Characterization of subgroup of neutral elements of reducible multiary group

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Statement

Consider a reducible multiary group, i.e., a n-ary group G with multiplication f:G^n \to G such that there exists a group structure on G such that f(a_1,a_2,\dots,a_n) = a_1a_2\dots a_n for all (possibly repeated) a_1,a_2,\dot,a_n \in G, with the multiplication on the right being as per the group structure.

Let C be the subset of G comprising those elements that are neutral for f. Consider C as a subset of G with its (ordinary 2-ary) group operation. The claim is that C is a subgroup of G, and in fact, it is the characteristic subgroup comprising those elements in the center of G whose order divides n - 1.

Note

Note that due to the equivalence of definitions of reducible multiary group, we know that neutral elements exist if and only if the n-ary group is reducible. Thus, for irreducible n-ary groups, we would get an empty set instead of a subgroup.

Related facts

Proof

Given: Group G with identity element e. C is the subset of G comprising those elements u of G for which u^iau^{n-1-i} = a for all a \in G and all i \in \{ 0,1,\dots, n-1\}.

To prove: C is precisely the subgroup of G comprising the elements u in the center of G with u^{n-1} = e.

Proof:

Step no. Assertion/construction Explanation
1 Every element of C is in the center of G. Set i = 0 to get au^{n-1} = a. Now set  i = 1 to get uau^{n-2} = a. Equating the two left sides gives au^{n-1} = uau^{n-2}. Cancel u^{n-2} from the right of both expressions to get au = ua for all a \in G.
2 For all u \in C, u^{n-1} = e. Set i = 0 and a = e to get eu^{n-1} = e which simplifies to u^{n-1} = e.
3 If u is in the center of G and u^{n-1} = e, then u \in C. Consider the product u^iau^{n-1-i}. Because u is central, we can commute it past a and get u^{n-1}a, which simplifies to a since u^{n-1} = e. This is true for all a \in G and all i \in \{0,1,\dots,n-1\}. Thus, u \in C.
4 The set of elements u such that u is in the center of G and u^{n-1} = e forms a characteristic subgroup of G. The center is already a subgroup, so we need only check that the condition u^{n-1} = e is a subgroup of it. Since abelian implies universal power map is endomorphism, it can be described as the kernel of the (n-1)-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.