How many cyclic subgroups of order p can there be? First I ask: How many elements are there of order p? Let this number be X. If group element 'a' is of order p, then so is a^2, a^3, ... a^{p-1}. I.e. elements of order p come in sets of size p-1. I.e. p-1 divides X. By the orbit stabilizer argument in the proof, the number of orbits of size one is divisible by p. But one of those orbits is the tuple of all identities, and every other orbit of size one is a tuple of an element of order p. There are X of these, by definition. I.e. p divides X+1. Given that p-1 divides X and p divides X+1, a little algebra gives that X is of the form X = (qp+1)(p-1) where q is a non-negative integer. The simplest case is q=0, then X=p-1. In this case, our original group has only one cyclic subgroup of order p. In general, there are X/(p-1) = qp+1 cyclic subgroups of order p.
Yes..indeed it is. First we move everything to the right by p-1 then we move everything by 1 which results in not moving at all...which is the 0 action
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How many cyclic subgroups of order p can there be?
First I ask: How many elements are there of order p? Let this number be X.
If group element 'a' is of order p, then so is a^2, a^3, ... a^{p-1}. I.e. elements of order p come in sets of size p-1. I.e. p-1 divides X.
By the orbit stabilizer argument in the proof, the number of orbits of size one is divisible by p. But one of those orbits is the tuple of all identities, and every other orbit of size one is a tuple of an element of order p. There are X of these, by definition. I.e. p divides X+1.
Given that p-1 divides X and p divides X+1, a little algebra gives that X is of the form
X = (qp+1)(p-1) where q is a non-negative integer.
The simplest case is q=0, then X=p-1. In this case, our original group has only one cyclic subgroup of order p.
In general, there are X/(p-1) = qp+1 cyclic subgroups of order p.
Do the colors denote something or do you just like coloring everything a random color?
does the group action cycle in the sense of 1*p-1 goes back to 0
or do I understand it wrongly?
Am with u 😶
Yes..indeed it is.
First we move everything to the right by p-1 then we move everything by 1 which results in not moving at all...which is the 0 action