Spring 2018, problem 63
For which commutative finite groups is the product of all elements equal to the unit element?
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naomievans
1 year ago
Let G be the ultimate Abelian band. Well, then: 1) a work of all elements of group G, the orders of which are different from 2, equal to a single item; 2) if group G contains a part of order 2, the work of all elements of group G shall be equal to the work of all elements of rule 2 of group G. Proof. If e≠x ∈ G, then O(x) = 2 then and only when x=x^(-1). If O(x) > 2, then O(x^(-1)) = O(x) > 2, then x ≠ x-1. Since G is an Abel band, then:
Claim The condition holds true for all groups but two cases:
(a) the cyclic groups of even order, Z╱(2p)Z; with additive notations the sum of all elements is p.
(b) the direct sum of cyclic groups, all but one having odd order
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The easiest examples and counterexamples are given by the cyclic groups Z╱nZ, that satisfy the wanted property if and only if n is odd. In general, denoting by n the order of the group:
(i) because of the Klein-four group, the oddness of n is not a necessary condition;
(ii) however, as we will show later on, the oddness of n is a sufficient condition.
In order to find a general necessary and sufficient condition, let us fix some notations. G will denote our finite commutative group, and e will be its unit element. We will call special any s∈G such that s∗s=e; S will denote the family of special elements. We want to prove:
Theorem S is a subgroup, whose order k can be 1,2,4,8,... The product of all elements of G coincides with the product of all elements of S; and such product is e if and only if k≠2.
The proof will be splitted in many parts.
The group-structure of S is obvious; its order will be evaluated later on. In the product ∏g∈Gg we can suppress any couple of factors like g∗g−1 provided g∉S; thus the product reduces to ∏g∈Sg. Of course the elements of S can not be coupled because g−1≡g.
Concerning the order k of S, we already encountered examples with k=1 (say S={e}; this appens e.g. for G=Z╱nZ with odd n) and examples with k=2, as in Z╱(2p)Z where, with additive notations, S={0,p}.
Now let us assume k>2; if e,a,b are elements of S, also a∗b∈S; and it is a fourth element because it is obviously different from e,a,b. Remark that the product of these four elements is e; in fact each non-unit element appears twice in the product. Now we work in a similar way: if a fifth element c belongs to S, the further elements a∗c,b∗c,a∗b∗c are all special and different from the previous four a,b,a∗b,c; thus the order could be 8. Remark that again the product of the 8 elements is e, because each non-unit element appears in the product an even number of times.
A trivial induction argument now suffices for the proof of the Theorem; in particular: {∏g∈Gg=e} ⟺ {k≠2}
Final remark Till now we just used elementary results about groups. However, by using the Kronecker decomposition theorem, we can prove that the condition k=2 holds true if and only if the group is a cyclic group of even order or, more generally, a direct sum of cyclic groups having all but one odd order.