Let G be a group, p a prime dividing |G| and X = {(x_0,...,x_p-1) elementof G^p
ID: 1720749 • Letter: L
Question
Let G be a group, p a prime dividing |G| and X = {(x_0,...,x_p-1) elementof G^p : ||_i x_i = 1}. Let E be the relation defined on X by (x_0,...,x_p-1).E(y_o,..., y_p-0 if there exists k elementof Z/pZ such that y_i+k = x_i (the indexes are considered to live in Z/pZ, so if i + k greaterthanequalto p, what we mean is y_i+k-p = x_i). Show that |X| = |G|^p-1. Show that E is an equivalence relation. Show that an equivalence class of E contains a single element if and only if this element is of the form (x,...,x) where x^p = 1. Show that the equivalence classes of E that are not singletons have cardinal p. Show that p divides the number of equivalence classes of E that are singletons. Show that there exists an element x elementof G of order p.Explanation / Answer
X = {g=(x[0],x[1],......x[p-1]) | Product of all the p elements =1}
1) As the last element x[p-1] is determined by the first p-1 elements (it being the inverse of their product). These p-1 places can be filled by any of the elements of G , so it follows that |X| = |G|p-1 .
(2) Clearly it is reflexive and symmetric (change k to -k ).
If g Rh (as tuples) and h Rz with k and m , then g Rz will be related by k+m.
(3) If element g =(x[0],x[1],.....x[p-1]) has such a property (its equivalence class consists only a singleton) means every x[i] = x[0] =x which implies xp =1, by the definition of the set X.
(4) From (3) , at least two of the tuples defining g should be unequal. we can cyclically change the indices and obtain p element which belong to the same equivalence class of g. Hence the result.
(5) This follows from 4 as the total number of elements in X is divisible by p.(by (1))
(6) So the set of elements of order p is divisible by p (it is greater than 1, clearly , because (e,e,...e) belongs to X, which implies the existence of an element of order p
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