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Spring 2018, problem 63

For which commutative finite groups is the product of all elements equal to the unit element?

Comments

Claudio
3 years ago

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

The easiest examples and counterexamples are given by the cyclic groups ZnZ, 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 sG such that ss=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 k2.

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 gGg we can suppress any couple of factors like gg1 provided gS; thus the product reduces to gSg. Of course the elements of S can not be coupled because g1g.

Concerning the order k of S, we already encountered examples with k=1 (say S={e}; this appens e.g. for G=ZnZ 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 abS; 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 ac,bc,abc are all special and different from the previous four a,b,ab,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: {gGg=e}  {k2}

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.

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:

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