisms is . These groups include , , , , , , , and others.
Case : There are homomorphisms , but only up to the action of . They are in bijection with subgroups of . The corresponding groups of order are , , , , and .
Case : Any homomorphism factors through . Just as in the case of there are homomorphisms up to the action of . The groups of order include and in which acts nontrivially on by conjugation.
Case : Any homomorphism factors through . In this case acts on as , so as in the case of there are just homomorphisms up to automorphisms of . Thus there are semidirect products of the form .
Now consider the case in which is nonabelian. Since is a characteristic subgroup of (being the only subgroup of order ), we have
where is the subgroup of consisting of compatible pairs. Consider the compatibility relation . Since is abelian, is trivial, so the relation reduces to . Since is injective, this implies that is trivial. Therefore . Meanwhile one checks that . Therefore . Now if has order then any homomorphism takes values in a Sylow -subgroup isomorphic to , which are all conjugate. Tabulating homomorphisms up to the natural action of is therefore equivalent to tabulating homomorphisms up to automorphisms of , which just amounts to tabulating automorphism classes of subgroups of of index at most . This case is therefore somewhat easier than the previous one.
Case : There are just two. The corresponding groups of order have the forms and .
Case : There are three, corresponding to subgroups isomorphic to , , and . The corresponding groups of order include , , .
Case : There are two, since all index- subgroups are essentially the same. The corresponding groups of order are and .
Case : There are three, since there is a unique subgroup isomorphic to while the two subgroups isomorphic to are equivalent by an automorphism. There are three corresponding groups with structure .
Case : There are two, since the three subgroups of isomorphic to are equivalent by automorphisms. Thus there are just two groups with the structure .
Thus altogether there are groups of order .