How do you prove that a group of order 3 is cyclic?

Answer 1

Any group must have an identity element e. As it has order 3, it must have two other elements, a and b. Now, clearly, ab = e, for if ab = b, then a = e:

abb-1 = bb-1, so ae = e, or a = e.

This contradicts the givens, so ab != b. Similarly, ab != a, leaving only possibility: ab = e. Multiplying by a-1, b = a-1. So our group has three elements: e, a, a-1.

What is a2? It cannot be a, because that would imply a = e, a contradiction of the givens. Nor can it be e, because then a = a-1, and these were shown to be distinct. One possibility remains: a2 = a-1.

That means that a3 = e, and the powers of a are: a0 = e, a, a2 = a-1, a3 = e, a4 = a, etc. Thus, the cyclic group generated by a is given by: = {e, a, a-1}.

QED.

Let g be any element other than the identity. Consider , the subgroup generated by g. By Lagrange's Theorem, the order of is either 1 or 3. Which is it? contains at least two distinct elements (e and a). Therefore it has 3 elements, and so is the whole group. In other words, g generates the group.

QED

In fact here is the proof that any group of order p where p is a Prime number is cyclic. It follow precisely from the proof given for order 3.

Let p be any prime number and let the order of a group G be p. We denote this as

|G|=p. We know G has more than one element, so let g be an element of the group and g is not the identity element in G. We also know contains more than one element and ||<|G|, so by Lagrange (as above)

|| divided |G|. Therefore || divides a prime p=|G| which tells us

||=G from which it follows that G is cyclic.

QED

Notation...Order of an element is the number of elements in the subgroup generated by that element. It is also the min n>1 such that gn =1 if such an n exists.