Characterization of subgroup of neutral elements of reducible multiary group
Consider a reducible multiary group, i.e., a -ary group with multiplication such that there exists a group structure on such that for all (possibly repeated) , with the multiplication on the right being as per the group structure.
Let be the subset of comprising those elements that are neutral for . Consider as a subset of with its (ordinary 2-ary) group operation. The claim is that is a subgroup of , and in fact, it is the characteristic subgroup comprising those elements in the center of whose order divides .
Note that due to the equivalence of definitions of reducible multiary group, we know that neutral elements exist if and only if the -ary group is reducible. Thus, for irreducible -ary groups, we would get an empty set instead of a subgroup.
- Equivalence of definitions of reducible multiary group
- Groups giving same reducible multiary group are isomorphic
Given: Group with identity element . is the subset of comprising those elements of for which for all and all .
To prove: is precisely the subgroup of comprising the elements in the center of with .
|1||Every element of is in the center of .||Set to get . Now set to get . Equating the two left sides gives . Cancel from the right of both expressions to get for all .|
|2||For all , .||Set and to get which simplifies to .|
|3||If is in the center of and , then .||Consider the product . Because is central, we can commute it past and get , which simplifies to since . This is true for all and all . Thus, .|
|4||The set of elements such that is in the center of and forms a characteristic subgroup of .||The center is already a subgroup, so we need only check that the condition is a subgroup of it. Since abelian implies universal power map is endomorphism, it can be described as the kernel of the -power map, which is a subgroup. It is characteristic because of the unique manner of definition.|
|5||We are done||Steps (1) and (2) show one direction, Step (3) shows the reverse direction. Step (4) confirms that it is a subgroup.|