Let p be a prime. If a group has more than p-1 elements of order p, why can\'t the group be cyclic? Solution This follows from the Linear Congruence Theorem. Let\'s assume the group (say G) in question is cyclic - every cyclic group is isomorphic either to the additive group of Z = {0, ±1, ±2, . .} if G is infinite, or to the additive group Z_{n} = {0, 1, 2,..., n-1} of the classes by modulo n if |G| = n /n - natural number/. The first case is trivial (no elements of order p), let\'s consider the second one. Let x is an element of Z_{n} of order p (p - prime), that means px = 0 in Z_{n}, the latter equality is equivalent to the congruence px 0 (mod p) in Z. According the cited theorem it has exactly GCD(p, n) solutions in the set of residues {0,1,2,...,n-1}. Since p is prime, we have: 1) GCD(p, n) = 1, if p doesn\'t divide n, so the only solution is 0, but 0 is of order 1, so no elements of order p exist in G; 2) GCD(p, n) = p, if p is a divisor of n. Again 0 is a solution, but the neutral element of each group is of 1st order, so the elements of order p can be maximum p-1. We arrived to a contradiction with the condition that elements of order p are more than p-1, so G cannot be cyclic. .