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.
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Let p be a prime- If a group has more than p-1 elements of order p- wh.docx
1. 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.
